probability the definition – probability of an event applies only to the special case when 1.the...
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Probability
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The definition – probability of an Event
no. of outcomes in =
total no. of outcomes
n E n E EP E
n S N
Applies only to the special case when
1. The sample space has a finite no.of outcomes, and
2. Each outcome is equi-probable
If this is not true a more general definition of probability is required.
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Summary of the Rules of Probability
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The additive rule
P[A B] = P[A] + P[B] – P[A B]
and
if P[A B] = P[A B] = P[A] + P[B]
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The Rule for complements
for any event E
1P E P E
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P A BP A B
P B
Conditional probability
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if 0
if 0
P A P B A P AP A B
P B P A B P B
The multiplicative rule of probability
and
P A B P A P B
if A and B are independent.
This is the definition of independent
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Counting techniques
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Summary of counting results
Rule 1
n(A1 A2 A3 …. ) = n(A1) + n(A2) + n(A3) + …
if the sets A1, A2, A3, … are pairwise mutually exclusive
(i.e. Ai Aj = )
Rule 2
n1 = the number of ways the first operation can be performed
n2 = the number of ways the second operation can be performed once the first operation has been completed.
N = n1 n2 = the number of ways that two operations can be performed in sequence if
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Rule 3
n1 = the number of ways the first operation can be performed
ni = the number of ways the ith operation can be performed once the first (i - 1) operations have been completed. i = 2, 3, … , k
N = n1n2 … nk
= the number of ways the k operations can be performed in sequence if
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Basic counting formulae
!
!n k
nP
n k
1. Orderings
! the number of ways you can order objectsn n
2. Permutations
The number of ways that you can choose k objects from n in a specific order
!
! !n k
n nC
k k n k
3. Combinations
The number of ways that you can choose k objects from n (order of selection irrelevant)
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Applications to some counting problems
• The trick is to use the basic counting formulae together with the Rules
• We will illustrate this with examples• Counting problems are not easy. The more practice
better the techniques
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Random Variables
Numerical Quantities whose values are determine by the outcome of a random
experiment
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Random variables are either• Discrete
– Integer valued – The set of possible values for X are integers
• Continuous– The set of possible values for X are all real
numbers – Range over a continuum.
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Examples• Discrete
– A die is rolled and X = number of spots showing on the upper face.
– Two dice are rolled and X = Total number of spots showing on the two upper faces.
– A coin is tossed n = 100 times and X = number of times the coin toss resulted in a head.
– We observe X, the number of hurricanes in the Carribean from April 1 to September 30 for a given year
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Examples• Continuous
– A person is selected at random from a population and X = weight of that individual.
– A patient who has received who has revieved a kidney transplant is measured for his serum creatinine level, X, 7 days after transplant.
– A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected) . X = the average weight of the 100 individuals.
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The Probability distribution of A random variable
A Mathematical description of the possible values of the random variable together with
the probabilities of those values
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The probability distribution of a discrete random variable is describe by its :
probability function p(x).
p(x) = the probability that X takes on the value x.
This can be given in either a tabular form or in the form of an equation.
It can also be displayed in a graph.
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Example 1• Discrete
– A die is rolled and X = number of spots showing on the upper face.
x 1 2 3 4 5 6
p(x) 1/6 1/6 1/6 1/6 1/6 1/6
formula
– p(x) = 1/6 if x = 1, 2, 3, 4, 5, 6
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Graphs
To plot a graph of p(x), draw bars of height p(x) above each value of x.
Rolling a die
0
1 2 3 4 5 6
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Example 2
– Two dice are rolled and X = Total number of spots showing on the two upper faces.
x 2 3 4 5 6 7 8 9 10 11 12
p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
12,3,4,5,6
36( )13
7,8,9,10,11,1226
xx
p xx
x
Formula:
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Rolling two dice
0
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36 possible outcome for rolling two dice
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Comments:Every probability function must satisfy:
1)(0 xp
1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive:
x
xp
1)(
2. The sum of the probabilities assigned to all the values of the random variable must equal 1:
b
ax
xpbXaP )(3.
