Chapter 4. Discrete Probability Distributions

Chapter 4: Discrete Probability Distributions 89
4-2 Random Variables
Chapter 4. Discrete Probability Distributions
Identifying Discrete and Continuous Random Variables. In Exercises 1 and 2, identify the given random variable
as being discrete or continuous.
1. a. Height of giraffe is a continuous random variable, since the observation can be a fractional value.
b. Number of bald eagles is a discrete random variable, since the observation can take on only whole
numbers.
c. Exact gestation time is a continuous random variable, since the observation can be a fractional value.
d. Number of blue whales is a discrete random variable, since the observation can take on only whole
numbers.
e. Number of manatees killed is a discrete random variable, since the observation can take on only whole
numbers.
2. a. Cost of conducting a genetics experiment is a discrete random variable, since the observation can be a
fractional value, but not less than 0.01 dollars, so there is a discrete unit of measurement..
b. Number of enkaryotic cells in an ant is a discrete random variable, since the observation can take on only
whole numbers.
c. Exact life span of a koala bear is a continuous random variable, since the observation can be a fractional
value.
d. Number of monkeys in Gibraltar is a discrete random variable, since the observation can take on only
whole numbers.
e. Weight of an elephant is a continuous random variable, since the observation can be a fractional value.
Identifying Probability Distributions. In Exercises 3 – 8, determine whether a probability distribution is given. In
those cases where a probability distribution is not described, identify the requirements that are not satisfied. In
those cases where a probability is described, find its mean and standard deviation.
There are two requirements for a probability distribution:
1. Each P(x) must be equal to or greater than 0 and equal to or less than 1.
2. The sum of probabilities, ΣP(x), must be equal to 1.
3. Gender Selection This is a probability distribution since for each x, 0 ≤ P(x) ≤ 1 and
ΣP(x) = 0.125 + 0.375 + 0.375 + 0.125= 1.000
Mean (µ)= Σ[ x ∗ P(x)] = (0 ∗ 0.125) + (1 ∗ 0.375) + (2 ∗ 0.375) + (3 ∗ 0.125)= 1.50
Standard Deviation
3.00 −1.5
2
[ x ∗ P(
x)
]
2
σ = ∑ − µ
2
=
=
(0
3.00 − 2.25 =
2
∗ 0.125) + (1
0.75 = 0.866
2
∗ 0.375) + (2
2
∗ 0.375) + (3
2
∗ 0.125) −1.5
2
=
4. Numbers of Girls This is not a probability distribution since ΣP(x) = 0.977 ≠ 1.000
5. Genetics Experiment This is not a probability distribution since ΣP(x) = 0.94 ≠ 1.00
6. Mortality Study This is a probability distribution since for each x, 0 ≤ P(x) ≤ 1 and
ΣP(x) = 0.0000 + 0.0001 + 0.0006 + 0.0387 + 0.9606 = 1.0000
Mean, (µ)= Σ[ x ∗ P(x)] = (0 ∗ 0.0000) + (1 ∗ 0.0001) + (2 ∗ 0.0006) + (3 ∗ 0.0387) +
(4 ∗ 0.9606) = 3.96

90 Chapter 4: Discrete Probability Distributions
Standard Deviation
(0
2
2
[ x ∗ P(
x)
]
2
σ = ∑ − µ
∗0.0000)
+ (1
2
=
∗0.0001)
+ (2
2
∗0.0006)
+ (3
2
∗0.0387)
+ (4
2
∗0.9606)
−3.96
2
=
15.72−3.96
2
=
15.72−15.68
=
0.04 = 0.201
7. Genetic Disorder This is a probability distribution since for each x, 0 ≤ P(x) ≤ 1 and
ΣP(x) = 0.4219 + 0.4219 + 0.1406 + 0.0156 = 1.0000
Mean, µ= Σ[ x ∗ P(x)] = (0 ∗ 0.4219) + (1 ∗ 0.4219) + (2 ∗ 0.1406) + (3 ∗ 0.0156) = 0.75
Standard Deviation
2
[ x ∗ P(
x)
]
2
σ = ∑ − µ
1.1247−
0.75
2
=
=
(0
1.1247−
0.5625 =
2
2
∗0.4219)
+ (1 ∗0.4219)
+ (2
0.5622 = 0.75
2
∗0.1406)
+ (3
2
∗0.0156)
− 0.75
2
=
8. Diseased Seedlings This is not a probability distribution since ΣP(x) = 0.986 ≠ 1.00
9. Gender Selection Technique Effectiveness From Table 4-1
a. P(9)= 0.122
b. P(9 or more)= 0.122 + 0.061 + 0.022 + 0.006 + 0.001 + 0.000= 0.212
c. The probability from part (b) since any outcome 9 or above achieves the same criterion of
being unusually high.
d. No, P(9 or more) is not unusual, P(9 or more) > 0.05. This would happen by chance about one time out of
every five samples. We would conclude there is not sufficient evidence to conclude that the technique is
effective.
10. Gender Selection Technique Effectiveness From Table 4-1
a. P(12)= 0.006
b. P(12 or more)= 0.006 + 0.001 + 0.000= 0.007
c. The probability from part (b) since any outcome 12 or above achieves the same criterion
of being unusually high.
d. Yes, P(12 or more) is unusual, P(12 or more) < 0.05. This would happen by chance only
about seven times out of every 1000 samples. We would conclude there is sufficient
evidence to conclude that the technique is effective.
11. Gender Selection Technique Effectiveness From Table 4-1
a. Include probabilities of 11 or more, P(11 or more)= 0.022 + 0.006 + 0.001= 0.029
b. Yes, P(11 or more) is unusual, P(11 or more) < 0.05. This would happen by chance only
about three times out of every 100 samples. We would conclude there is sufficient
evidence to conclude that the technique is effective.
12. Gender Selection Technique Effectiveness From Table 4-1
a. Include probabilities of 10 or more, P(10 or more)= 0.061 + 0.022 + 0.006 + 0.001= 0.090
b. No, P(10 or more) is not unusual, P(10 or more) < 0.05. This would happen by chance
about nine times out of every 100 samples. We would conclude there is not sufficient
evidence to conclude that the technique is effective.

