pre-calculus / math notes (unit 12 of 22)

7/30/2019 Pre-calculus / math notes (unit 12 of 22)

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Unit 12

Probability and Calculus II

Continuous Distributions

So far, we have only been concerned with discrete random variables. Re-call that a discrete random variable is one that can take any one of (only)a finite number of values. In this section, we study another kind of randomvariable, called a continuous random variable.

Definition 12.1. A Continuous Random Variable is one which may assumeany of an infinite number of values on some interval of the real number line.

Because a continuous R.V. can have any of an infinite number of values,events are defined as sub-intervals in which the value of the R.V. could fall(i.e. we never define events as specific values of the R.V.).

Example 1. Consider the experiment: randomly pick a number between 0and 1 inclusive. Let X be the number picked, so that X is a continuousrandom variable. Identify the sample space, S, for this experiment and someevents on that sample space.

Solution: Here, X can take on any of the infinitely many numbers whichlie in the interval [0, 1]. So it is this interval that is the sample space, i.e.,S =

{0

X

1

}or S =

{x

|0

x

1

}.

Notice: The notation {x| some condition } says the set of all xs suchthat the specified condition is satisfied. That is, the | means such that.

Any sub-interval of the sample space is a possible event. For instance, wemight be interested in the event A, that the number chosen is bigger than.5, or in event B, that a number between 1

4and 1

3is chosen. We have

A = {0.5 < X 1} and B = 14

X 13

.

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For a continuous random variable, because there are so many possible

values, the probability of any exact particular value occurring is effectively0. That is:

Theorem: For a continuous R.V., X, prob{X = x} = 0 for all values x.(This is why we dont bother with events which correspond to specific valuesof X.)

Example 2. Identify the following random variables as either Discrete or Con-tinuous.

(a) X is the sum of 2 dice when rolled ;(b) Y is the height of a randomly selected calculus student.

Solution:(a) This is a discrete random variable because there are only 11 possiblevalues (the integers 2 through 12).

(b) A persons height, when measured precisely, can have any real valuewithin some range. That is, we could identify limits on the possible heightvalues which might be observed - an interval outside of which a personsheight could not possibly be. For instance, it would appear reasonable to

assume that no calculus student could ever be less that 1 foot tall, or morethan 10 feet tall. However, within a certain range, any value could be ob-served. So in this case, Y is a continuous random variable.

Notice: This means that the probability of any particular height is 0. Forinstance, measuring in feet, prob{X = 6} = 0, i.e., the probability that some-one is exactly 6 feet tall, not 6.001 feet or 5.999 feet or 5.9999982 feet, is 0.However, the probability that someone is within, say, 0.1 feet of being 6 feetin height, i.e., prob{5.9 X 6.1} is not 0.

Definition 12.2. Associated with any Continuous Random Variable X,there is a function, f(x), called its probability density function, which satisfies

prob{c X d} =dc

f(x)dx

Notice: This probability is the area under the curve y = f(x), between x = cand x = d.

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In order to be a probability density function (often abbreviated as pdf),

f(x) must satisfy both of the following conditions:

1. f(x) 0 for all values of x and2. The total area under the curve y = f(x) must be 1 i.e.

f(x)dx = 1

That is, we know that probabilities must always be between 0 and 1. Inorder to ensure that

dc

f(x)dx is non-negative for all possible c and d values,the curve y = f(x) must never fall below the x-axis. Also, we must have (asalways) Prob(S) = 1, so the total area under the curve must be 1.

Note: If f(x) is a probability density function, we consider its domain tobe all real values of x. Typically, f(x) is defined as some particular func-tion in some interval (the sample space), and is defined as f(x) = 0 elsewhere.

Notice also: Recognizing that the probability of an event involving a contin-uous R.V., X, is an area helps us to understand why prob{X = x} = 0. Evenif c is a possible value of X, we have prob{X = c} = c

cf(x)dx = 0. That

is, the probability that X is exactly c is the area under the curve y = f(x)right at x = c. But this region is just a line segment, and its area is 0.