)()1()( bpapap
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x 0 1 2 3
p(x) 6/14 4/14 3/14 1/14
ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:
P X( )the random variable equals 2 p ( ) 23
14
Note: This chart implies the only values x takes on are 0, 1, 2, and 3.
If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.
2least at is variablerandom the XP 32 pp 14
4
14
1
14
3
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A Bar Graph
No. of persons on base when a home run is hit
0.429
0.286
0.214
0.071
0.000
0.100
0.200
0.300
0.400
0.500
0 1 2 3
# on base
p(x)
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Discrete Random Variables
Discrete Random Variable: A random variable usually assuming an integer value.
• a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable
Note: Usually associated with counting
• The number of times a head occurs in 10 tosses of a coin
• The number of auto accidents occurring on a weekend
• The size of a family
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Continuous Random Variables
Continuous Random Variable: A quantitative random variable that can vary over a continuum
• A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line
Note: Usually associated with a measurement
• Blood Pressure
• Weight gain
• Height
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Probability Distributionsof Continuous Random Variables
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Probability Density Function
The probability distribution of a continuous random variable is describe by probability density curve f(x).
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Notes:
The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to
b is P[a < X < b].
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xa b
P a x b( )
Normal Probability Distributions(Bell shaped curve)
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Mean and Variance (standard deviation) of a
Discrete Probability Distribution
• Describe the center and spread of a probability distribution
• The mean (denoted by greek letter (mu)), measures the centre of the distribution.
• The variance (2) and the standard deviation () measure the spread of the distribution.
is the greek letter for s.
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Mean of a Discrete Random Variable
• The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together:
Notes: The mean is a weighted average of the values of X.
x
xxp
kk xpxxpxxpx 2211
The mean is the long-run average value of the random variable.
The mean is centre of gravity of the probability distribution of the random variable
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-
0.1
0.2
0.3
1 2 3 4 5 6 7 8 9 10 11
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2
Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:
Standard Deviation of a Discrete Random Variable: The positive square root of the variance:
x
xpx 22
2
2
xx
xxpxpx
22 x
xpx
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Example
The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.
x p (x ) xp(x) x 2 x 2 p(x)
0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643
Total 1.000 0.929 1.786
)(xp )(xxp )(2 xpx
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• Computing the mean:
Note:
• 0.929 is the long-run average value of the random variable
• 0.929 is the centre of gravity value of the probability distribution of the random variable
929.0x
xxp
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• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
923.0929.786.1 2
• Computing the standard deviation:
2
961.0923.0
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Random Variables
Numerical Quantities whose values are determine by the outcome of a random
experiment
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Random variables are either• Discrete
– Integer valued – The set of possible values for X are integers
• Continuous– The set of possible values for X are all real
numbers – Range over a continuum.
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The Probability distribution of A random variable
A Mathematical description of the possible values of the random variable together with
the probabilities of those values
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The probability distribution of a discrete random variable is describe by its :
probability function p(x).
p(x) = the probability that X takes on the value x.
This can be given in either a tabular form or in the form of an equation.
It can also be displayed in a graph.
![Page 44: Probability The definition – probability of an Event Applies only to the special case when 1.The sample space has a finite no.of outcomes, and 2.Each](https://reader036.vdocuments.site/reader036/viewer/2022081519/56649e8f5503460f94b92bed/html5/thumbnails/44.jpg)
x 0 1 2 3
p(x) 6/14 4/14 3/14 1/14
ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:
P X( )the random variable equals 2 p ( ) 23
14
Note: This chart implies the only values x takes on are 0, 1, 2, and 3.
If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.