Chapter 4: Discrete Probability Distributions 91
13. Finding Mean and Standard Deviation Possible outcomes of gender of four children:
Outcome # girls (x) Outcome # girls (x)
BBBB
BBBG
BBGB
BGBB
GBBB
BBGG
BGGB
GGBB
GBGB
BGBG
GBBG
0
1
1
1
1
2
2
2
2
2
2
GGGB
GGBG
GBGG
BGGG
GGGG
Number of 0
Number of 1
Number of 2
Number of 3
Number of 4
3
3
3
3
4
1
4
6
4
1
There are 16 possible different outcomes (permutations).
Number of
Girls (x) out P(x)= # girls/16 x ∗ P(x) x 2 x 2 * P(x)
of four
0 1/16= 0.0625 0.0000 0 0.0000
1 4/16= 0.2500 0.2500 1 0.2500
2 6/16= 0.3750 0.7500 4 1.5000
3 4/16= 0.2500 0.7500 9 2.2500
4 1/16= 0.0625 0.2500 16 1.0000
Total 1.0000 ∑[x ∗P(x)] = 2.0 ∑[x 2 ∗P(x)]= 5.0
Mean, µ = Σ
[( x) * P( x)
]
Standard deviation, σ =
= 2.00
4-3 Binomial Probability Distributions
Σ
2
2
2
[ x * P( x)
] − µ = 5.0 − 2.0 = 5.0 − 4.0 = 1.00 = 1. 00
Identifying Binomial Distributions. In Exercises 1 – 8, determine whether the given procedure results in a
binomial distribution. For those that are not binomial, identify at least one requirement that is not satisfied.
There are four requirements for a binomial distribution:
1. There are a fixed number of trials or observations
2. The trials are independent
3. Each trial has two possible outcomes
4. The probabilities remain constant for each trial
1. This is not a binomial distribution. The number of trials (people) is not fixed. Possible outcomes are not
classified into two categories; they could be any number. There could be any number of different answers to the
question.
2. This is a binomial distribution, all requirements are met, with only two possible outcomes (answers), “yes or
no”.
3. This is not a binomial distribution. The outcomes (answers) are not classified into two categories.

92 Chapter 4: Discrete Probability Distributions
4. This is a binomial distribution, all requirements are met. The question has a “yes or no” answer.
5. This is a binomial distribution, all requirements are met. The outcomes are “male or female”.
6. This is a binomial distribution, all requirements are met. The outcomes are “acceptable or defective”.
7. This is not a binomial distribution. The outcomes (answers) are not classified into two categories. The couples
could have any number of children.
8. This is a binomial distribution, all requirements are met. The outcomes are “yes or no”.
9. Finding Probabilities when Guessing Answers P(wrong) = 4/5 = 0.8, P(correct) =1/5 = 0.2
a. P(WWC) = 0.8 ∗ 0.8 ∗ 0.2 = 0.128
b. Three possible arrangements: WWC, WCW, CWW
P(WWC) = 0.8 ∗ 0.8 ∗ 0.2 = 0.128
P(WCW) = 0.8 ∗ 0.2 ∗ 0.8 = 0.128
P(CWW) = 0.2 ∗ 0.8 ∗ 0.8 = 0.128
c. P( exactly one correct answer in 3 guesses) = P(WWC) + P(WCW) + P(CWW)= 0.128 + 0.128 + 0.128=
0.384
10. Finding Probabilities when Guessing Answers P(W) = 0.75, P(C) = 0.25
a. P(WWCCCC) = 0.75 ∗ 0.75 ∗ 0.25 ∗ 0.25 ∗ 0.25 ∗ 0.25 = 0.0022
6!
(6 − 2)!2!
6 ∗5
∗ 4!
4! 2!
30
2
b. Combinations= P = = = 15
6 2
=
List of two wrong and four correct answers:
WWCCCC, WCWCCC, WCCWCC, WCCCWC, WCCCCW, CWWCCC, CWCWCC,
CWCCWC, CWCCCW, CCWWCC, CCWCWC, CCWCCW, CCCWWC, CCCWCW,
CCCCWW
. P(each outcome)= 0.75 ∗ 0.75 ∗ 0.25 ∗ 0.25 ∗ 0.25 ∗ 0.25 = 0.0022
c. P(four correct out of six guesses)= 15 ∗ 0.0022= 0.0330
Using Table A-1. In Exercises 11 – 16, assume that a procedure yields a binomial distribution with a trial
repeated n times. Use Table A-1 to find the probabilities of x successes given the probability p of success on a
given trial.
11. n= 2, x= 0, p=0.01 0.01 Column, 2-0 Row P(0)= 0.980
12. n= 7, x= 2, p=0.10 0.10 Column, 7-2 Row P(2)= 0.124
13. n= 4, x= 3, p= 0.95 0.95 Column, 4-3 Row P(3)= 0.171
14. n= 6, x= 5, p= 0.99 0.99 Column, 6-5 Row P(5)= 0.006
15. n=10, x= 4, p= 0.95 0.95 Column, 10-4 Row P(4)= 0.000 +
16. n=11, x= 7, p= 0.05 0.05 Column, 11-7 Row P(7)= 0.000 +
Note: for answers 15 and 16, the numbers are greater than zero, but when rounded to three decimal places, they are
0.000.

Chapter 4: Discrete Probability Distributions 93
Using the Binomial Probability Formula. In Exercises 17-20, assume that a procedure yields a binomial
distribution with a trial repeated n times. Use the binomial probability formula to find the probabilities of x
successes given the probability p of success on a given trial.
17. n=6, x= 4, p= 0.55
P(
x)
=
P(4)
=
( n − x)
6!
∗ 0.55
!4!
( 6 − 4)
n!
∗ p
! x!
∗ q
∗ 0.45
15 ∗ 0.0915∗
0.2025 = 0.280
18. n= 6, x= 2, p= 0.45
P(
x)
=
P(2)
=
( n − x)
( 6 − 2)
n!
∗ p
! x!
x
6!
∗ 0.45
!2!
4
∗ q
n−x
∗ 0.55
15 ∗ 0.2025∗
0.0915 = 0.280
x
2
n−x
6−4
6−2
6 ∗5
∗ 4!
= ∗ 0.55
2!4!
6 ∗5
∗ 4!
= ∗ 0.45
4!2!
4
2
∗ 0.45
2
∗ 0.55
4
=
=
19. n= 8, x= 3, p= 0.25
P(
x)
=
P(3)
=
( n − x)
8!
3
∗ 0.25
!3!
( 8 − 3)
n!
∗ p
! x!
∗ q
n−x
∗ 0.75
56 ∗ 0.01563∗
0.2373 = 0.208
x
8−3
8 ∗ 7 ∗ 6 ∗ 5!
= ∗ 0.25
5!3!
3
∗ 0.75
5
=
20. n= 10, x= 8, p= 0.3333
P(
x)
=
( n − x)
n!
∗ p
! x!
∗ q
n−x
10!
P(8)
= ∗ 0.3333
(10 − 8)!8!
∗ 0.6667
45 ∗ 0.0001523∗
0.4445 = 0.00305
x
8
10−8
10 ∗ 9 ∗8!
= ∗ 0.3333
2!8!
8
∗ 0.6667
2
=
Using Computer Results. In Exercises 21-24, refer to the Minitab display in the margin. The probabilities were
obtained by entering the values of n= 6 and p= 0.167. In a clinical test of the drug Lipator (atorvastatin), 16.7%
of the subjects treated with 10 mg of atorvastatin experienced headaches (based on data from Parke-Davis). In
each case, assume that 6 subjects are randomly selected and treated with 10 mg of atorvastatin, then find the
indicated probability.
21. P(at least 5 have headache)= P(5 or 6)= P(5) + P(6)= 0.0006 + 0.0000= 0.0006
It would be very unusual for 5 out of 6 subjects to get a headache. This would happen only 6 times out of 1000
samples by chance.
22. P(2 or less have headache)= P(0 or 1 or 2)= P(0) + P(1) + P(2)= 0.3341 + 0.4019 + 0.2014= 0.9374. It would
not be unusual at all for up to 2 out of 6 subjects to get a headache. This would happen about 94 times out of
100 samples by chance.