And notice: Since prob{X = x} = 0 for all x, then there is no difference be-tween prob{X x} and prob{X < x}. That is, with continuous R.V.s, wedo not need to distinguish between the events {X x} and {X < x}. Thisis not true for discrete random variables. IfX is a discrete random variableand x is one of its possible values, then these 2 events are very different.

Example 3. Show that f(x) = 38x2 when 0 x 2 and f(x) = 0 otherwise

could be a pdf for some random variable.

Solution: We need to show that this function has the 2 properties requiredof a probability density function. First, we observe that 38x

2 0 for all xin the interval [0, 2]. So we have f(x) = 0 on (, 0) and on (2, ), andf(x) 0 on [0, 2] and hence f(x) 0 for all values of x, as required.

We must also verify that the total area under y = f(x) is 1. Of course, wehave:

f(x)dx =

0

f(x)dx +

20

f(x)dx +

2

f(x)dx

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So then

f(x)dx =

0

0dx +

20

3

8x2dx +

2

0dx

= 0 +3

8

20

x2dx + 0

=3

8

x3

3

20

=3

8

23 03

3

= 1

We see that both of the requirements of a pdf are satisfied, so this functioncould be the pdf for some random variable.

Example 4. Find the value of the constant k such that the function f(x) = kxon the interval [3,5] and f(x) = 0 elsewhere is a probability density function.

Solution: We need to have f(x) 0 for all values ofx. Clearly, we need k tobe some non-negative constant, in order to have kx 0 on [3,5], so we knowthat k 0. To find the specific value of k, we use the fact that we need thetotal area under y = f(x) to be 1, i.e. we need

f(x)dx = 1. Of course,

since f(x) = 0 everywhere outside the interval [3,5], then we have

f(x)dx =

53

kxdx

so we actually need 53

kxdx = 1

We get:

5

3 kxdx = k

5

3 xdx = kx

2

25

3

=k

2

52 32 = 16k

2= 8k

Thus we see that we need 8k = 1, so k = 18

. That is, f(x) is a pdf only whenk = 1

8.

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Example 5. The continuous random variable X has pdf f(x) = 4225(x3 x)

on the interval [1,4] and f(x) = 0 elsewhere. Find the probability that Xhas a value between 2 and 3.

Solution: We see that X has possible values which lie on the interval [1,4].Also, we know that the probability that X has a value between some num-bers a and b which are in the interval on which the pdf is non-zero is givenby prob{a X b} = b

af(x)dx. We are looking for prob{2 X 3}, and

both 2 and 3 lie in [1, 4], so f(x) = 4225(x3 x) on [2, 3]. Thus we have

prob

{2

X

3

}=

3

2

4

225

(x3

x)dx

=4

225

x4

4 x

2

2

32

=4

225

34

4 3

2

2

24

4 2

2

2

=4

225

81

4 9

2 16

4+

4

2

=

4

22555

4

=

11

45

We see that prob{2 X 3} = 1145 .

Example 6. For the continuous random variable X in the preceding example,find prob{X 2}.

Solution: Here, the possible values of X are the numbers in the interval[1,4], so {X 2} means {1 X 2}. That is, technically the event{X 2} is the event { X 2}, but since f(x) = 0 on (, 1) then

prob{ X 1} = 0 and we see thatprob{X 2} = prob{ X 2}

= prob{ X 1} + prob{1 X 2}= 0 +prob{1 X 2}= prob{1 X 2}

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Thus we have

prob{X 2} =

2

1

4225

(x3 x)dx = 4225

x4

4 x2

2

2

1

=4

225

24

4 2

2

2

14

4 1

2

2

=4

225

16

4 4

2

1

4 1

2

=4

225

2

1

4

=

4

225

9

4

=

9

225

Definition 12.3. Let X be a Continuous Random Variable with probabilitydensity function f(x). Then , the mean of X, is given by

=

x f(x)dx

and 2, the variance of X, is given by

2 =

(x )2 f(x)dx =

x2f(x)dx 2

Also, the standard deviation of X is defined to be the positive square root

of the variance, 2

, and is denoted by . That is, we define the standarddeviation ofX as: =

2

Notes:

1. Notice the similarities to the definitions of mean and variance of adiscrete random variable. Summing over all discrete possible valuesbecomes integrating over all continuous possible values, i.e. over theentire interval over which the density function is defined.