2least at is variablerandom the XP 32 pp 14
4
14
1
14
3
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A Bar Graph
No. of persons on base when a home run is hit
0.429
0.286
0.214
0.071
0.000
0.100
0.200
0.300
0.400
0.500
0 1 2 3
# on base
p(x)
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Probability Distributionsof Continuous Random Variables
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Probability Density Function
The probability distribution of a continuous random variable is describe by probability density curve f(x).
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Notes:
The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to
b is P[a < X < b].
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Mean, Variance and standard deviation of Random Variables
Numerical descriptors of the distribution of a Random Variable
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Mean of a Discrete Random Variable
• The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together:
Notes: The mean is a weighted average of the values of X.
x
xxp
kk xpxxpxxpx 2211
The mean is the long-run average value of the random variable.
The mean is centre of gravity of the probability distribution of the random variable
![Page 51: Probability The definition – probability of an Event Applies only to the special case when 1.The sample space has a finite no.of outcomes, and 2.Each](https://reader036.vdocuments.site/reader036/viewer/2022081519/56649e8f5503460f94b92bed/html5/thumbnails/51.jpg)
-
0.1
0.2
0.3
1 2 3 4 5 6 7 8 9 10 11
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2
Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:
Standard Deviation of a Discrete Random Variable: The positive square root of the variance:
x
xpx 22
2
2
xx
xxpxpx
22 x
xpx
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Example
The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.
x p (x ) xp(x) x 2 x 2 p(x)
0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643
Total 1.000 0.929 1.786
)(xp )(xxp )(2 xpx
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• Computing the mean:
Note:
• 0.929 is the long-run average value of the random variable
• 0.929 is the centre of gravity value of the probability distribution of the random variable
929.0x
xxp
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• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
923.0929.786.1 2
• Computing the standard deviation:
2
961.0923.0
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The Binomial distribution
An important discrete distribution
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Situation - in which the binomial distribution arises
• We have a random experiment that has two outcomes– Success (S) and failure (F)– p = P[S], q = 1 - p = P[F],
• The random experiment is repeated n times independently
• X = the number of times S occurs in the n repititions• Then X has a binomial distribution
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Example
• A coin is tosses n = 20 times
– X = the number of heads
– Success (S) = {head}, failure (F) = {tail
– p = P[S] = 0.50, q = 1 - p = P[F]= 0.50
• An eye operation has %85 chance of success. It is performed n =100 times
– X = the number of Sucesses (S)
– p = P[S] = 0.85, q = 1 - p = P[F]= 0.15
• In a large population %30 support the death penalty. A sample n =50 indiviuals are selected at random
– X = the number who support the death penalty (S)
– p = P[S] = 0.30, q = 1 - p = P[F]= 0.70
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The Binomial distribution1. We have an experiment with two outcomes –
Success(S) and Failure(F).
2. Let p denote the probability of S (Success).
3. In this case q=1-p denotes the probability of Failure(F).
4. This experiment is repeated n times independently.
5. X denote the number of successes occuring in the n repititions.
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The possible values of X are
0, 1, 2, 3, 4, … , (n – 2), (n – 1), n
and p(x) for any of the above values of x is given by:
xnxxnx qpx
npp
x
nxp
1
X is said to have the Binomial distribution with parameters n and p.
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Summary:
X is said to have the Binomial distribution with parameters n and p.
1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment.
2. The probability of success is p.
3. The probability function
xnx ppx
nxp
1
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Example:
1. A coin is tossed n = 5 times. X is the number of heads occurring in the 5 tosses of the coin. In this case p = ½ and
3215
215
21
21
555
xxxxp xx
x 0 1 2 3 4 5
p(x)321
325
325
321
3210
3210
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Note:
5 5!
! 5 !x x x
5 5!
10 0! 5 0 !
5 5! 5!
51 1! 5 1 ! 4!
5 5 45!10
2 2!3! 2 1
5 5 45!10
3 3!2! 2 1
5 5!5
4 4!1!
5 5!1
5 0!5!
