statistics and probabilities

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15 Probability Distributions

Assessment statements

Concept of discrete random variables and their probability distributions. Expected value (mean), E( x) fordiscrete data.

Binomial distribution.

ean and variance of the binomial distribution.

!ormal distribution and curves. Properties of the normal distribution. "tandardi#ation of normal variables.

Introduction

Investing in securities, calculating premiums for insurance policies or overbooking policies used in the

airline industry are only a few of the many applications of probability and statistics. Actuaries, forexample, calculate the expected ‘loss’ or ‘gain’ that an insurance company will incur and decide on howhigh the premiums should be. These applications depend mainly on what we call probabilitydistributions. A probability distribution describes the behaviour of a population in the sense that it liststhe distribution of possible outcomes to an event, along with the probability of each potential outcome.This can be done by a table of values with their corresponding probabilities or by using a mathematicalmodel.

In this chapter, you will get an understanding of the basic ideas of distributions and will study twospecific ones the binomial and normal distributions.



$%.$ Random variables

In !hapter ", variables were defined as characteristics that change or vary over time and#or for differentob$ects under consideration. A numerically valued variable x will vary or change depending on the outcome of the experiment we are performing. %or example, suppose you are counting the number of mobile phonesfamilies in a certain city own. The variable of interest, x, can take any of the values &, ', (, ), etc. dependingon the random outcome of the experiment. %or this reason, we call the variable x a random variable.

Random variable

A random variable is a variable that ta&es on numerical values determined by the outcome of a randomexperiment.

*hen a probability experiment is performed, often we are not interested in all the details of theoutcomes, but rather in the value of some numerical

518

($, $)' x %

(, $)' x %

(, $)' x % *

(*, $)' x % %

(%, $)' x % +

(+, $)' x %

($, )' x %

(, )' x % *

(, )' x % %

(*, )' x % +

(%, )' x %

(+, )' x % -

($, )' x % *

(, )' x % %

(, )' x % +

(*, )' x %

(%, )' x % -

(+, )' x %

($, *)' x % %

(, *)' x % +

(, *)' x %

(*, *)' x % -

(%, *)' x %

(+, *)' x % $/

($, %)' x % +



(, %)' x %

(, %)' x % -

(*, %)' x %

(%, %)' x % $/

(+, %)' x % $$

($, +)' x %

(, +)' x % -

(, +)' x %

(*, +)' x % $/

(%, +)' x % $$

(+, +)' x % $

otice that events can be more accurately and concisely defined interms of the random variable x

- for example, the event of tossing asum at least e+ual to but less than " can be replaced by 0 x , ".

*e can think of many examples of random variables

X % the number of calls received by a household on a %riday night.

X % the number of free beds available at hotels in a large city.

X % the number of customers a sales person contacts on a workingday.

X % the length of a metal bar produced by a certain machine.

X % the weight of newborn babies in a large hospital.

As you have seen in!hapter ", these variablesare classified as discrete or continuous, according tothe values that x can

assume. In the examples

above, the first three arediscrete and the last twoare continuous. Therandom variable is discreteif its set of possible valuesis isolated points on thenumber line, i.e. there is acountable number of possible values for thevariable. The variable is

continuous if its set of possible values is an entireinterval on the numberline, i.e. it can take anyvalue in an interval.!onsider the number oftimes you toss a coin untilthe head side appears. The possible values are x % ',(, ), / . This is a discrete

variable, even though thenumber of times may beinfinite0 1n the other hand,consider the time it takes astudent at your school toeat#have his#her lunch.This

can be anywhere between2ero and & minutes3given that the lunch

period at your school is& minutes4.



/

%/

discrete

continuous

Example 1

5tate whether each of the following is a discrete or a continuousrandom variable.

The number of hairs on a 5cottish Terrier

The height of a building

Table 15.1 "ample space

and the values of the random

variable x in the t1o2diceexperiment.

519



Probability Distributions

amount of fat in a steak

gh school student’s grade on a maths test

number of fish in the Atlantic 1cean

temperature of a wooden stove

tion

n though the number of hairs is ‘almost’ infinite, it is countable. 5o, it is a discrete random variable.

can be any real number. 6ven when you say this building is ' m high, the number could be '.' or '.&(, etc.

e, it is continuous.

is continuous, as the amount of fat could be 2ero or anything up to the maximum amount of fat that can bein one piece.

des are discrete. o matter how detailed a score the teacher gives, the grades are isolated points on a scale.

is almost infinite, but countable, hence discrete.

is continuous, as the temperature can take any value from room temperature to '&& degrees.

obability distribution



hapter ", you learned how to work with the fre+uency distribution and relative or percentage fre+uencyibution for a set of numerical measurements on a variable x. The distribution gave the following informationt x

e value of x that occurred.

w often each value occurred.

also learned how to use the mean and standard deviation to measure the centre and variability of the data set.

is an example of the fre+uency distribution of ( families in ;ower Austria that were polled in a marketing survey

t the number of litres of milk consumed during a particular week, reproduced below. As you will observe, the tablethe number of litres consumed along with the relative fre+uency with which that number is observed. As you recall

pter '&, one of the interpretations of probability is that it is understood to be the long<term relative fre+uencye event.

e 15.2

ber of litres

ve fre4uency



ble like this, where we replace the relative fre+uency with probability, is called a probability distribution ofandom variable.



5he probability distribution for a discrete random variable is a table, 6raph or formula that 6ives the possible values of x, and the probability P( x)

associated 1ith each value of x.

;etting x be the number of litres of milk consumed by a family above, the probability distribution of x would b

as follows

x

/

$

*

%

Table 15.3

P ( x) /./- /./ /.+ /./ /.$ /./*

The other form of representing the probability distribution is with a histogram, as shown below. 6very columncorresponds to the probability of the associated value of x. The values of x naturally represent mutually

exclusive events. 5umming =3 x4 over all values of x is e+uivalent to adding all probabilities of all simple events in the sampspace, and hence the total is '.

/.*/



/.%

Probability

/./

/.%



/./

/.$%



/.$/

/./%

/.//

/

$

*

%



7itres of mil&

The result above can be generali2ed for all probability distributions

Reuired properties o! probability distribution !unctions o! discrete random variables

;et x be a discrete random variable with probability distribution function, =3 x4. Then

& 0 =3 x4 0 ', for any value x.

The individual probabilities sum to '- that is, > =3 x4 % ' where the

x

notation indicates summation over all possible values x.

Example 2



?adon is a ma$or cause of lung cancer after smoking. It is a radioactive gas produced by the natural decay ofradium in the ground. 5tudies in areas rich with radium revealed that one<third of houses in these areas havedangerous levels of this gas. 5uppose that two houses are randomly selected and we define the random variable x

to be the number of houses with dangerous levels. %ind the probability distribution of x by a table, a graph and aformula.

Solution

5ince two houses are selected, the possible values of x are &, ' or (. To find their probabilities, we utili2e what welearned in !hapter '&. The

521



% Probability Distributions

sumption here is that we areoosing the houses randomly

d independently of eachher0

x % (4 % =3(4 % =3'st houseth gas and (nd house withs4

=3'st house with gas4

(nd house with gas4 % ') '

)

"

x % &4 % =3&4 % =3'st housethout gas and (nd housethout gas4

'st house without gas4 (nd house without gas4

() % @

"

x % '4 % ' B=3&4 $ =3(4C % '

B@" $ '" C% @"

ble



x)



raph

bability

$

/

/

$

!umber of houses

ny type of graph can be usedgive the probabilitytribution, as long as it shows

e possible values of x and therresponding probabilities.e probability here is

aphically displayed as theight of a rectangle.oreover, the rectanglerresponding to each value of

has an area e+ual to theobability =3 x4. The histogram

the preferred tool due to itsnnection to the continuoustributions discussed later in

e chapter.



rmula/rule

e probability distribution ofan also be given by thelowing rule. Eon’t bencerned now with how we

me up with this formula, aswill discuss it later in the

apter. The only reason we areoking at it now is to illustratee fact that a formula#rule canmetimes be used to give theobability distribution.

x

( ( x

4

3 4 represents the binomial

efficient you saw in !hapter



otice that when x is replaced&, ' or ( we obtain theults we are looking for

&4

&



'4

'



(4

(



4

4



2



Example 3

Dany universities have the policy of posting the grade distributions fortheir courses. 5everal of the universities have a grade<point average that

codes the grades in the following manner A % @, G % ), ! % (, E % ' and %% &. Euring the spring term at a certain large university, ')H of thestudents in an introductory statistics course received A’s, )H G’s, @H!’s, @H received E’s and 'H received %’s. The experiment here is tochoose a student at random and mark down his#her grade. The student’sgrade on the @<point scale is a random variable x.

7ere is the probability distribution of x

x

/

$

*

P( x)

/./$

/./*

/.*%

/.

/.$



Is this a probability distributionJ

Solution

:es, it is. 6ach probability is between & and ', and the sum of all probabilities is '.

*hat is the probability that a randomly chosen student receives a G or betterJ

=3 x 8 )4 % =3 x % )4 $ =3 x % @4 % &.) $ &.') % &.@&

Example "

In the codes example in !hapter '&, we saw the probability with which people choose the first digits for the codes for their cellphones. The probability distribution is copied below for reference.

9irst di6it

/

$



*

%

+

-

Probability

/.//

/.//

/.$*

/.$

/./+

/./-

/./+

/./%-

/./%$

/./*%



7ere, x is the first digit chosen.

*hat is the probability that you pick a first digit and it is more than J5how a probability histogram for the distribution.

Solution

=3 x K 4 % =3 x % L4 $ =3 x % 4 $ =3 x % M4 $ =3 x% "4 % &.(('

ote that the height of each bar shows the probability of the outcome at its base. Theheights add up to ', of course. The bars in thishistogram have the same width, namely '. 5o,the areas also display the probabilityassignments

of the outcomes. Think of such histograms

3probability histograms4 as ideali2ed picturesof the results of very many repeated trials.