2. Just as in the case of discrete random variables, we have 2 formulas

for variance here. The first is, technically, the definition of variance,while the second (which can easily be shown to be equivalent) is easierto use.

3. Because of the way that variance is defined, variance is always non-negative. That is, (x )2 0 and also f(x) 0, so it must be truethat

(x )2f(x)dx 0. Therefore, =

2 is always defined,

for any continuous random variable.

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4. As we have seen before, if the pdf ofX is only non-zero on some interval

[a, b], then for both mean and variance we need only integrate from ato b, rather than from to .

Example 7. Let X be a continuous random variable with pdff(x) = 38x2 on

[0, 2] and f(x) = 0 elsewhere. Find the mean and standard deviation of X.

Solution: We have =

xf(x)dx, but f(x) = 0 everywhere outside of the

interval [0, 2]. Thus we get

= 2

0

x

3

8x2

dx =3

8 2

0

x3dx =3

8

x4

4

20

=3

8

24

4 0

4

4

=

3

8

(4) =

3

2

To find the standard deviation, , we must first find the variance, 2. Wehave

2 =

x2f(x)dx 2 =20

(x2)

3

8x2

dx

3

2

2

=

3

820 x

4

dx 9

4 =

3

8x5

52

0 9

4 =

3

825

5 05

5

9

4

=3

8

32

5

9

4=

12

5 9

4=

48 4520

=3

20

We see that the variance of X is 2 = 320 and so the standard deviation of X

is =

2 =

320

.

Example 8. X is a continuous random variable with pdf f(x) = 12 sin x on

[0, ] and f(x) = 0 elsewhere. Find the mean and variance of X.

Solution: First of all, notice that f(x) = 12

sin x on [0, ] and f(x) = 0 el-swhere does meet the requirements of a pdf. From x = 0 to x = (i.e.,in the first and second quadrants), the function y = sin x increases from 0to 1 (at x =

2) and then decreases back to 0. Therefore for the function

f(x) = 12

sin x on [0, ] and f(x) = 0 elswhere, we have f(x) 0 everywhere.

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As well, we see that

f(x)dx =

0

0dx +

0

1

2sin xdx +

0dx

=1

2

0

sin xdx =1

2[ cos x]0

= 12

[cos cos 0] = 12

[(1) (1)] = 12

(2) = 1

Now, we need to find the mean of X. We have

=

xf(x)dx =

0

x12

sin x dx =

0

x sin x

2dx

To evaluate this, we use integration by parts, letting u = x2

and dv = sin xdxso that du = 12dx and v = cos x. We get:

=

0

x sin x

2dx = x cos x

2

0

0

cos x2

dx

=

0

cos x

2dx x cos x

2

0

=sin x

2

0

x cos x2

0

= (sin sin0) ( cos 0cos0)2

=

0 0 (1) + 0(1)

2

=

2

Therefore the mean of X is = 2

. For the variance of X we need to find

2 =

x2f(x)dx 2 =0

x2

1

2sin x

dx

2

2

To evaluate the integral here, we must use integration by parts twice. First,

let u =x2

2 and dv = sin xdx, so that du = xdx and v = cos x. Then we getx2 sin x

2dx = x

2 cos x

2

(x cos x)dx = x2 cos x

2+

x cos xdx

Now, to find

x cos xdx, we use u = x and dv = cos x, so that du = dx andv = sin x. We get

x cos xdx = x sin x

sin xdx = x sin x( cos x)+C = x sin x+cos x+C

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Thus we have x2 sin x

2dx = x2 cos x

2+ x sin x + cos x + C

This gives

2 =

0

x2 sin x

2dx

2

2

=

x

2 cos x

2+ x sin x + cos x

0

2

2

=

2 cos

2 +

sin

+ cos

02 cos0

2 + 0 sin 0 + cos0

2

4

=

2(1)2

+ (0) + (1)

0(1)

2+ 0(0) + (1)

2

4

=

2

2 2

2

4=

2

4 2

The Exponential Distribution

Next, we look at two useful types of continuous distributions which oftenarise in practical situations. For the first one, consider the following question.