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0.0
0.1
0.2
0.3
0.4
1 2 3 4 5 6
number of heads
p(x
)
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Computing the summary parameters for the distribution – , 2,
x p (x ) xp(x) x 2 x 2 p(x)
0 0.03125 0.000 0 0.0001 0.15625 0.156 1 0.1562 0.31250 0.625 4 1.2503 0.31250 0.938 9 2.8134 0.15625 0.625 16 2.5005 0.03125 0.156 25 0.781
Total 1.000 2.500 7.500
)(xp )(xxp )(2 xpx
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• Computing the mean:
5.2x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
25.15.25.7 2
• Computing the standard deviation:
2
118.125.1
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Example:• A surgeon performs a difficult operation n =
10 times.
• X is the number of times that the operation is a success.
• The success rate for the operation is 80%. In this case p = 0.80 and
• X has a Binomial distribution with n = 10 and p = 0.80.
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xx
xxp
1020.080.0
10
x 0 1 2 3 4 5p (x ) 0.0000 0.0000 0.0001 0.0008 0.0055 0.0264
x 6 7 8 9 10p (x ) 0.0881 0.2013 0.3020 0.2684 0.1074
Computing p(x) for x = 0, 1, 2, 3, … , 10
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The Graph
-
0.1
0.2
0.3
0.4
0 1 2 3 4 5 6 7 8 9 10
Number of successes, x
p(x
)
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Computing the summary parameters for the distribution – , 2,
)(xxp )(2 xpx
x p (x ) xp(x) x 2 x 2 p(x)
0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0001 0.000 4 0.0003 0.0008 0.002 9 0.0074 0.0055 0.022 16 0.0885 0.0264 0.132 25 0.6616 0.0881 0.528 36 3.1717 0.2013 1.409 49 9.8658 0.3020 2.416 64 19.3279 0.2684 2.416 81 21.743
10 0.1074 1.074 100 10.737Total 1.000 8.000 65.600
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• Computing the mean:
0.8x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
60.10.86.65 2
• Computing the standard deviation:
2 118.125.1
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Notes The value of many binomial probabilities are found in Tables posted
on the Stats 245 site.
The value that is tabulated for n = 1, 2, 3, …,20; 25 and various values of p is:
Hence
c
x
c
x
xx xpppx
ncXP
00
101
cpppp 210
1for valueTabledfor valueTabled cccp The other table, tabulates p(x). Thus when using this
table you will have to sum up the values
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Example Suppose n = 8 and p = 0.70 and we want to
compute P[X = 5] = p(5)
Table value for n = 8, p = 0.70 and c =5 is 0.448 = P[X ≤ 5]
P[X = 5] = p(5) = P[X ≤ 5] - P[X ≤ 4] = 0.448 – 0.194 = .254
c 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95
n = 5 0 0.663 0.430 0.168 0.058 0.017 0.004 0.001 0.000 0.000 0.000 0.000
1 0.943 0.813 0.503 0.255 0.106 0.035 0.009 0.001 0.000 0.000 0.000
2 0.994 0.962 0.797 0.552 0.315 0.145 0.050 0.011 0.001 0.000 0.000
3 1.000 0.995 0.944 0.806 0.594 0.363 0.174 0.058 0.010 0.000 0.000
4 1.000 1.000 0.990 0.942 0.826 0.637 0.406 0.194 0.056 0.005 0.000
5 1.000 1.000 0.999 0.989 0.950 0.855 0.685 0.448 0.203 0.038 0.006
6 1.000 1.000 1.000 0.999 0.991 0.965 0.894 0.745 0.497 0.187 0.057
7 1.000 1.000 1.000 1.000 0.999 0.996 0.983 0.942 0.832 0.570 0.337
8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
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We can also compute Binomial probabilities using Excel
=BINOMDIST(x, n, p, FALSE)
The function
will compute p(x).