94 Chapter 4: Discrete Probability Distributions
23. P(more than 1 have headache)= P(2 or 3 or 4 or 5 or 6)= P(2) + P(3) + P(4) + P(5) + P(6)=
0.2014 + 0.0538 + 0.0081 + 0.0006 + 0.0000= 0.2639
P(not having more than one headache out of 6) = 1 – P(2, 3, 4, 5, or 6)= 1.0000 – 0.2639= 0.7361). It would
not be unusual to not have more than 1 headache out of 6 patients.
24. P(at least 1 has headache)= P(1) + P(2) + P(3) + P(4) + P(5) + P(6)=
0.4019 + 0.2014 + 0.0538 + 0.0081 + 0.0006 + 0.0000= 0.2639= 0.6658
P(not having at least one headache out of 6) = 1 – P(1, 2, 3, 4, 5, or 6)= 1.0000 – 0.6658= 0.3342). It would not
be unusual to not have at least 1 headache out of 6 patients.
25. Drug Reaction n= 8, p= 0.04
a. x= 3, P(3 out of 8)
P(
x)
=
( n − x)
n!
∗ p
! x!
8!
P(8)
= ∗ 0.04
(8 − 3)!3!
∗ q
∗ 0.96
8−3
56 ∗ 0.000064 ∗ 0.8154 = 0.00292
x
3
n−x
8 ∗ 7 ∗ 6 ∗5!
=
∗ 0.04
5!3!
3
∗ 0.96
5
=
b. x= 8, P(8 out of 8)
P(
x)
=
( n − x)
n!
∗ p
! x!
8!
P(8)
= ∗ 0.04
(8 − 8)!8!
x
∗ q
8
n−x
∗ 0.96
∗ 0.96
1∗
0.00000000000655∗1
= 0.00000000000655 = 0.000
0
=
8!
∗ 0.04
0!8!
c. If all 8 experienced headaches, this would be evidence that this placebo group was certainly different than
the 4% group. It is highly unlikely the all 8 in the placebo group would have a headache.
8
0
=
÷
26. Color Blindness n= 6, x= 2, p= 0.09, P(2 out of 6)
P(
x)
=
( n − x)
n!
∗ p
! x!
∗ q
6!
P(2)
= ∗ 0.09
(6 − 2)!2!
∗ 0.91
15 ∗ 0.0081∗
0.6857 = 0.0833
x
2
n−x
6−2
6 ∗ 5∗
4!
= ∗ 0.09
4!2!
2
∗ 0.91
4
=
27. Acceptance Sampling n= 24, p= 0.04, x= 1 or 0, P(0 or 1)= P(0) + P(1)
P(
x)
=
( n − x)
n!
∗ p
! x!
∗ q
24!
P(0)
= ∗ 0.04
(24 − 0)!0!
n−x
∗ 0.96
24!
1
P(1)
= ∗ 0.04 ∗ 0.96
(24 −1)!1!
x
0
24−0
24−1
24!
∗ 0.04
24!0!
24 ∗ 23! 1
= ∗ 0.04 ∗ 0.96
23!1!
P(0 or 1) = P(0)
+ P(1)
= 0.3754 + 0.3754 = 0.7508
The probability of this shipment being accepted is 0.751.
=
0
∗ 0.96
24
23
= 1∗1∗
0.3754 = 0.3754
= 24 ∗ 0.04 ∗ 0.3911 = 0.3754

Chapter 4: Discrete Probability Distributions 95
28. Affirmative Action Programs n=10, x= 9 or 10, p= 0.94,
a. P(9 or 10 out of 10)
P(
x)
=
( n − x)
n!
∗ p
! x!
∗ q
10!
P(9)
= ∗ 0.94
(10 − 9)!9!
10!
P(10)
=
∗ 0.94
(10 −10)!10!
x
9
n−x
∗ 0.06
∗ 0.06
10 ∗ 9!
= ∗ 0.94
1!9!
10−10
10!
∗ 0.94
0!10!
P(9 or 10) = P(9)
+ P(10)
= 0.3438 + 0.5386 = 0.8824
10
10−9
Probability that 9 or more graduated is 0.8824.
=
9
∗ 0.06
10
1
∗ 0.06
= 10 ∗ 0.5730 ∗ 0.06 = 0.3438
0
= 1∗
0.5386 ∗1
= 0.5386
b. P(x ≤ 7)= 1 – P(x > 7)= 1 – [P(8) + P(9) + P(10)]
P(
x)
=
( n − x)
n!
∗ p
! x!
∗ q
10!
P(8)
= ∗ 0.94
(10 − 8)!8!
∗ 0.06
45 ∗ 0.6096 ∗ 0.0036 = 0.0988
x
8
n−x
10−8
10 ∗ 9 ∗8!
= ∗ 0.94
2!8!
8
∗ 0.06
2
=
P(x ≤ 7)= 1 – P(x > 7)= 1 – [0.0988 + 0.3438 + .5386]= 1 – 0.9812= 0.0188
Since P(x ≤ 7) < 0.05, this would be unusual to have only 7 graduate.
29. Identifying Gender Discrimination n= 20, p= 0.5, P(2 or less)
P(
x)
=
0.0001812
( n − x)
n!
∗ p
! x!
∗ q
20!
P(0)
= ∗ 0.5
(20 − 0)!0!
20!
1
P(1)
= ∗ 0.5 ∗ 0.5
(20 −1)!1!
20!
P(2)
= ∗ 0.5
(20 − 2)!2!
x
0
2
n−x
∗ 0.5
20−1
∗ 0.5
20−0
20−2
=
20!
∗ 0.5
20!0!
∗ 0.5
20 ∗19!
1
= ∗ 0.5 ∗ 0.5
19!1!
20 ∗19
∗18!
=
∗ 0.5
18!2!
= 1∗1∗
0.000000953 = 0.000000953
= 20 ∗ 0.5 ∗ 0.00000191 = 0.0000191
∗ 0.5
= 190 ∗ 0.25∗
0.00000381 =
P(2 or less) = P(0)
+ P(1)
+ P(2)
= 0.000000953 + 0.0000191+
0.00018120 = 0.000201
Yes, this result would tend to support evidence that discrimination occurred since by chance this would have
happened about 2 times out of 1000 samples.
0
20
19
2
18