Pr o

bability

/

.%

/./

/.%

/./

/.$%

/.$/

/./%

/ / $ * % + -

9irst di6it of a cellphone code



523




xpected values

e probability distribution for andom variable looks verymilar to the relative fre+uency

tribution discussed in !hapterThe difference is that theative fre+uency distributionscribes a sample ofasurements, whereas the

obability distribution isnstructed as a model for theire

population. Nust as thean and standard deviation

ve you measures for

e centre and spread of themple data, you can calculatemilar measures to describee centre and spread of thepulation.

e population mean, whicheasures the average value ofn the population, is alsoled the expected value of

e random variable x. It is thelue that you would expect toserve onaverage if youpeat the experiment anfinite number of times. Thermula we use to determinee expected value can bemply understood with anample.

t’s revisit the milknsumption example. ;et x bee number of litres consumed.



re is the table ofobabilities again

x)

-

/

+

/

*

ppose we choose a largember of families, say '&& &&&.uitively, using the relative+uency concept of probability,u would expect to observe&& families consuming nolk, (& &&& consuming ' litre,d the rest similarly done )L&, (& &&&, '( &&& and @&&&.

e average 3mean4 value of x,defined in !hapter ", would

en be e+ual to

m of all measurements

FFFF



M&&& $ 'O(& &&& $ (O)L &&& $(& &&& $ @O'( &&& $ O@&&&

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

& &&&

M&&&

(& &&&

L &&&

& &&&

( &&&

&&&



& &&&

& &&&

& &&&

& &&&

& &&&

& &&&

&.&M $ 'O&.(& $ (O&.)L $ &.(& $ @O&.'( $ O&.&@

=3&4 $ 'O=3'4 $ (O=3(4 $ =3)4 $ @O=3@4 $ O=34 % (.(

at is, we expect to seemilies, onaverage, nsuming (.( litres of milk0is does not mean that weow what a family will nsume, but we can say whatexpect to happen.

x be a discrete randomiable 1ith probability distribution

x). 5he mean or expected value

x is 6iven by

% E( x) % Ʃ x P( x)



surance companies maketensive use of expected valueculations. 7ere is a

mplified example.

insurance company offers alicy that pays you :'& &&&en you totally damage your

or :&&& for ma$or damages&H4. They charge you :& perar for this service. Theestion is, how can they makeofitJ

"



To understand how they can afford this, suppose that the ‘total damage’car accident rate, in any year, is ' out of every '&&& cars, and that another( out of '&&& will have serious damages. Then we can display the

probability model for this policy in a table like this

5ype of accident

Amount paid x

Probability P( X % x)

5otal dama6e

$/ ///$

$///

a;or dama6e

%///

$///



inor or no dama6e

/

$///

The expected amount the insurance company pays is given by

m % 63 X 4 %

Ʃ

x=3 x4 % :'& &&&

'

$ :&&&

(



3'&&& 4

3'&&& 4

$ :&3 FFFF "" 4 % :(& '&&&

This means that the insurance company expects to pay, on average, anamount of :(& per insured car. 5ince it is charging people :& for the

policy, the company expects to make a profit of :)& per car. Thinkingabout the problem in a different perspective, suppose they insure '&&&cars, then the company would expect to pay :'& &&& for ' car and :&&&to each of two cars with ma$or damage. This is a total of :(& &&& for all

cars, or an average of FFFFFF (&'&&&&&& % :(& per car.

1f course, this expected value is not what actually happens to any particular policy. o individual policy actually costs the insurance

company :(&. *e are dealing with random events, so a few car ownersmay re+uire a payment of :'& &&& or :&&&, many others receivenothing 0 Gecause of the need to anticipate such variability, the insurancecompany needs to know a measure of this variability, which is nothing butthe standard deviation.



<ariance and standard deviation

%or data in !hapter ", we calculated the variance by computing thedeviation from the mean, x m, and then s+uaring it. *e do that withrandom variables as well.

*e can use similar arguments to $ustify the formulae for the population

variance s( and, conse+uently, the population standard deviation s. These

measures describe the spread of the values of the random variable aroundthe centre. *e similarly use the idea of the ‘average’ or ‘expected’ valueof the s+uared deviations of the x<values from the mean m or 63 x4.

7et x be a discrete random variable 1ith probability distribution P( x) and mean m.

5he variance o! x is 6iven by

s

% E3( x m) 4 % Ʃ( x m) P( x) (5his is sometimes called <ar( x).)

525




ote It can also be shown,milar to what you saw in

apter ", that you haveother ‘computation’ formular the variance

% Ʃ3 x m4( =3 x4 % Ʃ x

(

x4 m( % Ʃ x

( =3 x4

3x4C(

Ʃ x ( =3 x4 BƩ x=3 x4 C(

e standard deviation s of a

dom variable x is e4ual to thesitive s4uare root of its variance.

t us go back to the milknsumption example. ?ecallt we calculated the expectedue, mean, to be (.( litres. In

der to calculate the variance,can tabulate our work toke the manual calculation

mple.

x)

viation ( x m)

uared deviation ( x m)

m) P ( x)



-

*

-

/

*

--/

+

*

$**

/

-

*

-/

-

*



---

*

-

*

$+

tal

x )

m)

, the variance of the milk

nsumption is '.( litres(, ore standard deviation is '.())res.

$% notes

e above calculations, alongth the expected value



culation, can be easily doneng your 8E!.

rst, store x and =3 x4 into ;'d ;(.

L2 L3 2

6

(1)=.08

en, to find x =3 x4, weultiply ;' and ;( and storee result in ;).

*L2 L3

0 .2 .72 .6 .4…

find the expected value, youmply get the sum of thetries in ;), since they

rrespond to Ʃ x =3 x4.

*L2 L3

.2 .72 .6 .4…

m(L3)



2

&



L1-2.2 L4

To find the variance, we need to find the deviations from the mean- so

(-2.2 -1.2 -.2…

we make ;@ that deviation, i.e. we store ;' (.( into ;@. Then, to get the

(L4)2*L2 L5

s+uared deviations multiplied by the corresponding probability, we set

(.3872 .228 .01…

up ; to be ;@ s+uared multiplied by ;(, the probability. ow, to find the

sum(L5)

variance, $ust add the terms of ;.

'o!t(are note

In the comfort of home#class, the above calculation can be performed on a computer with a simplespreadsheet like the following one

x

P( x )

x P( x )

x m( x m)



( x m)P( x )

/

/./-

/

.

*.-*

/.-

$

/.

/.

$.

$.**

/.--

/.+

/.

/.

/./*

/./$**

/.

/.+

/.-

/.+*

/.$-

*

/.$

/.*-



$.-

.*

/.---

E*=B*

%

/./*

/.

.-

.-*

/.$+

5otals

$

.

$.%

A .

E+>

A=B

"?(C@C)



Example 5

A computer store sells a particular type of laptop. The daily demand for the laptops is given in the table below. x is the number of laptops in demand. They have only @ laptops left in stock and would like to

know how well they are prepared for all eventualities. *ork out the expected value of the demand andthe standard deviation.

x

/

$

*

%

P( x) /./- /.*/ /.* /.$% /./- /./%

Solution

63 x4 % Ʃ x =3 x4 % & &.&M $ ' &.@& $ ( &.(@ $ ) &.' $ @ &.&M $ &.& % '."&

Par3 x4 % s(

% Ʃ 3 x m4(

=3 x4

3& '."4( &.&M $ 3' '."4( &.@& $ 3( '."4( &.(@ $ 3) '."4( &.' $ 3@ '."4( &.&M $ 3 '."4( &.& %'.L)

s % '.(M

5preadsheet output is also given.

x



P( x )

x P ( x )

x m

( x m)

( x m)P( x )

/

/./-

/

$.

.+$

/.---

$

/.*

/.*

/.

/.-$

/.*

/.*



/.*-

/.$

/./$

/.//*

/.$%

/.*%

$.$

$.$

/.$-$%

*

/./-

/.

.$

*.*$

/.%-



%

/./%

/.%

.$

.+$

/.*-/%

5otals

$

$.

$.+

52)




e graph of the probabilitytribution is given below.

%

%



bability

%



%

%



mber of laptops



an approximation, we cane the empiricalrule to seehere most of the demand ispected to be. ?ecall that the

mpirical rule tells us thatout "H of the values wouldwithin ( standard deviations

m the mean. In this case m +% '." + ( '.(M ⇒ 3&.LL,@L4. This interval does notntain the units of demand.e can say that it is unlikelyat or more customers of thisop will want to buy a laptopday.

um(L1*L2)

DC

ter entering the demand in ;'

d the probabilities in ;(, it is

ough to

9

d the sum of their product.

r the variance, we follow theme procedure as described ine previous example, see right.

otice here that we combinedveral steps in one.

L2 L3



.4 .48 .45 .…

1-1.9)2*L2 L5

2888 .324 .00…

xpected valueroperties

um(L5)

63

e will start this sub<sectionth some examples.

ample &

e random variable Q has aobability distribution



nction 3=E%4 given in thelowing table.

X % x)

%

%

%

nd 63 X 4 and Par3 X 4.

lution

X 4 % > xP 3 x4 % & &.( $ '

& $ ( &. $ ) &.' $ @ &.' $

& % (

r 3 X 4 % > 3 x E 3 X 44( P 3 x4 %

(4( &.( $ 3' (4(

&.& $

% '.M

8



Example )

%or the data given in 6xample L, find each of the following.

a4

63) x4

b4

Par3) x4

c4

63) x $ 4

d4

Par3) x $ 4

Solution

a4 E 3) X 4 % > 3) x4 P 3) x4

& &.( $ ) &.& $ L &. $ " &.' $ '( &.' $ ' &.& % L

Var 3) X 4 % > 3) x E 3) X 44( P 3) x4

3& L4( &.( $ 3) L4( &.& $ / % 'L.(

E 3) X $ 4 % > 3) x $ 4 P 3) x$ 4

%



&.( $ M &.&

$ '' &. $ '@ &.' $ ' &.' $ (& &.&

% ''

d4 Var 3) X $ 4 %

> 3) x $ E 3) X $ 44( P 3) x $

4

%

3 ''4( &.(

$ 3M ''4( &.& $

/

%

'L.(



In the previous examples you can observe that 63) X 4 % L % )63 X 4 and that63) X $ 4 % '' % )63 X 4 $ . *e can generali2e this result.