=BINOMDIST(c, n, p, TRUE)
The function
will compute
c
x
c
x
xx xpppx
ncXP
00
101
cpppp 210
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Mean, Variance and standard deviation of
Binomial Random Variables
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Mean of a Discrete Random Variable
• The mean, , of a discrete random variable x
Notes: The mean is a weighted average of the values of X.
x
xxp
kk xpxxpxxpx 2211
The mean is the long-run average value of the random variable.
The mean is centre of gravity of the probability distribution of the random variable
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2
Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x
Standard Deviation of a Discrete Random Variable: The positive square root of the variance:
x
xpx 22
2
2
xx
xxpxpx
22 x
xpx
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The Binomial ditribution
X is said to have the Binomial distribution with parameters n and p.
1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment.
2. The probability of success is p.
3. The probability function
xnx ppx
nxp
1
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Mean,Variance & Standard Deviation of the Binomial
Ditribution• The mean, variance and standard deviation of the
binomial distribution can be found by using the following three formulas:
np 1.
pnpnpq 1 2. 2
pnpnpq 1 3.
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Solutions:
1) n = 20, p = 0.75, q = 1 - 0.75 = 0.25
np ( )(0. )20 75 15
npq ( )(0. )(0. ) . .20 75 25 375 1936
Example:Find the mean and standard deviation of the binomial distribution when n = 20 and p = 0.75
p xx
xx x( ) (0. ) (0. )
2075 25 20 for 0, 1, 2, ... , 20
2)These values can also be calculated using the probability function:
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Table of probabilitiesx p (x ) xp(x) x 2 x 2 p(x)
0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0000 0.000 4 0.0003 0.0000 0.000 9 0.0004 0.0000 0.000 16 0.0005 0.0000 0.000 25 0.0006 0.0000 0.000 36 0.0017 0.0002 0.001 49 0.0088 0.0008 0.006 64 0.0489 0.0030 0.027 81 0.24410 0.0099 0.099 100 0.99211 0.0271 0.298 121 3.27412 0.0609 0.731 144 8.76813 0.1124 1.461 169 18.99714 0.1686 2.361 196 33.04715 0.2023 3.035 225 45.52516 0.1897 3.035 256 48.55917 0.1339 2.276 289 38.69618 0.0669 1.205 324 21.69119 0.0211 0.402 361 7.63220 0.0032 0.063 400 1.268
Total 1.000 15.000 228.750
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• Computing the mean:
0.15x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
75.30.1575.228 2
• Computing the standard deviation:
2 936.175.3
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Histogram
-0.1
0.1
0.2
0.3
no. of successes
p(x)
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Probability Distributionsof Continuous Random Variables
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Probability Density Function
The probability distribution of a continuous random variable is describe by probability density curve f(x).
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Notes:
The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to
b is P[a < X < b].
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xa b
P a x b( )
Normal Probability Distributions
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Normal Probability Distributions
• The normal probability distribution is the most important distribution in all of statistics
• Many continuous random variables have normal or approximately normal distributions
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2 3 2 3
The Normal Probability Distribution
Points of Inflection
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Main characteristics of the Normal Distribution
• Bell Shaped, symmetric
• Points of inflection on the bell shaped curve are at – and + That is one standard deviation from the mean
• Area under the bell shaped curve between – and + is approximately 2/3.
• Area under the bell shaped curve between – 2 and + 2is approximately 95%.
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There are many Normal distributions
depending on by and
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
Normal = 100, = 40 Normal = 140, =20
Normal = 100, =20
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The Standard Normal Distribution = 0, = 1
0
0.1
0.2
0.3
0.4
-3 -2 -1 0 1 2 3
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• There are infinitely many normal probability distributions (differing in and )
• Area under the Normal distribution with mean and standard deviation can be converted to area under the standard normal distribution
• If X has a Normal distribution with mean and standard deviation than
has a standard normal distribution.
• z is called the standard score (z-score) of X.