96 Chapter 4: Discrete Probability Distributions
30. Testing Effectiveness of Gender Selection Technique n=12, p= 0.5
Values for the Table can be found in Table A-1 for n= 12 and p= 0.5
x (girls) P(x)
0 0.000
1 0.003
2 0.016
3 0.054
4 0.121
5 0.193
6 0.226
7 0.193
8 0.121
9 0.054
10 0.016
11 0.003
12 0.000
The probability of getting 9 girls and 3 boys is 0.054. If we use the criterion that the probability need to be <
0.05 to be considered “unusual” or different than what would have been resulting from chance, we would have
to conclude that the technique was not effective.
31. Geometric Distribution x= 7, p= 0.2
P(x) =
p
x−1
7−1
6
( 1−
p) = 0.
2( 1-0.
2) = 0.2 ∗ 0.8 = 0.
0524
32. Hypergeometric Distribution A= 6, B= 54 – 6) = 48, n=6
A!
B!
( A + B)!
P(
x)
= ∗
÷
( A − x)!
x!
( B − n + x)!(
n − x)!
( A + B − n)!
n!
Where x is the number of events of type A and n – x is the number of events of type B
a. P(all six winning numbers)
P(6)
=
6!
1∗1
=
25,827,165
b. P(5 out of 6 winners)
P( 5 ) =
6 ∗ 48
=
25,827,165
*
( 6 − 6) !6! ( 48 − 6 + 6) ! ( 6 − 6)
6!
c. P(3 out of 6 winners)
1
25,827,165
*
48!
÷
!
= 0.00000003872
( 6 − 5) !5! ( 48 − 6 + 5) ! ( 6 − )
288
25,827,165
48!
÷
5 !
= 0.00001115
( 6 + 48 )!
( 6 + 48 − 6)
( 6 + 48 )!
( 6 + 48 − 6)
=
!6!
=
!6!
6!
0!6!
6!
1!5!
∗
∗
48!
48!0!
48!
47!1!
÷
÷
54!
48!6!
54!
48!6!
=
=

Chapter 4: Discrete Probability Distributions 97
P(3)
=
6!
20 ∗17296
=
25,827,165
*
( 6 − 3) !3! ( 48 − 6 + 3) ! ( 6 − 3)
d. P(no winning numbers)
P(0)
=
6!
1*12,271,512
25,827,165
345,920
25,827,165
*
48!
= 0.0134
( 6 − 0) !0! ( 48 − 6 + 0) ! ( 6 − 0)
=
12,271,512
25,827,165
48!
= 0.475
÷
!
÷
!
( 6 + 48 )!
( 6 + 48 − 6)
( 6 + 48 )!
( 6 + 48 − 6)
=
!6!
=
!6!
6!
3!3!
6!
6!0!
+
∗
48!
45!3!
48!
42!6!
33. Multinomial Distribution, n= 20, Six categories (genetic genotypes): A, B, C, D, E, and F
Find P(5A’s and 4B’s and 3C’s and 2D’s and 3E’s)
P(A)= p 1 , P(B)= p 2 , P(C)= p 3 , P(D)= p 4 , P(E)= p 5 , P(F)= p 6
÷
÷
54!
48!6!
54!
48!6!
=
=
p
1
( x !)( x
1
=
p
2
2
=
!)( x
20!
n!
!)( x
!)( x
( 5! )( 4! )( 3! )( 2! )( 3! )( 3! )
1.95546 E12
∗ 0.000129
p
3
3
=
p
4
4
!)( x
∗ 0.
16667
∗ 0.
16667
1.95546 E12
∗ 2.73818 E − 16 = 0.000535
=
p
5
5
=
6
p
6
!)
5
=
p
x1
1
1
6
p
x2
2
= 0.16667
p
x3
3
p
4
x4
4
p
x5
5
p
x6
6
∗ 0.
16667
=
3
∗ 0.
16667
2
∗ 0.
16667
3
∗ 0.
16667
∗ 0.000772 ∗ 0.004630 ∗ 0.027779 ∗ 0.004630 ∗ 0.004630 =
3
=
4-4 Mean, Variance, and Standard Deviation for the Binomial Distribution
Finding µ, σ, and Unusual Values. In Exercises 1 – 4, assume that a procedure yields a binomial distribution
with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean µ
and standard deviation σ. Also, use the range rule of thumb to find the minimal usual value µ − 2σ and the
maximum usual value µ + 2σ.
1. n= 400, p= 0.2, q= 1 – p= 1 – 0.2= 0.8
µ = np = 400∗0.2
= 80.0
σ =
npq=
400∗0.2
∗0.8
=
64.0 = 8.0
maximumusualvalue=
µ + 2σ
= 80.0 + 2∗8.0
= 96.0
minimumusualvalue=
µ −2σ
= 80.0 −2∗8.0
= 64.0
2. n= 250, p= 0.45, q= 1 – p= 1 – 0.45= 0.55
µ = np = 250∗0.45=
112.5
σ =
npq =
250∗0.45∗0.55
= 7.87
maximumusualvalue=
µ + 2σ
= 112.5+
2∗7.87=
128.2
minimunmusualvalue=
µ - 2σ
= 112.5- 2∗7.87=
96.8