9or a random variable X and constants a and b,

E(aX ) % aE( X ) and E(aX $ b) % aE( X ) $ b

Also, you can observe that Par3) X 4 % 'L.( % "Par3 X 4 as well as Par3) X $ 4 % 'L.( % " Par3 X 4.

*e can also generali2e this result.

9or a random variable X and constants a and b@

<ar(aX ) % a <ar( X ) and <ar(aX $ b) % a

<ar( X )

5he proof of the statements in this section is beyond the scope of the syllabus, buthere they are for interest.

E (aX $ b) % > (a x $ b) p( x) % > (a x p( x) $ bp( x))

> a x p( x) % > bp( x) % a > x p( x) $ b > p( x)

aE ( X ) $ b($) % aE ( X ) $ b

Var (aX $ b) % Var (Y ) % > (Y E (Y )) p(y )



> (aX $ b aE ( X ) b) p( x)

> (a( X E ( X ))) p( x)

a > ( X E ( X )) p( x) % a

Var ( X )

529




ample 8

has =E% as shown in thele below.

x)

$



nd the following.



63 X 4

63( X )4

Par3( X )4

lution

Q4 % (&)

$ '& )&

$ '

' $ (&

'&

) $ (

L&

') % 'L

ing a 8E!, we can store thelues of x in a list, L' forample, and the probabilitiesL(. Then multiply L' and L(d store them in L), andally get the sum of L). 7ere

a TI<M@$ output

L3

sum(L1*L2)

6



333

667



(6)

a !asiofx<8(& the output

Norm1 d/c

1

2

3 L$s% 4

4

20

0.3

25 0.2166

&''L &

L& L-+LL

N!R&

,a% Rad Norm1

d/c Rea

!um (L$s%

1L$s% 2)

16

!um rod um

L$s%

e variance can be calculatedng the 8E!

Norm1 d/c Rea

ar$ae



6

05

nce, Var 3Q4 % ( % @"

( X )4 % ( E 3Q4 ) % ( S ) % ("

r 3(Q )4 % (( Var 3 X 4 % @ S

% '"L

ample 9

e random variable X has theE% defined by the rule

X % x4 % k 3( x

(4, for x ',

), @, U.



nd thelue of k .

nd 63 X 4d Par3 X 4.

b4 %ind=3' V x )4.

d4 %ind63) X $ (4.

0



Solution

a4 *e make a table summari2ing the =E% of the variable.

x

$

*

%

P ( x)

*k

$k

$+k

k

/

5ince X is a random variable, then the sum of all its probabilities must be'.

7ence, > p3 x4 % (@k $ ('k $ 'Lk $ "k $ & % '⇒ k % &'

b4 P 3' V x )4

% P 3 X % (4 $ P 3 X % )4 % )k % &) .

c4 E 3 X 4 % > xP 3 x4 % ' (@k $ ( ('k $ ) 'Lk $ @ "k $ &

'&k

'



Var 3 X 4 % >3 x E 3 x44( P 3 x4

3' ' 4( (@k $ 3( '

4( ('k $ 3) ' 4( 'Lk $ 3@

' 4( "k $ 3

' 4( &

((k %

((@"& % (@

(L'

E 3) X $ (4 % ) '

$ ( % "

Exercise $%.$

(Problems mar&ed 1ith (=) are optional.)

Classify each of the follo1in6 as discrete or continuous random variables.

5he number of 1ords spelled correctly by a student on a spellin6 test.

5he amount of 1ater flo1in6 throu6h the !ia6ara 9alls per year.

5he len6th of time a student is late to class.

5he number of bacteria per cc of drin&in6 1ater in eneva.

5he amount of C produced per litre of unleaded 6as.

f ) 5he amount of a flu vaccine in a syrin6e.

5he heart rate of a lab mouse.



5he barometric pressure at ount Everest.

5he distance travelled by a taxi driver per day.

5otal score of football teams in national lea6ues.

ei6ht of ocean tides on the shores of Portu6al.

5ensile brea&in6 stren6th (in !e1tons per s4uare metre) of a % cm diametersteel cable.

!umber of overdue boo&s in a public library.

2 A random variable y has this probability distribution@

y

/

$

*

%



P( y)

/.$

/.

/.$

/./%

/./%

531




d P().

nstruct a probability histo6ramthis distribution.

d m and s.(=)

cate the interval m + s as 1ellm + s on the histo6ram.(=)

discrete random variable x cansume five possible values@ $, $%, $-, and /. Fts probabilitytribution is sho1n belo1.

x)



*

$

+

hat is P($%)H

hat is the probability that x uals $ or /H

hat is P( x 0 $-)H

d E( x).

d <( x).(=)

dical research has sho1n that atain type of chemotherapy is

ccessful /I of the time 1hened to treat s&in cancer. Fn ady to chec& the validity of suchlaim, researchers choseerent treatment centres and

ose five of their patients at

dom. ere is the probabilitytribution of the number ofccessful treatments for 6roups of e@



x)

/

/

+/

+-

d the probability that at leasto patients 1ould benefit from treatment.

d the probability that the;ority of the 6roup does notnefit from the treatment.

d E ( x) and interpret thesult.

o1 that s

( x) % $./.(=)

aph P( x). 7ocate m, m + s and+ s on the 6raph. ?se thepirical rule to approximate the

obability that x falls in thiserval. Compare this 1ith theual probability.

5he probability function of acrete random variable X is

en by P( X % x) % k

x

, for x

, $*, $+, $-.

t up the table sho1in6 thebability distribution and find theue of k .

has probability distribution aso1n in the table.



x)

d the value of k .

d P( x . $/).

d P(% , x 0 /).



d the expected value and thendard deviation.

he discrete random variable Y

s a probability density function

Y % y) % k ($+ y), for y % /, $,

, *.

d the value of the constant k .

a1 a histo6ram to illustrate thetribution.

2



9ind P($ 0 y 0 ).

9ind the mean and variance.

5he probability distribution of students cate6ori#ed by a6e that visit acertain movie house on 1ee&ends is 6iven belo1. 5he probabilities for $-2and $2year2olds are missin6. Ge &no1 that

P( x % $-) % P( x % $)

/.%

/.*

/.

/.

/.$

/

$%

$+

$

$-

$



Complete the histo6ram and describe the distribution.

9ind the expected value and the variance.

Fn a small to1n, a computer store sells laptops to the local residents.o1ever, due to lo1 demand, they li&e to &eep their stoc& at a mana6eablelevel. 5he data they have indicate that the 1ee&ly demand for the laptopsthey sell follo1s the distribution 6iven in the table belo1.

X @ number of laptops bou6ht

/

$

*

%

P( X % x)



/.$/

/.*/

/./

/.$%

/.$/

/./%

9ind the mean and standard deviation of this distribution.

?se the empirical rule to find the approximate number of laptops that is soldabout %I of the time.

Fs it li&ely that % or more customers buy a laptop in any 1ee&H

10 5he discrete random variable x has probability function 6iven by

P( x) %

T

3*$ 4

x ( $

x % , , *, %, +

k

x %

/



other1ise

1here k is a constant. Determine the value of k and the expected value of x.

11 5he follo1in6 is a probability distribution for a random variable y.

y

/

$

P(Y % y)

/.$

/.$$

k

(k J $)

9ind the value of k .

9ind the expected value.

A closed box contains ei6ht red balls and four 1hite ones. A ball is ta&en outat random, its colour noted, and then returned. 5his is done three times. 7et

X represent the number of red balls dra1n.



"et up a table to sho1 the probability distribution of X .

Ghat is the expected number of red balls in this experimentH

A discrete random variable Y has the follo1in6 probability distribution function

P(Y % y) % k (* J y ), for y % /, $, , and *.

533




9ind the value of k . b) 9ind P($ 0 y , ).

5he PD9 of a random variable X is 6iven in the follo1in6 table.

x)

( AA a

a

lculate a as 1ell as E ( X ).

e random variable X has the follo1in6 PD9@ P ( X % x) %

%k x, for x % $, , , *, %.

d k .



d E ( X ) and Var ( X ).

d E ( X $ %).

*e binomial distribution

amples of discrete random variables are abundant in everyday situations. 7owever, there are a few discreteobability distributions that are widely applied and serve as models for a great number of the applications. In thisok, we will study one of them only the binomial distribution.

e will start our discussion of the binomial distribution with an example.

ppose a cereal company puts miniature figures in boxes of cornflakes to make them attractive for children and thusost sales. The manufacturer claims that (&H of the boxes contain a figure. :ou buy three boxes of this cereal. *hathe probability that you’ll get exactly three figuresJ

get three figures means that the first box contains a figure 3&.(& chance4, as does the second 3also &.(&4, ande third 3&.(&4. :ou want three figures- therefore, this is the intersection of three events and the probability is

mply &.(&) % &.&&M.

ouwanttocalculateteprobabilityofgettingexactlytwofigures , the situation becomes more complicated. A treeagram can help you visuali2e it better.

x $

x

x



/-



-

-



-

f

$

n

"



;et f stand for figure and n for no figure. There are three events of interestto us. 5ince we are interested in two figures, we want to see ffn, which has

a probability of &.( &.( &.M % &.(( &.M % &.&)(, and the other eventsof interest are fnf and nff , with probabilities &.( &.M &.( % &.&)( and&.M &.( &.( % &.&)(.

5ince the order of multiplication is not important, you see that three probabilities are the same. These three events are dis$oint, as can beclearly seen from the tree diagram, and hence the probability of exactlytwo figures is the sum of the three numbers &.&)( $ &.&)( $ &.&)(. 1fcourse, you may reali2e by now that it would be much simpler if youwrote )3&.&)(4, since there are three events with the same probability.

*hat if you have five boxesJ

The situation is similar, of course. 7owever, a tree diagram would not beuseful in this case, as there is too much information to assemble to see thesolution. As you have seen above, no matter how you succeed in finding afigure, whether it is in the first box, the second or the third, it has the same probability, &.(. 5o, to have two successes 3finding figures4 in the five boxes,

you need the other three to be failures 3no figures4, with a probability of &.Mfor each failure. Therefore, the chance of having a case like ffnnn is

&.(( &.M). 7owever, this can happen in several dis$oint ways. 7owmanyJ If you count them, you will find '&. This means the probability of

having exactly two figures in five boxes is '& &.(( &.M) % &.(&@M.