X
z
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Converting Area
under the Normal distribution with mean and standard deviation
to
Area under the standard normal distribution
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Perform the z-transformation
then
Area under the Normal distribution with mean and standard deviation
X
z
P a X b
a X bP
a bP z
Area under the standard normal distribution
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P a X b
Area under the Normal distribution with mean and standard deviation
a b
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a bP z
Area under the standard normal distribution
a b
0
1
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Using the tables for the Standard Normal distribution
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Table, Posted on stats 245 web site
• The table contains the area under the standard normal curve between -∞ and a specific value of z
0 z0 z
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Example
Find the area under the standard normal curve between z = -∞ and z = 1.45
9265.0)45.1( zP
• A portion of Table 3:
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06
1.4 0.9265
...
...
0 145.
9265.0
z0 145.
9265.0
z
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P z( 0. ) 98 .01635
Example
Find the area to the left of -0.98; P(z < -0.98)
0000.98
Area asked for
0.980.98
Area asked for
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0735.09265.00000.1)45.1( zP
Example
Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)
9265.0
145.
Area asked for
0 z
9265.0
145.
Area asked for
145.
Area asked forArea asked for
0 z0 z
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(0 1.45) 0.9265 0.5000 0.4265P z
ExampleFind the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45)
• Area between two points = differences in two tabled areas
145.0 z145.145.0 z0 z
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Notes Use the fact that the area above zero and the area
below zero is 0.5000
the area above zero is 0.5000
When finding normal distribution probabilities, a sketch is always helpful
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3962.01038.05000.0)026.1( zP
Example:Find the area between the mean (z = 0) and z = -1.26
0 z 1 2 6.
A r e a a s k e d f o r
0 z0 z 1 2 6.
A r e a a s k e d f o r
1 2 6.
A r e a a s k e d f o rA r e a a s k e d f o r
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Example: Find the area between z = -2.30 and z = 1.80
( 2.30 1.80) 0.9641 0.0107 0.9534P z
0 .. - 2 . 3 0
Area Required
1 . 8 00 .. 0 .. 0 .. - 2 . 3 0
Area Required
1 . 8 0- 2 . 3 0
Area Required Area Required
1 . 8 0
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Example: Find the area between z = -1.40 and z = -0.50
2277.00808.03085.0)50.040.1( zP
Area asked for
-1.40 -0.500
Area asked for
-1.40 -0.50
Area asked for
-1.40 -0.5000
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Computing Areas under the general Normal Distributions
(mean , standard deviation )
1. Convert the random variable, X, to its z-score.
Approach:
3. Convert area under the distribution of X to area under the standard normal distribution.
2. Convert the limits on random variable, X, to their z-scores.
X
z
b
za
PbXaP
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Example 1: Suppose a man aged 40-45 is selected at random from a population.
• X is the Blood Pressure of the man.
• Assume that X has a Normal distribution with mean =180 and a standard deviation = 15.
• X is random variable.
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The probability density of X is plotted in the graph below.
• Suppose that we are interested in the probability that X between 170 and 210.
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Let
15
180
XXz
667.015
180170170
a
000.215
180210210
b
000.2667.210170 zPXP
Hence
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000.2667.210170 zPXP
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000.2667.210170 zPXP
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Example 2
A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally distributed and a bottle is selected at random:
1) Find the probability the bottle contains between 32.00 oz and 32.025 oz
2) Find the probability the bottle contains more than 31.97 oz
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When x = 32.00
Solution part 1)
32.00 32.00 320.00
0.02z
When x = 32.025
32.025 32.025 321.25
0.02z
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P X PX
P z
( . )0. 0.
.
0.
( . ) .
32.0 32 02532.0 32.0
02
32.0
02
32 025 32.0
02
0 125 0 3944
Graphical Illustration:
0 1 2 5. z3 2 . 0 x3 2 0 2 5.
A r e a a s k e d f o r
0 1 2 5. z0 1 2 5. z3 2 . 0 x3 2 0 2 5.
A r e a a s k e d f o r
3 2 . 0 x3 2 . 0 x3 2 0 2 5.