Example 9. For what values ofk is the function f(x) = kekx on the interval[0, ) and f(x) = 0 elsewhere a probability density function?

Solution: For f(x) to be a probability density function, we need f(x) 0 forall x. Since ekx > 0 for all values of k and all values of x, then kekx 0only for k 0. That is, we cannot have k < 0.

As well, we require that

f(x)dx = 1, so we must have

0kekxdx = 1.

If k = 0, then

0 kekxdx =

0 0dx = 0, so we cannot have k = 0.

For any k > 0, we have

0

kekxdx = limb

b0

kekxdx

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And so

0

kekxdx = limb

k

ekx

kb

0

= limb

ekxb0

= limb

ekb + e0

= 1 limb

ekb

But of course limb

ekb = limb

1ekb

, and since k > 0 then as b we havekb and so ekb so that 1

ekb 0.

Therefore

0 kekxdx = 1 0 = 1 for any k > 0 and so the given function

is a probability density function for any k > 0.

This gives us an important family of continuous random variables.

Definition 12.4. A random variable, X, whose probability density functionhas the form

f(x) = kekx on [0, ) for some k > 0= 0 on (

, 0)

is called an exponentially distributed random variable. The value k, whichdistinguishes one member of this family from another, is called a parameter.

Example 10. If the continuous random variable X has pdf f(x) = 2e2x on[0, ) and f(x) = 0 elsewhere, find prob{X 1}.

Solution: Notice that X is an exponential random variable. The parameter

value is k = 2. Here, as with any exponential random variable, f(x) is onlynon-zero for x 0. So the event {X 1} is actually the event {0 X 1}.Thus we get:

prob{X 1} = prob{0 X 1}

=

10

f(x)dx =

10

2e2xdx

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So we have

prob{X 1} = 2

e2x

210

= e2x10

= e2 (e0) = e2 + 1

= 1 1e2

.86466

Notice: Since

kekxdx = ekx + C, in general for any exponential R.V.

X, we have

prob{0 X a} =a0

keax = ekxa0

= eak + e0 = 1 eak

Fact: Exponentially distributed R.V.s arise in situations involving randomarrivals. That is, if X measures the time between occurrences of someevent that occurs randomly, then X is a random variable with an exponen-tial distribution. Examples: arrivals of customers dialling up to an internetprovider, arrivals of patients at an emergency room. (We refer to the timebetween 2 consecutive arrivals as the interarrival time.)

Theorem 12.5. If X is a Random Variable with probability density functionf(x) = kekx on [0, ) and f(x) = 0 elsewhere, then the mean and standardderivation of X are respectively:

=1

k, =

1

k

That is, if X has an exponential distribution, then the mean and thestandard deviation of X are both given by 1

k, where k is the parameter ap-

pearing in the pdf of X. (Note: Exponential R.V.s are the only R.V.s forwhich = .)

The fact that = 1k

can be turned around to tell us what the parametervalue is, if we know the average time between arrivals. For instance, if weknow that the average time between phone calls arriving at a switchboardis 4 minutes (where the phone calls are assumed to arrive at random), thenwe have = 4. But we know that = 1

k, so then k = 1

= 14 . So if X is

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the time between arrivals of phone calls, then X is exponentially distributed

with pdf f(x) =14e

1

4

x

on [0, ) (and f(x) = 0 elsewhere). Notice that forexponentially distributed interarrival times, the parameter value is the meanarrival rate. For instance, if the mean time between calls is 4 minutes, thencalls arrive at an average rate of 1

4per minute.