37ere are the '& possibilities ffnnn, fnfnn, fnnfn, fnnnf , nffnn, nnffn, nnnff ,nfnfn, nnfnf , nfnnf .4

The number '& is nothing but the binomial coefficient 3=ascal’s entry4 yousaw in !hapter ). This is also the ‘combination’ of three events out offive. 3The proof of this result is beyond the scope of this book.4

The previous result can be written as 3( 4&.(( &.M), where 3

( 4 is the

binomial coefficient.



:ou can find experiments like this one in many situations. !oin<tossing isonly a simple example of this. Another very common example is opinion

polls which are conducted before elections and used to predict voter preferences. 6ach sampled person can be compared to a coin but a biased coin0 A voter you sample in favour of your candidate cancorrespond to either a ‘head’ or a ‘tail’ on a coin. 5uch experiments allexhibit the typical characteristics of the binomial experiment.

A binomial experiment is one that has the follo1in6 five characteristics@

5he experiment consists of n identical trials.

Each trial has one of t1o outcomes. Ge call one of them success, S, and the other failure, F .

5he probability of success on a sin6le trial, p, is constant throu6hout the 1holeexperiment. 5he probability of failure is $ p, 1hich is sometimes denoted by q.5hat is, p $ q % $.

5he trials are independent.

Ge are interested in the number of successes x that are possible durin6 the n

trials. 5hat is, x % /, $, , K, n.

535




the cereal company’sample above, we started with

% ) and p % &.( and asked fore probability of twoccesses, i.e. x % (. In thecond part, we have n % .

t us imagine repeating aomial experiment n times. Ifprobability of success is p, the

obability of having x successes

pppp!,x times 3 p

x4, becauseorder is not important, as we

w before. 7owever, in order

have exactly x successes, thet, 3n x4 trials, must belures, that is, with probability

""""!,

3n

x4 times 3"n

x

4.is is only one orderombination4 where theccesses happen the first x

mes and the rest are failures. Inder to cater for ‘all orders’, weve to count the number of

ers 3combinations4 possible. This

given by the binomial coefficient

4. x

e will state the followingult without proof.

e binomial distribution

ppose that a random experiment canult in t1o possible mutuallylusive and collectively exhaustive



comes, LsuccessM and Lfailure,M andt p is the probability of a successultin6 in a sin6le trial. Ff n ependent trials are carried out, theribution of the number of successes

resultin6 is called the binomial

tribution. Fts probability distributionction for the binomial random

able x is@

x successes in n independentls)

p)

x

x

r x %

, , K n.



tation,

e notation used to indicate that aiable has a binomial probabilitytribution 1ith n trials and

ccess probability of p is@ x ∼

n, p).

ample &

e computer shop orders itstebooks from a supplier,

hich like many suppliers hasate of defective items ofH. The shop usually takes ample of '& computers and

ecks them for defects. If theyd two computers defective,ey return the shipment. *hatthe probability that theirndom sample will containo defective computersJ

lution

e will consider this to be andom sample and thepment large enough to

nder the trials independent ofch other. The probability ofding two defectivemputers in a sample of '& isven by



x % (4 % 3'& ( 4&.'(&."'& (

@ &.&' &.@)&@ % &.'"@

course, it is a daunting taskdo all the calculations bynd. A 8E! can do thisculation for you in two

fferent ways.

e first possibility is to let theculator do all the calculationsthe formula above 8o to theth menu, then choose =?G,n go to W).

", R#

+&6 N", 0/.

0 r 2)*.12*.

&

#



d

8

ac

c

r

937102445

r

N(



:

N

rad%(

;,$(

radNorm(

;,a(

rad#$(



&



The second one is direct. *e go to the ‘EI5T?’ button, then scroll down to‘binompdf’ and write down the two parameters followed by the number ofsuccesses

R+<

$omd;(10>.1>2

!&R

1

ormad;(

)

.1937102445

2ormacd;(

3$?Norm(



4$?&(

5%d;(

6%cd;(

7

2d;(



Rsing a spreadsheet, you can also produce this result or even a set of probabilities covering all the possible values. The command used here for6xcel is 3GI1DEI5T3G'8','&,&.',%A;5644 which produced the table

below

x

/.//

$.//

.//

.//

*.//

%.//

+.//

.//

-.//

.//

$/.//

P ( x)

/.*

/.-

/.$*

/./%

/./$$

/.//$

/.///

/.///



/.///

/.///

/.///

5imilarly, the 8E! can also give you a list of the probabilities

$omd;(10>.1>L

L1

L2

L3 2

1) L2

0

.34868

(.3486784401 .3…

1

.38742



2

.19371

3

.0574

4

.01116

5

.00149

L2(1)=.3486784401…

;ike other distributions, when you look at the binomial distribution, youwant to look at its expected value and standard deviation.

Rsing the formula we developed for the expected value, Ʃ x= 3 x4, we canof course add x= 3 x4 for all the values involved in the experiment. The

process would be long and tedious for something we can intuitively know.%or example, in the defective items sample, if we know that the defective

rate of the computer manufacturer is '&H, it is natural to expect to have '& &.' % ' defective computer0 If we have '&& computers with a defectiverate of '&H, how many would you expect to be defectiveJ !an you thinkof a reason why it would not be '&J



This is so simple that many people would not even consider it. Theexpected value of the successes in the binomial is actually nothing but thenumber of trials n multiplied by the probability of success, i.e. np0

T*e binomial probability model n % number of trials

p % probability of success, q % $ p probability of failure x % number of successesin n trials

n

x

n x

%

n

x

n ( x

, for x %

/, $, , K n

P( x) % 3

4 p ($ p)

3

4 p q

x



x

Expected value % m % np

<ariance % s

% npq, s %

Xnpq

53)




, in the defective notebooksse, the expected number of

fective items in the sample of is np % '& &.' % '0

F FFFFFFFFFFFFF

nd the standard deviation is s Xnp" % X'& &.' &." % "@".

ample )

mong the studies carried outexamine the effectiveness ofvertising methods, a studyported that @ out of '& webrfers remembervertisement banners after

ey have seen them.

(& web surfers are chosen atndom and shown an ad, whatthe expected number ofrfers that would remembere adJ

hat is the chance that ofose (& will remember the adJ

hat is the probability that atost ' surfer would remembere adJ

hat is the chance that at leasto surfers would remembere adJ



lution

Y 3(&, &[email protected] expectedmber is

mply (& &.@

M. *e expectof the surfersremember the otice on

e histogramlow that theea in redrresponds to

theexpected valueM.

$om

d;(20>.4>

5

)

.

07464

70195

4 %3 (&

4&.@

L4' % &.&@L, or e the output frome 8E! to theht. 8raphically,s area is shownthe histogram as

e green area.

x 0 '4 % =3 x % &4=3 x % '4% &.&&&@

x 8 (4 % ' =3 x 0

' &.&&& (@ %""" @

$om

d;(20>

.4>0

)

3.6561

5844-

5

$omd;(20>

.4>1

)

4.874877

92-4

4.874877

92-4



to.ram o! (eb sur!ers

/

bability

%

/



*+

%

/



mber of surfers

8



Exercise $%.

(Problemsmar&ed 1ith

(=) areoptional.)

1 Consider thefollo1in6binomial

distribution P( x)

%3 x

% 4(/.+)

x(/.*) %

x, x %

/, $, K, %

a&e a tablefor thisdistribution.

raph thisdistribution.

9ind themean andstandarddeviation int1o 1ays@

by formula

by usin6 thetable ofvalues youcreated inpart a).

7ocate the

mean m andthe t1ointervals m +s and m + s on the 6raph.



9ind theactualprobabilities

for x to lie

1ithin eachof theintervals m +s and m + s

and comparethem to theempiricalrule.

A poll of /adults is ta&enin a lar6e city.5he purposeis to determine1hether

they supportbannin6smo&in6 in

restaurants.Ft is &no1nthatapproximately +/I of thepopulationsupports thedecision. 7et

x representthe numberof

respondentsin favour ofthe decision.

Ghat is theprobabilitythat %respondentssupport the

decisionH



Ghat is theprobabilitythat none ofthe /supports thedecisionH

Ghat is theprobabilitythat at least $respondentsupports thedecisionH

Ghat is theprobabilitythat at leastt1orespondentssupport thedecisionH

9ind the

mean andstandarddeviation ofthedistribution.

Consider thebinomialrandomvariable 1ithn % + and p %/..

9ill in theprobabilitiesbelo1.

k

/

$



*

%

+

P( x 0 k )

b) 9ill in thetable belo1."ome cellshave beenfilled for youto 6uide you.

!umber of

7ist the



Grite the

Explain it, if

9ind the

probability

re4uired

successes x

values of x

needed

statement

probability

At most



At least

ore than

*, %, +

P( x . )

$ P( x 0 )

/.// *



9e1er than

Bet1een

and

% (inclusive)



Exactly

3epeat4uestion 1ith n % and p % /.*.

A box

contains -balls@ % are6reen and are 1hite,red andyello1.5hree ballsare chosenat random1ithoutreplacement

and thenumber of

6reen balls y

is recorded.



Explain 1hy

y is not abinomialrandomvariable.

Explain 1hy,1hen 1erepeat theexperiment1ithreplacement,

then y is abinomial.

ive the valuesof n and p anddisplay theprobabilitydistribution intabular form.

Ghat is theprobabilitythat at most

6reen ballsare dra1nH

Ghat is theexpectednumber of6reen ballsdra1nH

f ) Ghat isthe varianceof thenumber ofballs dra1nH

6) Ghat isthe

probabilitythat some6reen balls1ill bedra1nH



539




a multiple choice test, theree $/ 4uestions, each 1ith %ssible ans1ers, one of 1hichcorrect. !ic& is una1are of thentent of the material andesses on all 4uestions.

d the probability that !ic&es not ans1er any 4uestionrrectly.

d the probability that !ic&s1ers at most half of the 4uestionsrectly.

d the probability that !ic&s1ers at least one 4uestionrrectly.