98 Chapter 4: Discrete Probability Distributions
3. n= 1984, p= 0.75, q= 1 – p= 1 – 0.75= 0.25
µ = np = 1984∗0.75
= 1488
σ =
npq =
1984∗0.75∗0.25
= 19.29
maximumusualvalue=
µ + 2σ
= 1488+
2∗19.29
= 1526.6
minimumusualvalue=
µ - 2σ
= 1488−
2∗19.29
= 1449.4
4. n= 767, p= 0.1667, q= 1 – p= 1 – 0.1667= 0.8333
µ = np = 767∗0.17
= 130.39
σ =
npq =
767∗0.17∗0.83
= 10.40
maximumusualvalue=
µ + 2σ
= 130.39+
2∗10.40
= 151.1
minimumusualvalue=
µ - 2σ
= 130.39−
2∗10.40
= 109.6
5. Guessing Answers, guessing 7 out of 10 correct when n= 10 and p= 0.5
a. n=10, p= ½= 0.5, q= 1 – p = 1 – 0.5= 0.5
µ = np = 10 ∗ 0.5 = 5
σ =
npq =
10∗
0.5 ∗ 0.5 =
maximum usual value
2.50 = 1.58
= µ + 2σ
= 5 + 2 ∗1.58
= 8.16
minimum usual value = µ - 2σ
= 5 − 2 ∗1.58
= 1.84
b. No, the maximum usual value is 8.16, so a value of 7 correct out of 10 would not be considered unusual
since it is between 1.84 and 8.16.
6. Guessing Answers Guessing 7 out of 10 correct when n= 10 and p= 0.2
a. n=10, p= 1/5= 0.2, q= 1 – p = 1 – 0.2= 0.8
µ = np = 10∗
0.2 = 2
σ =
npq =
10∗
0.2 ∗ 0.8 =
1.60 = 1.26
maximumusualvalue=
µ + 2σ
= 2 + 2 ∗1.26
= 4.53
minimumusualvalue=
µ - 2σ
= 2 − 2 ∗1.26
= −0.53
b. Yes, the maximum usual value is 4.53, so a value of 7 correct out of 10 would be considered unusual since
it is higher than 4.53.
7. Playing Roulette P(winning)= 1/38= 0.0263, number of trials= n= 100
a. n=100, p= 0.0263, q= 1 – p = 1 – 0.263= 0.974
µ = np = 100 ∗ 0.0263 = 2.63
σ = npq = 100 ∗ 0.0263 ∗ 0.974 = 2.562 = 1.60
2.63 2 1.60 2.63 3.20 0.57
2σ µ minimum usual value = - = − ∗ = − = −
maximum usual value = + = 2.63 + 2 ∗1.60
= 2.63
2ο µ
+ 3.20 = 5.83
b. No, it would not be unusual to not win at least once in the 100 rotations of the wheel since 0 is in the range
of minimum usual value to maximum usual values.

Chapter 4: Discrete Probability Distributions 99
8. Left-Handed People P(left-handed)= 0.10, n= 25 students in biology class
a. n=25, p= 0.10, q= 1 – p = 1 – 0.10= 0.90
µ = np = 25 ∗ 0.10 = 2.50
σ =
npq =
25 ∗ 0.10 ∗ 0.90 =
2.25 = 1.50
maximum usual value = µ + 2σ
= 2.50 + 2 ∗1.50
= 5.50
minimum usual value = µ - 2σ
= 2.50 − 2 ∗1.50
= −0.50
b. No, it would not be unusual to have 5 out of 25 students being left-handed out of 25
students since 5 is in the range of minimum usual value to maximum usual values.
9. Amazing Results of Experiments in Gender Selection 15 couples in control group and 15 couples in an
experimental group, each group has one child
a. Possible outcomes of variable x, number of girls, P(girl)= 0.50
Using Table A-1, n= 15, p= 0.50
x(number
P(x)
of girls)
0 0.000
1 0.000
2 0.003
3 0.014
4 0.042
5 0.092
6 0.153
7 0.196
8 0.196
9 0.153
10 0.092
11 0.042
12 0.014
13 0.003
14 0.000
15 0.000
b. n= 15, p= 0.50, q= 0.50
µ = np = 15∗
0.50 = 7.50
σ = npq = 15 ∗ 0.50 ∗ 0.50 =
c. Is 10 girls and 5 boys unusual
3.75 = 1.94
maximum usual value = µ + 2σ
= 7.5 + 2 ∗1.94
= 11.38
mimmum usual value = µ - 2σ
= 7.5 - 2 ∗1.94
= 3.62
No, 10 girls out of 15 would not be unusual since 10 is in the range of minimum usual value to maximum
usual values.
10. Cell Phones and Brain Cancer 420,095 cell phone users, P(cancer)= 0.000340
a. n= 420,095, p= 0.000340, q= 0.999666
µ = np = 420095∗0.000340
= 142.83
σ = npq = 420095∗0.000340∗0.999666
= 142.785 = 11.95
b. Is 135 cases unusual
maximum usual value = µ + 2σ
= 142.83+
2∗11.95
= 166.73
minimum usual value = µ − 2σ
= 142.83−
2∗11.95
= 118.93
No, 135 cases out of 420,095 would not be unusual since 135 is in the range of minimum usual value to
maximum usual values.

100 Chapter 4: Discrete Probability Distributions
c. The concern is not supported by the evidence. Since 135 cases were in the range, this number of cases of
brain cancer is not unusual or unlikely to have occurred by chance.
11. Cholesterol-Reducing Drug 863 patients given Lipitor, 19 experienced flu symptoms,
P(flu for not treated)= 0.019
a. n= 863, p= 0.019, q= 0.981
µ = np = 863∗0.019
= 16.40
σ = npq = 863∗0.019∗0.981
= 16.085 = 4.01
b. Is it unusual to find 19 out of 863 with flu symptoms
maximum usual value = µ + 2σ
= 16.40 + 2∗4.01
= 24.42
minimum usual value = µ − 2σ
= 16.40 − 2∗
4.01 = 8.38
No, it is not unusual to have 19 patients with flu symptoms out of 863 patients since 19 is in the range of
the minimum usual value to maximum usual values.
c. Flu symptoms do not appear to be an adverse reaction that should be of concern to users of Lipitor.
12. Opinions About Cloning 1012 randomly selected adults, 89% indicated cloning of humans should not be
allowed
a. Number indicating cloning should not be allowed= 0.89 ∗ 1012= 900.68 ≈ 901 (rounded)
b. n= 1012, p= 0.50, q= 0.50
µ = np = 1012∗0.5
= 506
σ = npq = 1012∗0.50∗0.50
= 253.00 = 15.91
c. Is 89% unusually high compared with a 50% rate of not being in favor
maximum usual value = µ + 2σ
= 506 + 2∗15.91
= 537.82
minimum usual value = µ − 2σ
= 506 − 2∗15.91
= 474.18
Yes, it appears that an overwhelming majority of adults, assuming a random sample, are not in favor of
human cloning since 901 (89%) of the 1012 respondents are not in the range of the minimum and
maximum usual values. In fact, 901 is well above the maximum usual value, by almost 23 standard
deviations.
13. Car Crashes 34% of those in age 20-24 have car crashes in one year, in a sample of 500 randomly selected
New Your City drivers age 20-24, 42% had accidents
a. Number in sample having accidents= 42/100 ∗ 500= 0.42 ∗ 500= 210
b. n= 500, p= 0.34, q= 0.66
µ = np = 500∗0.34
= 170
σ = npq = 500∗0.34∗0.66
= 112.20 = 10.59
c. Is 42% unusually high for the NYC drivers in this age group
maximum usual value = µ + 2σ
= 170 + 2∗10.59
= 191.18
minimum usual value = µ − 2σ
= 170 − 2∗10.59
= 148.82
Yes, the result of 42% or 210 drivers is unusually high compared with 34% rate in the general population
of 20-24 year olds since 210 is not in the range of usual minimum and maximum values. 210 accidents is
higher than the usual maximum value.
14. Using the Empirical Rule and Chebyshev’s Theorem Effectiveness of MircoSort gender selection method,
100 couples trying to have girls as birth child, range rule of thumb would be 40 to 60 girl births out of 100
a. Is binomial approximately bell-shaped
We have reason to believe it is for several reasons. Since the mean of this distribution would be in the
middle of the scale, µ= 50, with a standard deviation,σ, of 5, most of the scores (about 95%) will fall
between 40 and 60, clearly the middle of the scale. The number can range from 0 to 100 so there appears to
be a difference in the accumulation rate, an increase as values go from 0 to 50 and a decrease in
accumulation rate as values got from 50 to 100. In addition, we have seen distributions with fewer trials