Example 11. During a particularly busy period, patients arrive at the emer-gency room at University Hospital at an average rate of 1 patient every5 minutes. Assuming that the time between arrivals is exponentially dis-tributed, find the probability that 2 patients arrive within 1 minute.

Solution: Let T be the time between (consecutive) arrivals of patients. ThenT is an exponentially distributed random variable. Were told that the aver-age time between arrivals is 5 minutes, so we have = 5. And we know that = 1

k, so we have 1

k= 5 k = 15 = .2 (that is, patients arrive at an average

rate of .2 per minute). Therefore the pdf of T is f(t) = .2e.2t on [0, ) andf(t) = 0 elsewhere. We need to find prob{T 1}.

We can use our previous observation that for an exponentially distributedR.V. with parameter k, we have prob{0 X a} = 1 eak. In this case,for the exponentially distributed R.V. T with parameter value k = .2, we get

prob{T 1} = prob{0 T 1}

= 1 e(1)(.2) = 1 1e.2

.18127

Example 12. For the period described in the previous example, what fractionof patients will arrive at the emergency room at UH between 3 and 6 minutesafter the previous patient arrived?

Solution: Notice that What fraction of patients will arrive between 3 and6 minutes after the previous patient arrived? means what fraction of in-terarrival times are between 3 and 6 minutes?, which is another way ofsaying what is the probability that the time between consecutive arrivals isbetween 3 and 6 minutes?. That is, we have T being the random variabledefined in the previous example and we want to find prob{3 T 6}.

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We get:

prob{3 T 6} =

6

3

f(t)dt =

6

3

.2e.2tdt

= e.2t63

= e.2(6) (e.2(3))

= e.6 e1.2 .2476We see that approximately 1

4of patients will arrive between 3 and 6 minutes

after the previous patient arrived.

The Normal Distribution

Definition 12.6. A Random Variable X whose probability density functionhas the form

f(x) =1

2e

1

2(x )

2

, on (, )

for some values of and , with > 0, is called a normally distributed ran-dom variable.

The Normal Distribution is the most important, i.e., the most commonlyused, of all probability distributions. This is because many phenomena inthe real world are either normally distributed or at least approximately nor-mally distributed. Hence the normal distribution gives at least a very goodapproximation of the probabilities involved in many real situations.

Notice that the pdf of the normal distribution involves 2 parameters,shown above as and . The reason that we use and to represent these2 parameters is, of course, because the parameter values turn out to be themean and the standard deviation, respectively, of the distribution. That is,

it can be shown that:

Theorem 12.7. A normally distributed random variable X, with the pdf:

f(x) =1

2e

1

2(x )

2

on (, )

has mean and standard derivation .

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There are a whole family of Normal Distributions, with different values

of and . The pdf of a normally distributed random variable is the fa-miliar bell-curve or bell-shaped curve. The graph is symmetric aboutthe value , with the height and thickness of the bell determined by . (Asmaller value of gives a taller, thinner bell-shape, while a larger value of gives a shorter, squatter bell.) Of course, the total area under the curve (i.e.under the bell) is always 1.

There is one very special particular Normal Distribution which is veryuseful the one which has mean = 0 and standard deviation = 1. Anormal R.V. with this particular distribution is useful because any othernormally distributed R.V. can easily be expressed in terms of this StandardNormal Random Variable. This allows probabilities for any normally dis-tributed R.V. to be easily calculated, as long as we know probability valuesfor the standard normal random variable. By convention, this standard nor-mal random variable is represented by Z.

Definition 12.8. The standard normal random variable Z is normally dis-tributed with = 0, = 1. Thus, Z has the probability density function:

f(z) =12

e1

2(z2) on (, )

The graph ofy = f(z) is shown in Figure 1. (Notice: In spite of appear-ances in this picture, the tails of f(z) do not actually touch the z-axis, butcontinue out to infinity, getting closer and closer to the axis.) We see thatthe graph is a bell-shaped curve which is symmetric about the vertical axis(i.e., symmetric about the line z = , and for Z we have = 0). As for anypdf, the probability that the observed value of Z lies in the interval [c, d] isgiven by the area under the curve between z = c and z = d. For instance, thehighlighted region of the graph in Figure 1 corresponds to prob{0 Z z}for some value z.