1 many 4uestions should !ic&pect to ans1er correctlyH

uses in a lar6e city areuipped 1ith alarm systems totect them from bur6lary. A

mpany claims their system to beI reliable. 5hat is, it 1ill tri66eralarm in -I of the cases. Fn atain nei6hbourhood, $/ houses

uipped 1ith this systemperience an attempted bur6lary.

d the probability that all therms 1or& properly.

d the probability that at leastf of the houses tri66er anrm.



d the probability that at most -rms 1ill 1or& properly.

rry Potter boo&s arerchased by readers of all a6esOI of arry Potter boo&s 1ererchased by readers / yearsa6e or olderO $% readers areosen at random. 9ind theobability that

east $/ of them are / orer

of them are / or older

most $/ of them are youn6ern /.

actory ma&es computer hard&s. ver a lon6 period, $.%I ofm are found to be defective. Adom sample of %/ hard dis&s isted.

ite do1n the expected numberdefective hard dis&s in themple.

d the probability that threerd dis&s are defective.

d the probability that moren one hard dis& is defective.

r colour preferences chan6eer time and accordin6 to theea the customer lives in and car model heshe is

erested in. Fn a certain city, a6e dealer of BG cars noticed

t $/I of the cars he sells areetallic 6reyM. 51enty of hisstomers are selected atndom, and their car orders are



ec&ed for colour. 9ind theobability that

east five cars are LmetalliceyM

most + cars are Lmetallic 6reyM

re than % are Lmetallic 6reyM

t1een * and + are LmetalliceyM

re than $% are not LmetalliceyM.

a sample of $// customercords, find

the expected number of

etallic 6reyM car orders

the standard deviation ofetallic 6reyM car orders.(=)

cordin6 to the empirical rule,I of the Lmetallic 6reyM orderse bet1een a and b.

9ind a and b.(=)

6s have health insurance tooO1ners of do6s in manyuntries buy health insurancetheir do6s. I of all do6s

ve health insurance. Fn andom sample of $// do6s in a6e city, find



expected number of do6sh health insurance

probability that % of the do6sve health insurance

probability that more than $/

6s have health insurance.

balanced coin is tossed % times.

x be the number of headsserved.

in6 a table, construct the

obability distribution of x.

hat is the probability that noads are observedH

hat is the probability that allses are headsH

hat is the probability that atst one head is observedH

hat is the probability that atst one tail is observedH

iven that the coin is

balanced in such a 1ay that ito1s heads in every $/ses, ans1er the sameestions above.

0



$%. T*e normal distribution

Continuous random variables

*hen a random variable x is discrete, you assign a positive probability to eachvalue that x can take and get the probability distribution for x.

The sum of all the probabilities associated with the different values of x is '.

:ou have seen, in the discrete variable case, that we graphically represent the

probabilities corresponding to the different values of the random variable x with a probability histogram 3relative fre+uency histogram4, where the areaof each bar corresponds to the probability of the specific value it represents.

!onsider now a continuous random variable x, such as height and weight,and length of life of a particular product a TP set for example. Gecause it iscontinuous, the possible values of x are over an interval. Doreover, there arean infinite number of possible values of x. 7ence, we cannot find a

probability distribution function for x by listing all the possible values of xalong with their probabilities, as you see in the histogram below. If we try toassign probabilities to each of these uncountable values, the probabilities willno longer sum to ', as is the case with discrete variables. Therefore, you mustuse a different approach to generate the probability distribution for suchrandom variables.

5uppose that you have a set of measurements on a continuous randomvariable, and you create a relative fre+uency histogram to describe theirdistribution. %or a small number of measurements, you can use a smallnumber of classes, but as more and more measurements are collected, youcan use more classes and reduce the class width.

3elativ

efr e4uency

/./

/.$%

/.$/

/./%

/.//



3elativefr e

4uency

x

/./

/.$%

/.$/

/./%

/.//

x

The histogram will slightly change as the class width becomes smaller andsmaller, as shown in the diagrams on the next page. As the number ofmeasurements becomes very large and the class width becomes very narrow, the

relative fre+uency histogram appears more and more like the smooth curve yousee below. This is what happens in the continuous case, and the smooth curvedescribing the probability distribution of the continuous random variable becomes the PDF 3probability density

5"1




nction4 of x, represented by arve y

% f 3 x4. This curve is

ch that the entire area undere curve is ' and the areatween any two points is theobability that x falls betweenose two points.

/

/

bability

%

bability

%

/

/

%



%

/

/

robability densitynction

t x be a continuous randomriable. The probabilitynsity function, f 3 x4 , of thendom variable is a functionth the following properties

4 . & for all values of x.

e area under the probabilitynsity function f 3 x4 over alllues of

e random variable x is e+ual'.&.

) ). 5uppose this density

nction is graphed. ;et a and



be two possible values of thendom variable x, with a , b.en the probability that x liestween a and b B=3a , x , b4C ise area under the density

nction between these points.

tice that, based on this

finition, the probability that

uals any

int a is !" This is so because

area above a value, saya,

is a

tangle

hose width is &. 5o, for thentinuous case, regardless ofhether the



a x b)

dpoints a and b are

emselves included, the areacluded between a

a x b)

d b is the same.

a , x , b4 % =3a 0 x 0 b4 % =3a 0b4 % =3a , x 0 b4

ntinuous probabilitytributions can assume ariety of shapes. 7owever, for asons of staying within 3withme extensions4 theundaries of the IG syllabus,will focus on onetribution. In fact, a largember of random variablesserved in our surroundingsssess a fre+uencytribution that isproximately bell<shaped. *el that distribution thermal probability

stribution.

he normalstribution



e most important type ofntinuous random variable ise normal random variable.e probability density

nction of a normal randomriable x is determined by tworameters the mean orpected value m and thendard deviation s of the

riable.

e normal probability densitynction is a bell<shaped densityrve that is symmetric about thean m. Its variability isasured by s. The larger

2



the value of s the more variability there is in the curve. That is, the higher the probabilityof finding values of the random variable further away from the mean. %igure '.'represents three different normal density functions with the same mean but differentstandard deviations. ote how the curves ‘flatten’ as s increases. This is so because thearea under the curve has to stay e+ual to '.

/iure 15.1

σ /.%

σ

µ µ

σ *

µ

robability density !unction o! t*e normal distribution

5he probability density function for a normally distributed random variable x is

$



( x m)

$

$

x m

f ( x) %

s



3 s

4

for , x ,

X

AAA e

%

X

AAA e

s

p

s

p



1here m and s

are any number such that , m , and / , s

, , and 1here e and p are the 1ell2&no1n constants e % .$- -K and p % .$*$ %K.

+otation,

Ghen a variable is normally distributed, 1e 1rite X , ! (m, s

).

Although we will not make direct use of the formula above, it is interesting to note its properties, because they help us understand how the normal distribution works. oticethat the e+uation is completely determined by the mean m and the standard deviation s.

The graph of a normal probability distribution is shown in%igure '.(. As you notice, the mean or expected valuelocates the centre of the distribution, and the distributionis symmetric about this mean. 5ince the total area underthe curve is ', the symmetry of the curve implies that thearea to the right of the mean and the area to the left are both e+ual to &.. The shape, or how ‘flat’ it is, is

determined by s, as we have seen in %igure '.'. ;argevalues of s tend to reduce the height of the curve andincrease the spread, and small values

of s increase the height to compensate for thenarrowness of the distribution.

Area to the left Area to the ri6ht

of the mean is /.% of the mean is /.%

Q

/iure 15.2



5"3



ability Distributions

he normal distribution is fully determined by its mean, m, and



andard deviation, s. !hanging m without changing s moves the

mal curve along the hori2ontal axis without changing its spread.

ou have seen above, the standard deviation s controls the spread



e curve. :ou can also locate the standard deviation by eye on

urve. 1ne s to the right or left of the mean m marks the point

re the curvature of the curve changes. That is, as you move right

the mean, at the point where x % m $ s, the curve changes its



urvature from downwards to upwards. 5imilarly, as you move one

the left from the mean the curve changes its curvature from

re 15.3

nwards to upwards.

ough there are many normal curves, they all have common properties.

is one important one that you have seen in !hapter "

+-I



σ %I .I

re 15."

empirical rule

ted

e normal distribution 1ithn m and standardtion s@

oximately +-I of thervations fall 1ithin s ofean m.

oximately %I of thervations fall 1ithin s ofean m.

Approximately .I of the observations fall 1ithin s of the mean m.

%igure '.@ illustrates this rule. ;ater in this section, you will learn how to find thesareas from a table or from your 8E!.

Example 8

7eights of young 8erman men between 'M and '" years of age follow a distributionthat is approximately normal, with a mean of 'M' cm and a standard deviation of Mcm 3approximately4. Eescribe this population of young men.

Solution

According to the empirical rule, we find that approximately LMH of those young mehave a height between ') cm and 'M" cm, "H of them between 'L cm and '"cm, and "".H between ' cm and (& cm. ;ooking further, you can say that only&.'H are taller than (& cm, or shorter than ' cm.

As the empirical rule suggests, all normal distributions are the same if we measure iunits of si2e s about the mean m as centre. !hanging to these units is called

standardi#ing . To standardi2e a value, measure how far it is from the mean andexpress that distance in terms of s. This is how the calculation can be done

dardiin

a normal random variable, 1ith mean m and standard deviation s, the

rdi#ed value of x is x m

s

ndardi#ed value is also called the z score.

+uantity x m tells us how far our value is from the mean- dividing by s then tellsow many standard deviations that distance is e+ual to.