Chapter 4: Discrete Probability Distributions 101
(see n= 15, p= 0.5, in problem 4-4.9) have distributions that are symmetric and appear to be bell-shaped.
One would expect that when p= 0.5 with more trials the distribution would also be bell-shaped.
b. How likely to be in 40-60 range
As seen in a. 40 is 2σ below the mean of 50 and 60 is 2σ above the mean of 50. The empirical rule states
that about 95% of the scores would be between ± 2σ. Thus, it is very likely the number of girls born would
fall in the 40 to 60 range.
c. How likely is the number between 35 and 65
According to the empirical rule, between ±3σ, there would be 99% of the values. Since 35 is 3σ below the
mean and 65 is 3σ above the mean, we expect about 99% of the number of girls to be born would be in the
35 to 60 range.
d. Chebyshev’s theorem indicates that 1 – 1/K 2 of the values will fall in the range of K, where K is the
number of standard deviation. When K= 2, there will be at least 75% of the values between ±2 standard
deviation. We would conclude that, according to Chebychev’s Theorem, at least 75% or 75 of the 100
births of girls will be between 40 and 60.
4-5 The Poisson Distribution
Using a Poisson Distribution to Find Probability. In Exercises 1-4, assume the Poisson Distribution applies and
Proceed to use the given mean to find the indicated probability.
x −µ
3 −2
−2
µ ∗ e 2 ∗ e 8 ∗ 2.71828 1.0827
1. µ = 2, x = 3, P(3)
= = =
= = 0. 1804
x!
3! 6 6
2 −0.5
−0.5
0.5 ∗ e 0.25 ∗ 2.71828 0.1516
2. µ = 0.5, x = 2, P(2)
= =
= = 0. 0758
2!
2
2
99 −100
154
100 ∗ e 3.72E
100,
155
99! 9.333E
3. µ = x = 99,P(x) =
= = 0.0399 (Excel used for this computation)
4. µ=500, x= 512, P(512) Since numbers are too large for hand, calculator or Excel computations, an
approximation will be used
512 −500
500 ∗ e
µ = 500, x = 512, P(512)
=
512!
512! = 512.5 ∗ log512 − 0.39908993 − 0.43429446 ∗ 512 = 1166.541
500
e
−500
log
512
= loge
∗500
= 217.147
[ P(
x = 512) ]
P(512)
= 10
= 512 *log500 = 512 ∗ 2.6990 = 1381.873
−1.816
= 1381.873 − 217.147 −1166.541
= −1.816
= 0.0153
5. Radioactive Delay for Cessium 137, over 365 days decay was 1,000,000,000 down to 977,287
a. find mean decay per day
µ =
number of atoms 1,000,000 − 977,287
=
=
number of days 365
b. P(on a given day, 50 atoms lost)
22,713
365
= 62.227
50
62.227 ∗ e
P (50 on a given day) =
50!
−62.227
=
4.9997E89
∗9.4444E
− 28
3.0414E64
4.7219E62
=
3.0414E64
= 0.0155

102 Chapter 4: Discrete Probability Distributions
6. Births 11 babies born each year in Westport with population of 760
a. Mean number of births per day
number of births 11
µ =
= = 0.03014
number of days 365
b. P(no births in a given day)
0
0.03014 ∗e
P (0) =
0!
−0.03014
c. P(at least one birth on a given day)
1∗
2.71828
=
1
−0.03014
P ( x ≥ 1) = 1−
P(
x = 0) = 1−
0.9704 = 0.0296
= 0.9703
d. Based on these results, medical birthing personnel should be an as needed basis rather than on permanent
standby since they are not likely to be called more often than about 11 days a year. Yes, women in
Westport are probably not as likely to get immediate medical attention as they would likely get in a more
populated area where medical birthing personnel are more likely to be on permanent standby.
7. Deaths from Horse Kicks 196 horse kick deaths during 280 corps years
µ =
number of deaths
number of corps - years
196
= = 0.7
280
−0.7
−0.7
In the following: e = 2.71828 = 0. 4966
a.
0 −0.7
0.7 ∗ e 0.4966
P (0) = = = 0. 497
0! 1
b.
1 −0.7
0.7 ∗ e 0.7 ∗ 0.4966 0.3476
P (1) = =
= = 0. 348
1!
1 1
c.
2 −0.7
0.7 ∗ e 0.49 ∗ 0.4966 0.2433
P (2) = =
= = 0. 122
2!
2 2
d.
3 −0.7
0.7 ∗ e 0.343∗
0.4966 0.1703
P (3) = =
= = 0. 0284
3!
6 6
e.
4 −0.7
0.7 ∗ e 0.2401∗
0.4966 0.1192
P (4) = =
= = 0. 00497
4!
24 24
Comparison of actual results with Poisson Distribution for 280 Corps-years
Deaths
Expected number of corps-years Actual number of horse
from Poisson distribution kick deaths in corps-years
0 0.4973∗280= 139.24 144
1 0.3481∗280= 97.47 91
2 0.1218∗280= 34.10 32
3 0.0284∗280= 7.95 11
4 0.0050∗280= 1.40 2
Yes, the Poisson distribution serves as a good device for predicting the actual results. There are some deviations
that would be expected by chance. In general the shape and frequency of the values are very similar.
8. Homicide Deaths 116 homicide deaths in Richmond
µ =
number of deaths 116
= = 0.3178
number of days 365