A r e a a s k e d f o r
3 2 0 2 5.
A r e a a s k e d f o r
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P x Px
P z( . ).
( .
. . .
319732.0
0.023197 32.0
0.02150)
1 0000 00668 0 9332
Example 2, Part 2)
32.03197. x
0150. z32.03197. x32.03197. x
0150. z0150.150. z
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SummaryRandom Variables
Numerical Quantities whose values are determine by the outcome of a random
experiment
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Types of Random Variables
• Discrete
Possible values integers• Continuous
Possible values vary over a continuum
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The Probability distribution of a random variable
A Mathematical description of the possible values of the random variable together with
the probabilities of those values
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The probability distribution of a discrete random variable is describe by its :
probability function p(x).
p(x) = the probability that X takes on the value x.
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-
0.1
0.2
0.3
0.4
0 1 2 3 4 5 6 7 8 9 10
Number of successes, x
p(x
)
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The Binomial distribution
X is said to have the Binomial distribution with parameters n and p.
1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment.
2. The probability of success is p.
3. The probability function
xnx ppx
nxp
1
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Probability Distributionsof Continuous Random Variables
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Probability Density Function
The probability distribution of a continuous random variable is describe by probability density curve f(x).
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Notes:
The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to
b is P[a < X < b].
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2 3 2 3
The Normal Probability Distribution
Points of Inflection
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Normal approximation to the Binomial distribution
Using the Normal distribution to calculate Binomial probabilities
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-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
-0.5
Binomial distribution
Approximating
Normal distribution
Binomial distribution n = 20, p = 0.70
049.2
14
npq
np
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Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
2121 aYaPaXP
• Y has a Normal distribution
npq
np
correction continuity21
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-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
Binomial distribution
-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
Approximating
Normal distribution
P[X = a]
21a 2
1a
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-
0.0500
0.1000
0.1500
0.2000
0.2500
a-
-0.5
2121 aYaP
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-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
P[X = a]
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Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
13 want We XP
13 eexact valu The XP
1643.030.070.013
20 713
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Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
2121 131213 YPXP
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Hence
5.135.12 YP
049.2
145.13
049.2
14
049.2
145.12 YP
= 0.4052 - 0.2327 = 0.1725
24.073.0 ZP
Compare with 0.1643
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Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
2121 bYaP
• Y has a Normal distribution
npq
np
correction continuity21
)()1()( bpapapbXaP
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-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
bXaP
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-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
2121 bYaP
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Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
1411 want We XP 1411 eexact valu The XP
614911 30.070.014
2030.070.0
11
20
)14()13()12()11( pppp
5357.01916.01643.01144.00654.0
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Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
2121 14101411 YPXP
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Hence
5.145.10 YP
049.2
145.14
049.2
14
049.2
145.10 YP
= 0.5948 - 0.0436 = 0.5512
24.071.1 ZP
Compare with 0.5357
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Comment:
• The accuracy of the normal appoximation to the binomial increases with increasing values of n
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Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
2121 bYaP
• Y has a Normal distribution
npq
np
correction continuity21
)()1()( bpapapbXaP
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Example• The success rate for an Eye operation is 85%• The operation is performed n = 2000 times
Find the probability that1. The number of successful operations is
between 1650 and 1750.2. The number of successful operations is at
most 1800.
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Solution
• X has a Binomial distribution with parameters n = 2000 and p = 0.85
17201680 want We XP
5.17205.1679 YP
where Y has a Normal distribution with:
969.1515.85.200
1700)85.0(2000
npq
np
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17201680 Hence XP
969.15
17005.1720
969.15
1700
969.15
17005.1679 YP
= 0.9004 - 0.0436 = 0.8008
28.128.1 ZP
5.17205.1679 YP
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Solution – part 2.
1800 want We XP
5.1800 YP
969.15
17005.1800
969.15
1700YP
= 1.000
29.6 ZP
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Next topic: Sampling Theory