The standardi2ing process, as younotice, is a transformation of thenormal curve. %or discussion purposes, assume the mean m to



ive. The transformation x m shifts the graph back m units. 5o, the new centre ised from m back m units. That is, the new centre is &0 Eividing by s

ing to ‘scale’ the distances from the mean and express everything in terms of s.point that is one standard deviation from the mean is going to be ' unit above the

mean, i.e. it will be represented by $'. ow, if you look at the empirical rule weussed earlier, points that are within one standard deviation from the mean will be

in a distance of ' in the new distribution. Instead of being at m $ s and m s, theybe at & $ ' and & ' respectively, i.e. ' and $'. 35ee %igure '.L.4

new distribution we created by this transformation is called the standard normal

ibution. It has a mean of & and a standard deviation of '. It is a very helpfulibution because it will enable us to read the areas under any normal distributionugh the standardi2ation process, as will be demonstrated in the examples thatw.

ability density !unction o! t*e standard normal distribution

probability density function for standard normal distribution is

$ e $ (z ) for , z ,

e linear transformations can transform all normal functions to standard, thismes a very convenient and efficient way of finding the area under any normalibution.

us look at an example.

ung 8erman man with a height of '"( cm has a # <score of

m

'M'

# %

FFFFF

FFFFFFFFF

% '.)

s %

M

or '.) standard deviationsabove the mean. 5imilarly, ayoung man with a height of 'cm is

x m

' 'M'

# %

FFFFF

FFFFFFFFF

% &.

s %

M

or &. standard deviations belothe mean.



nd the probability that a normal variable x lies in the interval a to b, we need to find

rea under the normal curve 3m, s

(4 between the points a and b. 7owever, there is anitely large number of normal curves one for each mean and standard deviation. 35eere '..4

parate table of areas for each of these curves is obviously not practical. Instead, weone table for the standard normal distribution, which gives us the re+uired areas.

n you standardi2e a and b, you get two standard numbers # ' and # ( such that the

between # ' and # ( is the same as the area we need.

z

re 15.5

+-I

$

$

/



.I

/iure 15.&




ure 15.)

$ / $.%

$-$

/ z a Q b



a complete normalribution table visit1.pearsonhotlin&s.com,er the F"B! or title ofboo& and select

blin& *.

P (Z z )

z

/iure 15.8

In the example above, if we are interested in the proportion of young8erman men whose height is between ' cm and '"( cm, wecalculate the # <scores for these numbers and then read the area from thetable. 7ere is an abbreviated version of the table and instructions onhow to use it. 3There are many tables of the areas under normaldistributions. *e will use a table constructed in a similar way to the oneused on IG examinations.4

z

/.//

/./$

Z

/./%

/./+

/./

/./-

/./

/./

/.%///

/.%/*/

/.%$

/.%

/.%

/.%$

/.%%

/.$

/.%-

/.%*-

/.%%+

/.%++



+%

$*

%

-

-

/+

/+*

$/

$*$

-*

-+

**

+

-/

/$%

Z

[

$.

/./

/./*

/.$$%

/.$$

/.$*

0.91&2

/.$

$.*

/.$

/./

/.+%

/.

/.

/./+

/.$



*%

*

*/+

*$-

*

**$

*%

*+

/.%/%

/.%$%

/.%%

/.%%

/.%*%

The table, as constructed, gives you the areas under the normaldistribution to the left of some value # , as you see in %igure '.M.

The table starts at &, and gives the areas till # % ).". To read an area tothe left of a number # , say '.), you read the first column to find thefirst two digits of # . 5o, in the first column, we stop at the cellcontaining '.). To get the area for '.), we look at the first row andchoose the column corresponding to &.&. *here the row at '.) meetsthe column at &.& is

&



the area under the normal distribution corresponding to '.), namely, &."'@. That is, the probability of at most a heigwith # % '.) is &."'@. 5ince the table does not go to @ decimal places, our answers will not be very precise. 5o, to finthe probability corresponding to a height of '"( cm, we need a # of '.), which is not in the table. *e can use '.),'.)M, or take an average. If we want an average, we read the neighbouring area of &."'L(,

and get the average to be &."' @.

/.% /

Rnfortunately, due to limitations of space, this type of table does

not cater for negative values of # . The good news is that, due to the

/iure 15.9

symmetry of the distribution, the area to the left of a negative value

of # is the same as the area to the right of its absolute value. 5o, if

we are interested in the area to the left of &., we look for the

$ P(z /.%)

area to the right of &.I, i.e. ' =32 V &.I4 % ' &.II)@ % &.((LL



3see %igure '.'&4. 5o, in the example above, if we want to know the

P(z /.%)

probability of a young 8erman man, chosen at random, having a

P(z /.%)

height between 'I cm and '"( cm, we look up the corresponding

area under the standard normal distribution between &. and



/.% / /.%

'.). 5ince these two areas are cumulative, we need to subtract

/iure 15.10 P(z . /.%)

them, i.e. the re+uired area is &."' @ &.((LL % &.LMM.

% $ P(z , /.%).



/.++

/iure 15.11

/.+--%

/.% / $.% /.% / $.%

*hat is the chance that a young 8erman man is taller than ' cmJ



/iure 15.12

P(z /.%) P(z /.%)

P(z /.%)

/.% / /.% /.% / /.%

This means that we have to look at the area above &.. Eue to symmetry, the area in +uestion, which is to theright of &., is e+ual to the area below &., which in turn can be read directly from the table as &.)@.

These calculations are much easier to calculate using a 8E!, of course. Also, with the 8E!, you do not need tostandardi2e your variables either. 7owever, because there are cases where you need to understand

5")




r

d;

.

>

.

375).6888069418

standardi2ation andother cases whereyou are re"uiredto

useatable

, you needto know bothmethods.

7ere is how your8E! can give youyour answers.

:ou first go to the‘Eistribution’ menuand choose

‘normalcdf’. Thenyou enter thenumbers in thefollowing order

Lowerlimit,upper

limit,mean, and

standarddeviation.

The result will bethe area you need.5ee the screen

images below.

R+<

ormacd;(175>

19

!&R



o

a

;(

18

8)

r

;

8

6

1

$

or

$

(

%

(

6%cd;(

7

2d;(

If you want to usethe standard normal,your commands will

be the same, but youdo not need toinclude the meanand standarddeviation. They arethe default.

If you need the probability that ayoung man is taller

than ' cm, youcan also read iteither by looking atthe distribution withthe original data or

by standardi2ing.

Example 9

The age of graduatestudents inengineering programmesthroughout the R5 isnormally distributedwith mean m % (@.

and standarddeviation s % (..



adeisosat

ndm,

hattheobilit

#shsunr

an

arsdJ

hatopio

ofdeis

der an.ars

hatrceage

de

nts is between ((and (M years oldJ

what percentage ofthe ages falls within' standard deviationof the meanJ (standard deviationsJ) standarddeviationsJ

Solution

If we let X % age ofstudents, then X Y

3m % (@., s

(%

L.(4.

To answer this, wecan eitherstandardi2e and thenread the table for thearea left of &.L

=3 # ,

FFFFFFFF (L(@.

4 %

=3 # , &.L4 % &.(, or

use a 8E! (.

ormacd;(0>26

>2

4.5>2.5)

.7257469354



otice here that we put & as awer limit. :ou can put amber as a lower limit farough from the mean to makere you are receiving therrect cumulative distribution.

8



b4 This can be done similarly

=3 x . ().4 % = 3 # . FFFFFFFFFF ().(@. % &.)( 4 (.

5o, by symmetry we know that

=3 # . &.)(4 % =3 # , &.)(4 % &.L(

*ith a 8E! ormacd;(23.7>1 00>24.5>2.5

.6255157701

Also, notice here that we wrote '&& as an upper limit, which is anarbitrary number far enough to the right to be sure we include the whole

population.

(( (@.

(M (@.

c4 =3((

, x , (M4

FFFFFFFF

, # ,

FFFFFFFF

% = 3' , # , '.@4



% =

(.

(.4

3

*e find the area to the left of '.@ and to the left of ' and subtract them %&."'"( &.'M % &.L&L % L.&LH

*ith a 8E! ormacd;(22>28> 24.5>2.5)

.7605880293



This, as you know, is the empirical rule we talked about before. ;et us seewhat percentage of the approximately normal data will lie within ', ( or )standard deviations.

*e start with the traditional table

=3' 0 # 0 '4 % =3 # 0 '4 =3 # 0 '4 % &.M@') &.'M

% !"#$%#

This is the exact value corresponding to the empirical rule’s LMH0

=3( 0 # 0 (4 % =3 # 0 (4 =3 # 0 (4 % &."( &.&((M

% !"&'((

Again, this is the exact value corresponding to the empirical rule’s "H0

=3) 0 # 0 )4 % =3 # 0 )4 =3 # 0 )4 % &.""M &.&&')

% !"&&)*

And again, this is the exact value corresponding to the empirical rule’s"".H0

5he inverse normal distribution



Another type of problem arises in situations similar to the one above whenwe are given a cumulative probability and would like to find the value

in our data that has this cumulative probability. %or example, what age

5"9




rks the "th percentileJ That is, what age is higher or e+ual to "H of the populationJ To answer this +uestion, weed to reverse our steps. 5o far, we are given a value and then we look for the area corresponding to it. ow, we are

ven the area and we have to look for the number. That is why this is called the inverse normal distribution. Again,approach is to find the standardinversenormalnumberandtento$de%standardi#e&it.

at is, to find the value from the original data that corresponds to the # <value at hand.

ere is an inverse normal table available online 3see box on page @L4. *e

P (Z z )

ll produce a part of the inverse normal table here for explanation.



//

/$

/%

/+

/

/-

/

/

///

/%



$%

$%/

$%

/$

+

$

%$

+

+

*/$

*+

*%$

*+



%/

%

+

+%

+

/

-

%

-

--

/*

%*

ure 15.13



*

*

*+*

%--

+/

+%$

+-

$

%



*%

+

/

%

+

/$

+

/+

/%

%

%

/

%+

--

*$

%%*

%--

+$

+%%

+--

-

%+

-

+

+$

%

//

/+*

/



*

$/

*%

-$

e table gives a selection of probabilities above the mean and the body of the table gives the # <value correspondingthat area. :ou know that & has a cumulative probability of &.. ;ook at the table and observe the

ersection of the &.& row and the &.&&& column. It is &, the mean of the standard normal distribution.

we need to know what # <score the third +uartile \) is, for example, we need to look up &.. The # <score

rresponding to \) is &.L@ as you see.

ppose you want to find the # <score that leaves an area of &."' below it.

//

/%

%

/

///



$%

$

%$

+



$

*/-

ure 15.1"

In the first column, we choose &."', then at the intersection of the row at&."' and the column at &.&& the # <score corresponds to &."'. 5o,

P 3 # , '.)((4 % &."''