Chapter 4: Discrete Probability Distributions 103
−0.3178
−0.3178
In the following: e = 2.71828 = 0. 7277
a.
0 −0.3178
0.3178 ∗ e 1∗
0.7277 0.7277
P (0) =
= = = 0. 728
0!
1 1
b.
1 −0.3178
0.3178 ∗ e 0.3178 ∗ 0.7277 0.2313
P (1) =
=
= = 0. 231
1!
1
1
. c.
2 −0.3178
0.3178 ∗ e 0.1010 ∗ 0.7277 0.07350
P (2) =
=
= = 0. 0367
2!
2
2
d.
3 −0.3178
0.3178 ∗ e 0.03210 ∗ 0.7277 0.02336
P (3) =
=
= = 0. 00389
3!
6
6
e.
4 −0.3178
0.3178 ∗ e 0.01020 *0.7277 0.007423
P (4) =
=
= = 0. 000309
4!
24
24
Comparison of actual results with Poisson Distribution for 365 days
Homicides
Expected number of days with a
homicide from Poisson
distribution
Actual number of
days with a homicide
in a year
0 0.728∗365= 265.72 268
1 0.231∗365= 84.32 79
2 0.0367∗365= 13.40 17
3 0.00389∗365= 1.42 1
4 0.000309∗365= 0.11 0
Yes, the Poisson distribution serves as a good device for predicting the actual results of number of homicides per
day. There are some deviations that would be expected by chance. In general the shape and frequency of the values
are very similar.
9. Dandelions mean number of dandelions in a given area is 7 per square meter
µ= 7 per square meter
−7
−7
In the following: e = 2.71828 = 0. 0009119
0 −7
7 ∗ e 1∗
0.0009119 0.0009119
a. P (0) = =
= = 0. 000912
0! 1
1
b. P ( at least 1) = 1−
P(
0) = 1−
0.000912 = . 999088
1 −7
7 ∗ e 7 ∗ 0.0009119 0.0009119
c. P (1) = =
= = 0. 00638
1! 1
1
2 −7
7 ∗ e 49 ∗ 0.0009119 0.04468
P (2) = =
= = 0.0223
2!
2
2
P ( 2 at most) = P(0)
+ P(1)
+ P(2)
= 0.000912 + 0.00638 + 0.0223 = 0.0296
10. Earthquakes for 100 years, 93 major earthquakes in the world
number of earthquakes 93
µ =
= = 0.93 per year
number of years 100
−0.93
−0.93
In the following: = 2.71828 = 0. 3946
e
0 −0.93
0.93 ∗ e 1∗
0.3946 0.3946
a. P (0) =
= = = 0. 395
0! 1 1

104 Chapter 4: Discrete Probability Distributions
b.
1 −0.93
0.93 ∗ e 0.93∗
0.3946 0.3670
P (1) =
=
= = 0. 367
1!
1 1
c.
2 −0.93
0.93 ∗ e 0.8649 ∗ 0.3946 0.3413
P (2) =
=
= = 0. 171
2!
2
2
d.
3 −0.93
0.93 ∗ e 0.8044 ∗ 0.3946 0.3174
P (3) =
=
= = 0. 0529
3!
6
6
e.
4 −0.93
0.93 ∗ e 0.7481∗
0.3946 0.2952
P (4) =
=
= = 0. 0123
4!
24 24
f.
5 −0.93
0.93 ∗ e 0.6957 ∗ 0.3946 0.2745
P (5) =
=
= = 0. 00229
5!
120 120
g.
6 −0.93
0.93 ∗ e 0.6470 ∗ 0.3946 0.2553
P (6) =
=
= = 0. 000355
6!
720 720
h.
7 −0.93
0.93 ∗ e 0.6017 ∗ 0.3946 0.2374
P (7) =
=
= = 0. 0000471
7!
5040 5040
Comparison of actual results with Poisson Distribution for years with major earthquake per 100 years
Earthquakes
Expected number of years when a
major earthquake occurs from
Poisson distribution
Actual number of years
when a major earthquake
occurred over 100 years
0 0.395∗100= 39.50 47
1 0.367∗100= 36.70 31
2 0.171∗100= 17.10 13
3 0.0529∗100= 5.29 5
4 0.0123∗100= 1.23 2
5 0.00229∗100= 0.23 0
6 0.000355∗100= 0.04 1
7 0.0000471∗100= 0.00 1
Yes, the Poisson distribution serves as a good device for predicting the actual results of number of major
earthquakes in a 100 year period. There are some deviations that would be expected by chance. In general
the shape and frequency of the values are very similar.
Review Exercises
1. a. A random variable is a variable that has a single value, usually numeric, determined by chance, for each
outcome of a procedure or experiment.
b. A probability distribution is a description of the possible values a random variable can assume with the
associated probability for each possible outcome. A valid probability distribution has values of the
probabilities of 0 ≤ P(x) ≤ 1 for each value of x and the sum of the probabilities must equal 1.
c. The accompanying table describes a probability distribution because the sum of all the probabilities = 1,
and each probability value is 0 ≤ P(x) ≤ 1.
[ ]=
(0 ∗ 0.528) + (1 ∗ 0.360) + (2 ∗ 0.098) + (3∗
0.013) + (4 ∗ 0.001) + (5∗
0.000) =
0 + 0.360 + 0.196 + 0.039 + 0.004 + 0.000 = 0.599
d. Mean of this distribution, µ = Σ x∗P( x)
2
[ ] − µ =
2
e. Standard deviation of this distribution, σ = Σ x ∗ P( x)