0 Z

Figure 1: The pdf of the standard normal random variable Z, with areacorresponding to prob{0 Z z} shaded.

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Looking at the functional expression for the pdf ofZ quickly shows that

these areas involve an integral which we do not know how to evaluate. Thatis, we have

prob{c Z d} =dc

12

e1

2(z2)dz

and we find that neither substitution nor integration by parts is helpful inevaluating this.

Instead of evaluating this integral, we use a table to look up probabilitiesfor the standard normal random variable Z. The values ofprob{0 Z z}for z-values correct to 2 decimal places have been tabulated. We use these

values, and also the facts that the pdf is symmetric about the mean and thattherefore half of the area lies to the left of the mean (and half to the right)to calculate any probabilities involving Z that we need. The table of theseprobabilities (areas) is given on the next page (and also on p. 547 of the text).

The table gives values for prob{0 Z a.bc}, where a.bc represents anumber with 2 decimal places, from 0.00 to 3.09. The table has rows for a.band columns for 0.0c. That is, to find Prob{0 Z a.bc} from the table,we go to the row corresponding to the integer-part and first decimal place,and then find the column which corresponds to the second decimal place.The table entry in that row and column is prob

{0

Z

a.bc

}.

Example 13. Find prob{0 Z 1.51}.

Solution: In the table, find the row Z = 1.5 and look in the column 0.01(because we have a.bc = 1.51 in this case).

Z 0.00 0.01 0.02 0.03 ... 1.4 0.4192 0.4207 0.4222 0.4236 1.5 0.4332 0.4345 0.4357 0.4370

1.6 0.4452 0.4463 0.4474 0.4484 ...

We see that prob{0 Z 1.51} = 0.4345.

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To find other probabilities involving Z, we use our understanding of the

symmetry properties of the graph of the pdf and other basic properties ofareas. For instance, for any c > 0 and d > c, prob{c Z d} corresponds tothe area under f(z) from z = c to z = d. But this area can also be expressedas (the area under f(z) from z = 0 to z = d) minus (the area under f(z)

fromz = 0 to z = c). That is, we have

prob{c Z d} = prob{0 Z d} prob{0 Z c}

Also, because the graph is symmetric about z = 0,

prob{a Z 0} = prob{0 Z a}

for any positive value a. That is, the area under the curve from z = 0 to az-value a units away from 0 is the same, whether we go a units to the rightor a units to the left. And since the total area under the pdf is 1, we see that

prob{Z 0} = prob{Z 0} = 0.5. (Notice that in the table, as the z-valuegets large, the probability prob{0 Z z} approaches .5 .)

With these simple relationships we can find any probabilities involving Zwhich we need, using the tabulated values of prob{0 Z z}.

Example 14. IfZ is the standard normal random variable, find the following:(a) prob{1 Z 2} (b) prob{Z 0.5}(c) prob{0.32 Z 1.06} (d) prob{Z > 2.19}

Solution:(a) For prob{1 Z 2}, we are looking for the probability of the event{1 Z 2}. As noted above, we have

prob{1 Z 2} = prob{0 Z 2} prob{0 Z 1}

On the right hand side of this equation we just have 2 probabilities which wecan look up in the Z-table. We look up the table entries in row 2.0, column

0,00 and row 1.0, column 0.00 to see that

prob{1 Z 2} = .4772 .3413 = .1359

(b) To find prob{Z 0.5}, we use the symmetry of the Z-distribution tosee that prob{Z 0.5} = prob{Z 0.5}. (That is, the area under thecurve to the right of z = 0.5 is the same as the area under the curve tothe left of z = 0.5.) And of course {Z 0.5} = {Z < 0} {0 Z 0.5}