0



The 8E! can also be used in this case. The process is identical to the normalcalculation. The difference is in choosing ‘invorm’ instead.

$?Norm(.5)

$?Norm(.915)0 1.37220381

$?Norm(.75)

.6744897495

In the young 8erman men example, we would like to find what heightleaves "H of the population below it.

In this case, we look up the # <score corresponding to &." and we findthat it is # % '.L@@".

ow # % '.L@@" % FFFFFFF x'M'

⇒ x 'M' % M '.L@@" M

⇒ x % 'M' $ M '.L@@" % '"@.'L.

5o, "H of the young 8erman men are shorter

than '"@.'L cm.

The 8E! gives you thisnumber with less effort

$?Norm(.95>181

>

8)

194.158829

Example 10



5ince ovember (&&, the average time it takes fast trains 36urostar4 totravel between ;ondon and =aris is ( hours ' minutes, with a standarddeviation of @ minutes. Assume a normal distribution.

*hat is the probability that a randomly chosen trip will take longer than( hours and (& minutesJ

*hat is the probability that a randomly chosen trip will take less than (hours and '& minutesJ

*hat is the I\? of a trip on these trainsJ

Solution

*e will do each problem using a table and a 8E! to ac+uaint you with

both methods.



a4 The mean m % (.( and s % &.&L.

( hours (& minutes % ( FF '

]

FF hours- @ minutes ]

FFF @

]

FFF '

hour

)

)

L&



'.

I

=3 x

FF

(.(



)

. FF ) 4 % = #

. FFFFFFFF FFF '

4

% =3 # . '.(4



3

'

%rom the table =3 # . '.(4 % ' =3 # , '.(4 % ' &.M"@@ % &.'&L

Rsing your 8E! ormacd;(7/3>10 0>2.25>1/15)

.105649839

The number '&& is arbitrary/ $ust so long as it is a large number.

551




rm

cd

0>

/6

25

/1

56

83

b4 ( hours '& minutes%

')& hours ]

') hours

L&

L

= 3 x ,

')

4 % = #

')

(.(

4

L



FFFF

# ,.(

om

eble,d

mmy,s is

eme

=3 #

(4,hich

we found in part a4above.

GDC

<N'<

!adeNorm(0>13/

6

m$=2

>2.25>1/15)

ma=2.5

sc1=1

@m$=-2



a

c1

+rea=.10565

u=2.16667

res=1

oA=0

\' is the numberthat leaves (H ofthe data before it.5o, we need to findthe inverse normalvariable that has anarea of &.( beforeit.

%rom the table wecan only do sousing symmetry. 5o,we find the # <scorethat corresponds to&.( by finding its

symmetricalnumber, which isthe # <score with&.. 5o, we onlyneed to find # 3&.4.The table ofstandard inversenormal gives us # % &.L@.

5o, \' correspondsto &.L@.



ability to walk can be described by anormal model. It isknown that H of

babies learn how towalk by the age of'& months and (Hneed more than ')months. %ind the

mean and standarddeviation of thedistribution.

I

&in6 a6e

onths)

$

+*%

/.+*

Solution

;ooking at thediagram at leftwill help youvisuali2e thesolution. *e willshow twoapproaches tothis problem.

The firstapproach is toconsider thedistance between'& and ')months. In ourdata, thatdistance is )

months, but howmany standarddeviations doesthat representJ5ince we know

2



that '& months represents the lower H, and ') months represents the upper(H, we can obtain # <scores for those two data points without knowing themean and standard deviation. Rse the inverse table or a 8E!

$?Norm(.05)

-1.644853626 $?Norm(.75)

.6744897495

Therefore, the )<month distance is e+uivalent to &.L@ 3'.L@4 ] (.)'"standard deviations or

(.)'"s % ) ⇒ s % '.("@

And finally we use either of the data points 3'& months or ') months4 in

our # <score formula to find the mean

x m

') m

FFFFF

FFFFFF



# %

s ⇒ &.L@ %

'.("@

⇒ m %

'(.'(M

Thus the mean age that babies begin to walk is '(.' months with astandard deviation of '.(" months.

The second approach uses a bit of algebra instead. After obtaining the # <scores from the inverse table or 8E! 3as above4, we begin by writing twoe+uations using the # <score formula

'& m

'.L@ % FFFFFF

s ⇒ m '.L@s % '&,

') m

&.L@ % FFFFFF s ⇒ m $ &.L@s % ')

Aha0 *e have two linear e+uations with two unknowns. 5olve these toobtain

m % '(.'(M s % '.("@

Again, we conclude the mean age that babies begin to walk is '(.'months with a standard deviation of '.(" months.

Exercise $%.

5he time it ta&es to chan6e the batteries of your DC is approximately normal1ith mean %/ hours and standard deviation of .% hours.

9ind the probability that your ne1ly e4uipped DC 1ill last



a)

at least %/ hours

d)

bet1een *.% and %.% hours

b)

bet1een %/ and % hours

e)

more than +% hours

c)

less than *.% hours

f )

*.% hours

9ind each of the follo1in6 probabilities.

P(^ z ^ , $.)

P(^ z ^ . $.*)

P( x , .), 1here x Y ! (, )

P( x . .), 1here x Y ! (, )

A car manufacturer introduces a ne1 model that has an in2city milea6e of $$.*litres$// &ilometres. 5ests sho1 that this model has a standard deviation of$.+. 5he distribution is assumed to be normal.

A car is chosen at random from this model.

Ghat is the probability that it 1ill have a consumption less than -.* litres$//&ilometresH

553




hat is the probability that thensumption is bet1een -.* and* litres $// &ilometresH

d the value of z that 1ill beceeded only $/I of the time.

d the value of z % z / such that

I of the values of z lie bet1een

/ and

/.

e scores on a public schools

amination are normallytributed 1ith a mean of %%/d a standard deviation of $//.

hat is the probability that andomly chosen student froms population scores belo1/H

hat is the probability that adent 1ill score bet1een *%/ and/H

hat score should you have inder to be in the /thrcentileH

d the FR3 of this distribution.



company producin6 andc&a6in6 su6ar for homensumption put labels on their6ar ba6s notin6 the 1ei6ht to%// 6. 5heir machines are

o1n to fill the ba6s 1ithi6hts that are normallytributed 1ith a standardviation of %. 6. A ba6 thatntains less than %// 6 is

nsidered to be under1ei6htd is not appreciated bynsumers.

he company decides to setir machines to fill the ba6sh a mean of %$ 6, 1hatction 1ill be under1ei6htH

hey 1ish the percenta6e ofder1ei6ht ba6s to be at most

I, 1hat mean settin6 must theyveH

hey do not 1ant to set the meanhi6h as %$, but instead at %$/,at standard deviation 6ives

m at most *I under1ei6ht6sH

a lar6e school, hei6hts ofdents 1ho are $ years old

e normally

tributed 1ith a mean of $%$

and a standard deviation of -. 9ind the probability that a

ndomly chosen child is

shorter than $++ cm b)hin + cm of the avera6e.

e time it ta&es Sevin to 6et tohool every day is normallytributed 1ith a mean of $nutes and standard deviation



minutes. Estimate thember of days 1hen he ta&es

6er than $ minutes

s than $/ minutes

t1een and $ minutes.

ere are $-/ school days invinMs school.

as a normal distribution 1ithan $+. iven that the probabilityt X is less than $+.%+ is +*I,

d the standard deviation, s, ofs distribution.

has a normal distribution 1ithriance of . iven that theobability that X is more than% is .I, find the mean, m,

his distribution.

has a normal distribution sucht the probability that X is

6er than $*.+ is .%I and

x . .+) % .I. 9ind thean, m, and the standardviation, s, of this distribution.

Y !(m, s

), P( x . $.+) % /.$+d P( X , $.+) % /./$. 9ind m

d s.

tles of mineral 1ater sold by ampany are advertised to contain $

of 1ater. 5o 6uarantee customersfaction the company actually

usts its

n6 process to fill the bottlesh an avera6e of $/$ ml. 5he

ocess follo1s a normal



tribution 1ith standardviation of % ml.

d the probability that andomly chosen bottle containsre than $/$/ ml.

d the probability that a bottlentains less than the advertisedume.

a shipment of $/ /// bottles,at is the expected number ofderfilledM bottlesH

"



Cholesterol plays a ma;or role in a personMs heart health. i6h blood cholesterol is a ma;orris& factor for coronary heart disease and stro&e. 5he level of cholesterol in the blood ismeasured in milli6rams per decilitre of blood (m6dl). Accordin6 to the G, in 6eneral,less than // m6dl is a desirable level, // to is borderline hi6h, and above */ is ahi6h ris& level and a person 1ith this level has more than t1ice the ris& of heart disease asa person 1ith less than a // level. Fn a certain country, it is &no1n that the avera6echolesterol level of their adult

population is $-* m6dl 1ith a standard deviation of m6 dl. Ft can bemodelled by a normal distribution.

Ghat percenta6e do you expect to be borderline hi6hH

Ghat percenta6e do you consider are hi6h ris&H

Estimate the inter4uartile ran6e of the cholesterol levels in this country.

Above 1hat value are the hi6hest I of adultsM cholesterol levels in this countryH

A manufacturer of car tyres claims that the treadlife of its 1inter tyres can be

described by a normal model 1ith an avera6e life of % /// &m and astandard deviation of */// &m.

Tou buy a set of tyres from this manufacturer. Fs it reasonable for you tohope they last more than +* /// &mH

Ghat fraction of these tyres do you expect to last less than *- /// &mH

Ghat fraction of these tyres do you expect to last bet1een *- /// &m and%+ /// &mH

Ghat is the FR3 of the treadlife of this type of tyreH

5he company 1ants to 6uarantee a minimum life for these tyres. 5hat is,they 1ill refund customers 1hose tyres last less than a specific distance.Ghat should their minimum life 6uarantee be so that they do not end uprefundin6 more than I of their customersH



Chic&en e66s are 6raded by si#e for the purpose of sales. Fn Europe, modern e66 si#esare defined as follo1s@ very lar6e has a mass of 6 or more, lar6e is bet1een + and 6, medium is bet1een % and + 6, and small is less than % 6. 5he small si#e is usuallyconsidered as undesirable by consumers.

ature hens (older than $ year) produce e66s 1ith an avera6e mass of + 6. -I ofthe e66s produced by mature hens are above the minimum

desirable 1ei6ht. Ghat is the standard deviation if the e66 production canbe modelled by a normal distributionH

Toun6 hens produce e66s 1ith a mean masss of %$ 6. nly -I of their e66sexceed the desired minimum. Ghat is the standard deviationH

A farmer finds that I of his farmMs e66s are Lunder1ei6htM, and $I are very lar6e.Estimate the mean and standard deviation of this farmerMs e66s.