Chapter 4: Discrete Probability Distributions 105
(0
2
2
2
∗0.528)
+ (1 ∗0.360)
+ (2
2
2
∗0.098)
+ (3 ∗0.013)
+ (4
∗0.001)
+ (5
∗0.000)
− 0.599
0.885−
0.359 = 0.526 = 0.725
f. Yes, it would be very unusual to select 5 condoms at random that all failed the test. That would happen
very infrequently, less than 5 times out of 100 repetitions, a standard we often use to judge unusual events.
2. Employee Drug Testing 80% of the construction companies conduct drug testing, a random sample of 10
companies is chosen
n= 10, p= 0.80, x= 5, From Table A-1
a. ( 5) = 0.026
P
P ( at least 5) = P 5 + P 6 + P 7 + P 8 + P 9 + P 10
0.026 + 0.088 + 0.201+
0.302 + 0.268 + 0.107 = 0.992
µ = np = 10 ∗ 0.8 = 8.
b. () () () () () ( )=
c. 0
σ = npq
=
10 ∗ 0.8 ∗ 0.2 = 1.60 = 1.26
d. minimum usual value= µ − 2σ = 8.0 − 2 ∗ 1.26= 8.0 – 2.52= 5.48
maximum usual value= µ + 2σ = 8.0 + 2 ∗ 1.26= 8.0 + 2.52= 10.52
No, it would not be unusual to have 6 out of 10 companies test for substance abuse since 6 is in the range of
the minimum and maximum usual values.
3. Reasons for Being Fired 17% indicate “inability to get along with others” as reason for firing at the Kansas
Agriculture Company
n= 5, p= 0.17, q= 0.83, x= 4
a. P(at least 4 out of 5 cite this reason)
P(
x)
=
5!
P(4)
= 0.17
4!(5 − 4)!
0.00347
n!
∗ p
x! ( n − x)!
5!
P(5)
= 0.17
5!(5 − 5)!
∗ 0.83
∗ 0.83
5−5
P(4 or more) = P(4)
+ P(5)
=
x
5
4
∗ q
n−x
5−4
5 ∗ 4!
= ∗ 0.000835 ∗ 0.83
4!1!
5!
= ∗ 0.000142 ∗ 0.83
5!
0.00347 + 0.000142 = 0.00361
0
1
2
5
= ∗ 0.000835 ∗ 0.83 =
1
= 1∗
0.000142 ∗1
= 0.000142
b. If she finds four out of 5 firings due to “inability to get along with others” as the reason, then this company
would differ from other companies. The probability of this happening by chance would be very low, about
4 times out of 1000 and this would be considered to be very unusual compared with the other companies
where the percentage giving this as the reason is 17%.
4. Deaths 7 persons die per year in Westport which has a population of 760
number of deaths 7
a. µ =
= = 0. 0192
number of days 365
−0.0192
In the following: e = 0. 9810
x −µ
0 −0.0192
µ ∗ e 0.0192 ∗ e 1∗
0.9810
b. x = 0, P(
x)
= =
= = 0. 9810
x!
0!
1
1 −0.0192
0.0192 * e 0.0192 ∗ 0.9810 0.0188
c. x = 1, P(1)
=
=
= = 0. 0188
1!
1
1
P ( more than 1death) = 1 - P(0)
+ P(1)
= 1−
0.9810 + 0.0188 = 1−
0.9998 = 0.
d. [ ] [ ] 0002
2
=

106 Chapter 4: Discrete Probability Distributions
e. Based on these results, Westport would not need to have a contingency plan to handle more than one death
per day since that occurrence is not likely to happen. It would happen about 2 times in every 1000 days, or less
than once per every two years.
Cumulative Review Exercises
1. Weights: Analysis of Last Digits last digits in data, 0 – 9, distribution of frequency each digit is observed.
a. Mean and standard deviation
x f f ∗x x 2 f ∗ x 2
0 7 0 0 0
1 14 14 1 14
2 5 10 4 20
3 11 33 9 99
4 8 32 16 128
5 4 20 25 100
6 5 30 36 180
7 6 42 49 294
8 12 96 64 768
9 6 54 81 486
n= 78 Σx= 331 Σx 2 = 2089
µ =
σ =
∑
x 331
= = 4.244
n 78
∑
x
2
(
−
n
∑
x)
n
2
=
331
2089 −
78
78
2
=
109561
2089 −
78
78
=
2089 −1404.63
78
=
684.37
78
=
8.774
= 2.962
b. Relative Frequency Table
x f P(x) x∗ P(x) x 2 x 2 ∗ P(x)
0 7 0.0875 0.0000 0 0.0000
1 14 0.1750 0.1750 1 0.1750
2 5 0.0625 0.1250 4 0.2500
3 11 0.1375 0.4125 9 1.2375
4 8 0.1000 0.4000 16 1.6000
5 4 0.0500 0.2500 25 1.2500
6 5 0.0625 0.3750 36 2.2500
7 6 0.0750 0.5250 49 3.6750
8 12 0.1500 1.2000 64 9.6000
9 6 0.1000 0.9000 81 8.1000
n= 78 ∑P(x)= 1.00 4.3625 28.1375

Chapter 4: Discrete Probability Distributions 107
Mean
2
σ =
= ∑[ ( x ∗ P(
x)
]
2
∑ [ x ∗ P(
x)
] −
σ = 9.102 = 3.017
= 4.363
2
2
µ = 28.138 − 4.363
= 28.138 −19.036
= 9.102
c. Probability Distribution where each digit equally likely
x P(x) x∗ P(x) x 2 x 2 ∗ P(x)
0 0.1 0.0 0 0.0
1 0.1 0.1 1 0.1
2 0.1 0.2 4 0.4
3 0.1 0.3 9 0.9
4 0.1 0.4 16 1.6
5 0.1 0.5 25 2.5
6 0.1 0.6 36 3.6
7 0.1 0.7 49 4.9
8 0.1 0.8 64 6.4
9 0.1 0.9 81 8.1
Total Σ(x)=1 ∑[x∗P(x)] = 4.5 ∑[x 2 ∗ P(x)] = 28.5
. Mean, µ
= Σ
[ x ∗ P( x)
]
Standard deviation, σ =
= 4.5
Σ
2
2
[ x ∗ P(
x)
] − µ = 28.5 − 20.25 = 8.25 = 2. 872
d. The above table (c) describes what we would expect from a random selection of last digits-an equal
likelihood of each occurring. However, in Tables a and b, which describe our sample, we see some
deviation from this expected distribution. We would have expected the relative frequency to be as similar to
our probability distribution in Table c as possible. While the means and standard deviations do not vary by
a great deal, there are some digits that seem to occur more often than expected (1, 3, and 8) and a few that
seem to occur less often than expected (2, 5, and 6). This could be assessed more accurately with another
method that will be discussed later.
2. Determining the Effectiveness of an HIV Training Program 10% rate of HIV virus for the “at-risk”
population in New York State, intensive education program initiated to lower the rate, after the program 150 atrisk
individuals are studied in a follow-up Binomial distribution with n= 150, P(HIV)= p= 0.10, q= 0.90
a. Mean and Standard Deviation
Mean, µ = np = 150 ∗ 0.10 = 15
Standard deviation, σ = npq = 150 ∗ 0.10 ∗ 0.90 = 13.5 = 3.674
b. 8% or 12 of the 150 tested positive for the HIV Virus after the program
maximum usual value= µ + 2σ = 15.00 + 2 ∗ 3.674= 15.00 + 7.35= 22.35
minimum usual value= µ − 2σ = 15.00 – 2 ∗ 3.674= 15.00 – 7.35= 7.65
No, if the program is not effective, a value of 12 is not unusually low. It is in the range of the minimum
and maximum usual values, clearly not lower than the minimum usual value. No, this result suggests the
program was not effective in reducing the HIV virus rate.