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so that prob{Z 0.5} = prob{Z < 0} + prob{0 Z 0.5}. Using the factthat prob{Z 0} = .5, and looking up prob{0 Z 0.5} in the table (row0.5, column 0.00), we get

prob{Z 0.5} = prob{Z 0.5}= prob{Z 0} + prob{0 Z 0.5}= .5 + .1915 = .6915

Notice: In general, for any a > 0, we have

prob{Z a} = prob{Z a} = .5 + prob{0 Z a}

(c) To find prob{0.32 Z 1.06}, we need to realize that{0.32 Z 1.06} = {0.32 Z 0} {0 Z 1.06}

so that

prob{0.32 Z 1.06} = prob{0.32 Z 0} + prob{0 Z 1.06}Of course, we know that prob{0.32 Z 0} = prob{0 Z 0.32}, sowe have

prob

{0.32

Z

1.06

}= prob

{0.32

Z

0

}+ prob

{0

Z

1.06

}= prob{0 Z 0.32} + prob{0 Z 1.06}= .1255 + .3554 = .4809

Notice: In general, for a > 0 and b > 0, we have

prob{a Z b} = prob{0 Z a} + prob{0 Z b}

(d) One way that we can find prob{Z > 2.19} is to find the probability ofthe complementary event. We have

{Z > 2.19}C = {Z 2.19}and of course for any event A, prob(Ac) = 1 prob(A), so we have

prob{Z > 2.19} = 1 prob{Z 2.19}= 1 (prob{Z < 0} + prob{0 Z 2.19})= 1 (.5 + .4857) = 1 .9857 = .0143

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Alternatively, we could find this by recognizing that

prob{Z > 2.19} = prob{Z 0} prob{0 Z 2.19}so we have prob{Z > 2.19} = .5 .4857 = .0143.

We can also find probabilities involving other normal random variables,besides the standard normal random variable, by converting values of X,a normal R.V. with mean and standard deviation , to values of Z, thestandard normal R.V., using the relationship below.

Theorem 12.9. Given a normal random variable X with mean and stan-dard deviation , we can standardize values of X using the formula

Z =X

How do we use this formula? Suppose we have some normal random vari-able X, with mean known to be some value and standard deviation knownto be some value , and we want to find prob{a X b}.Consider the event

{a

X

b

}. (Recall: is always positive)

a X b a X b (subtract from each part)

a

X

b

(divide through by )

But we know that X

= Z, so we see that the event {a X b} isequivalent to the event {a

Z b

}, and so we have:

Theorem 12.10. If X is a normal random variable with mean and stan-dard deviation , then for any values a and b, with a

b,

prob{a X b} = prob

a

Z b

Example 15. Find prob{2.3 X 6.1} where X is a normal random vari-able with = 5 and = 2.

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Solution:

prob{2.3 X 6.1} = prob

2.3

X

6.1

= prob

2.3 5

2 Z 6.1 5

2

= prob

2.7

2 Z 1.1

2

= prob{1.35 Z 0.55}= prob

{1.35

Z

0}

+ prob

{0

Z

0.55

}= prob{0 Z 1.35} + prob{0 Z 0.55}= .4115 + .2088 = .6203

Example 16. The actual weight of a 180 gram bag of a certain brand of potatochips is normally distributed, with mean = 188 grams and standard devia-tion = 5 grams. What is the probability that a bag contains less than 180grams of chips?

Solution: Let X be the weight of a bag of chips. Then X is a normal random

variable with = 188 and = 5. We need to find prob{X < 180}. Ofcourse, since prob{X = 180} = 0, then this is the same as prob{X 180}.(Recall: for any continuous random variable, the event {X < x} is identicalto the event {X x} because prob{X = x} is infinitesimally small, i.e. is 0.)

So we have

prob{X < 180} = prob

X

180

= prob

Z 180 188

5

= probZ 8

5 = prob{

Z

1.60

}= prob{Z 1.60} = prob{Z 0} prob{0 Z 1.60}= .5 .4452 = .0548

We see that almost 5.5% of the bags contain less than the stated weight.

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pre-calculus / math notes (unit 12 of 22)

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