Practice 4uestions

1 3esidents of a small to1n have savin6s 1hich are normally distributed 1ith amean of U/// and a standard deviation of U%//.

Ghat percenta6e of to1nspeople have savin6s 6reater than U//H

51o to1nspeople are chosen at random. Ghat is the probability that both ofthem have savin6s bet1een U// and U//H

5he percenta6e of to1nspeople 1ith savin6s less than d dollars is *.I.9ind the value of d .

2 A box contains % red discs and % blac& discs. A disc is selected at randomand its colour noted. 5he disc is then replaced in the box.

Fn ei6ht such selections, 1hat is the probability that a blac& disc is selected



exactly onceH

at least onceH

555




process of selectin6 and replacin6 isried out *// times. Ghat is theected number of blac& discs that

uld be dra1nH

he 6raph sho1s a normal curve forrandom variable X , 1ith mean m

standard deviation s. y

s &no1n that P( X 8 $) % /.$.

e shaded re6ion A is the

6ion under the curve 1here x 8 Grite do1n the area of the

aded re6ion A.

s also &no1n that P( X 0 -) %.



d the value of m, explainin6ur method in full.

o1 that s % $.%+ to ancuracy of si6nificant fi6ures.

d P( X 0 $$).

fair coin is tossed ei6ht times.culate

probability of obtainin6actly * heads

probability of obtainin6actly heads

probability of obtainin6 , * or eads.

he lifespan of a particular species of ect is normally distributed 1ith aan of % hours and a standardiation of *.* hours.

e probability that the lifespan ofinsect of this species lies1een %% and +/ hours isresented by the shaded area infollo1in6 dia6ram. 5his

6ram represents the standardrmal curve.

ite do1n the values of a and

d the probability that thespan of an insect of thisecies is

re than %% hours



t1een %% and +/ hours.



I of the insects die after t urs.

4i 3epresent this information

tandard normal curve



6ram, similar to the oneo1n,

/

icatin6 clearly the arearesentin6 /I. 4ii 9ind theue of t .

n urban hi6h1ay has a speed limit of

&m h$. Ft is &no1n that the speeds of icles travellin6 on the hi6h1ay aremally distributed, 1ith a standardiation of

m h$, and that /I of the vehiclesn6 the hi6h1ay exceed the speedt. a "ho1 that the mean speed of

vehicles is approximately **.- &m.

he foo!"n# part "s opt"ona and

$%rrenty not "n the syab%s of

ths S'.)

e police conduct a L"afer Drivin6M

mpai6n intended to encoura6e1er drivin6, and 1ant to &no1ether the campai6n has beenective. Ft is found that a sample of



vehicles has a mean speed of

&m h$.

ven that the null hypothesis is

@ the mean speed has beenaffected by the campai6n state

te $, the alternativepothesis.

ate 1hether a one2tailed oro2tailed test is appropriate forse hypotheses, and explainy.

s the campai6n had si6nificantect at the %I levelH

&



) Fntelli6ence 4uotient (FR) in a certain population is normally distributed 1ith amean of $// and a standard deviation of $%.

Ghat percenta6e of the population has an FR bet1een / and $%H

Ff t1o persons are chosen at random from the population, 1hat is theprobability that both have an FR 6reater than $%H

(The foo!"n# part "s opt"ona and "s $%rrenty not "n the syab%s of &aths S'. )

5he mean FR of a random 6roup of % persons sufferin6 from a certain brain disorder 1asfound to be %.. Fs this sufficient evidence, at the /./% level of si6nificance, that peoplesufferin6 from the disorder have, on avera6e, a lo1er FR than the entire populationH "tateyour null hypothesis and your alternative hypothesis, and explain your reasonin6.

8 Ba6s of cement are labelled % &6. 5he ba6s are filled by machine and the actual 1ei6hts arenormally distributed 1ith mean %. &6 and standard deviation /.%/ &6.

Ghat is the probability a ba6 selected at random 1ill 1ei6h less than %./ &6H

Fn order to reduce the number of under1ei6ht ba6s (ba6s 1ei6hin6 less than % &6) to.%I of the total, the mean is increased 1ithout chan6in6 the standard deviation.

b "ho1 that the increased mean is +./ &6.

Ft is decided to purchase a more accurate machine for fillin6 the ba6s. 5here4uirements for this machine are that only .%I of ba6s be under % &6and that only .%I of ba6s be over + &6.

Calculate the mean and standard deviation that satisfy these re4uirements. 5he cost of the ne1machine is U%///. Cement sells for U/.-/ per &6.

Compared to the cost of operatin6 1ith a + &6 mean, ho1 many ba6s must be filledin order to recover the cost of the ne1 e4uipmentH



9 5he mass of pac&ets of a brea&fast cereal is normally distributed 1ith a meanof %/ 6 and standard deviation of % 6.

9ind the probability that a pac&et chosen at random has mass

less than */ 6

at least -/ 6

bet1een */ 6 and -/ 6.

51o pac&ets are chosen at random. Ghat is the probability that bothpac&ets have a mass that is less than */ 6H

5he mass of /I of the pac&ets is more than x 6rams. 9ind the value of x.

Fn a country called 5allopia, the hei6ht of adults is normally distributed 1itha mean of $-.% cm and a standard deviation of .% cm.

Ghat percenta6e of adults in 5allopia have a hei6ht 6reater than $ cmH

A standard door1ay in 5allopia is desi6ned so that I of adults have a space of at least$ cm over their heads 1hen 6oin6 throu6h a door1ay. 9ind the hei6ht of a standard

door1ay in 5allopia. ive your ans1er to the nearest cm.

Ft is claimed that the masses of a population of lions are normally distributed 1ith amean mass of $/ &6 and a standard deviation of / &6.

Calculate the probability that a lion selected at random 1ill have a mass of%/ &6 or more.

5he probability that the mass of a lion lies bet1een a and b is /.%, 1here a and b are symmetric about the mean. 9ind the values of a and b.



55)




action times of human bein6se normally distributed 1ith aan of /.+ seconds and andard deviation of /./+conds.

e 6raph belo1 is that of thendard normal curve. 5headed area represents theobability that the reaction time

a person chosen at random ist1een /./ and /. seconds.

/ b x

ite do1n the values of a and

culate the probability that thection time of a person chosen atdom is

6reater than /./ seconds



bet1een /./ and /.conds.

ee per cent (I) of the populatione a reaction time less than $ seconds. 3epresent this information on a

6ram similar to the one above.

icate clearly the areapresentin6 I.

9ind $ .

actory ma&es calculators.er a lon6 period, I of them

e found to be faulty. A randommple of $// calculators isted.

ite do1n the expected number faulty calculators in themple.

d the probability that threeculators are faulty.

d the probability that moren one calculator is faulty.

speeds of cars at a certain point on ai6ht road are normally distributed 1ithan m and standard deviation s. $%I ofcars travelled at speeds 6reater than

&m h$ and $I of them at speeds

s than */ &m h$. 9ind m and s.

6 A contains red balls and en balls. 51o balls are chosen

random from the ba6 1ithoutlacement. 7et X denote the

mber of red balls chosen. 5heo1in6 table sho1s thebability distribution for X .



X % x)

Calculate E( X ), the meanmber of red balls chosen.

6 B contains * red balls and 6reens. 51o balls are chosen at random

m ba6 B. b 4i Dra1 a tree dia6ramepresent the above information,udin6 the

obability of each event.

nce, find the probabilitytribution for Y , 1here Y is the

mber of red balls chosen.

standard die 1ith six faces ised. Ff a $ or + is obtained, t1ols are chosen from ba6 A,er1ise t1o balls are chosenm ba6 B.

lculate the probability that t1od balls are chosen.



ven that t1o red balls aretained, find the conditionalobability that a $ or + 1ased on the die.

8



Ball bearin6s are used in en6ines in lar6e 4uantities. A car manufacturer buysthese bearin6s from a factory. 5hey a6ree on the follo1in6 terms@ 5he carcompany chooses a sample of %/ ball bearin6s from the shipment. Ff they findmore than defective bearin6s, the shipment is re;ected. Ft is a fact that thefactory produces *I defective bearin6s.

Ghat is the probability that the sample is clear of defectsH

Ghat is the probability that the shipment is acceptedH

Ghat is the expected number of defective bearin6s in the sample of %/H

Each CD produced by a certain company is 6uaranteed to function properly 1ith aprobability of -I. 5he company sells these CDs in pac&a6es of $/ and offers amoney2bac& 6uarantee that all the CDs in a pac&a6e 1ill function.

Ghat is the probability that a pac&a6e is returnedH

Tou buy three pac&a6es. Ghat is the probability that exactly $ of them mustbe returnedH

5he table belo1 sho1s the probability distribution of a random variable X .

x

/

$

P( X % x)

k

(k (

k $ k



k $ k

Calculate the value of k .

9ind E( X ).

Ft is estimated that .I of the cherry tomato fruits produced on a certain farm areconsidered to be small and cannot be sold for commercial purposes. 5he farmers have toseparate such fruits and use them for domestic consumption instead.

$ tomatoes are randomly selected from the produce. Calculate

the probability that three are not fit for sellin6

the probability that at least four are not fit for sellin6.

Ft is &no1n that the si#es of such tomatoes are normally distributed 1ith a mean of cmand a standard deviation of /.% cm. 5omatoes that are cate6ori#ed as lar6e 1ill have tobe lar6er than .% cm. Ghat proportion of the produce is lar6eH

Ruestions $J$%@ V Fnternational Baccalaureate r6ani#ation

statistics and probabilities

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