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Chapter 5
Discrete Probability Distributions
5-2 Random Variables
1. As defined in the text, a random variable is a variable that takes on a single numerical value,
determined by chance, for each outcome of a procedure. In this exercise, the random variable
is the number of winning lottery tickets obtained over a 52 week period. The possible values
for this random variable are 0,1,2,…,52.
3. The probability distribution gives each possible non-overlapping outcome O_{i} of a procedure
and the probability P(O_{i} ) associated with each of those outcomes.
Since it is a certainty that one of the outcomes will occur,
P(O_{1} or O_{2} or O_{3} or…) = 1.
Since the outcomes are mutually exclusive,
P(O_{1} or O_{2} or O_{3} or…) = P(O_{1}) +P(O_{2}) +P(O_{3}) +….
Therefore,
P(O_{1}) +P(O_{2}) +P(O_{3}) +… = ΣP(O_{i}) = 1.
5. a. Discrete, since such a number is limited to certain values – viz., the non-negative integers.
b. Continuous, since weight can be any value on a specified continuum.
c. Continuous, since height can be any value on a specified continuum.
d. Discrete, since such a number is limited to certain values – viz., the non-negative integers.
e. Continuous, since time can be any value on a specified continuum.
NOTE: If one of the conditions for a probability distribution does not hold, the formulas do not apply – and they produce numbers that have no meaning. When working with probability distributions and formulas in the exercises that follow, always keep the following important facts in mind.
• ΣP(x) must always equal 1.000
• Σ[x∙P(x)] gives the mean of the x values and must be a number between the highest and lowest
x values.
• Σ[x^{2}∙P(x)] gives the mean of the x^{2} values and must be a number between the highest and
lowest x^{2} values.
• Σx and Σx^{2} have no meaning and should not be calculated
• The quantity “Σ[x^{2}∙P(x)] – μ^{2}” cannot possibly be negative – if it is, there is a mistake.
• Always be careful to use the unrounded mean in the calculation of the variance, and to take the
square root of the unrounded variance to find the standard deviation.
7. The given table is a probability distribution since 0P(x)1 for each x and ΣP(x)=1.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0.125 0 0 0 = 1.500, rounded to 1.5 children
1 0.375 0.375 1 0.375 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.375 0.750 4 1.500 = 3.000 – (1.500)^{2}
3 0.125 0.375 9 1.125 = 0.750
1.000 1.500 3.000 σ = 0.866, rounded to 0.9 children
9. The given table is not a probability distribution since ΣP(x) = 0.984 ≠ 1.
11. The given table is a probability distribution since 0P(x)1 for each x and ΣP(x)=1.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0.02 0 0 0 = 2.75, rounded to 2.8 TV sets
1 0.15 0.15 1 0.15 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.29 0.58 4 1.16 = 9.21 – (2.75)^{2}
3 0.26 0.78 9 2.34 = 1.6475
4 0.16 0.64 16 2.56 σ = 1.284, rounded to 1.3 TV sets
5 0.12 0.60 25 3.00
1.00 2.75 9.21
13. The given table is a probability distribution since 0P(x)1 for each x and ΣP(x)=1.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0+ 0 0 0 = 6.000, rounded to 6.0 peas
1 0+ 0 1 0 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.004 0.008 4 0.016 = 37.484 – (6.000)^{2}
3 0.023 0.069 9 0.207 = 1.484
4 0.087 0.348 16 1.382 σ = 1.218, rounded to 1.2 peas
5 0.208 1.040 25 5.200
6 0.311 1.866 36 11.196
7 0.267 1.869 49 13.083
8 0.100 0.800 64 6.400
1.000 6.000 37.484
15. a. P(x=7) = 0.267
b. P(x 7) = P(x=7 or x=8)
= P(x=7) + P(x=8)
= 0.267 + 0.100
= 0.367
c. The answer in part (b) is the one which determines whether 7 is an unusually high result.
When there are several different possible outcomes, the probability of getting any one of
them exactly (even the more likely ones near the middle of the distribution) may be small.
An outcome is unusually high if it is in the upper tail of the distribution – i.e., if the
probability of getting it or a higher value is small.
d. No; since P(x 7) = 0.367 > 0.05, 7 is not an unusually high number of peas with green
pods.
17. a. The given table is a probability distribution since 0P(x)1 for each x and ΣP(x)=1.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
4 0.1919 0.7676 16 3.0704 = 5.7772, rounded to 5.8
5 0.2121 1.0605 25 5.3025 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
6 0.2222 1.3332 36 7.9992 = 34.6834 – (5.7772)^{2}
7 0.3737 2.6159 49 18.3113 = 1.3074
0.9999 5.7772 34.6834 σ = 1.1434, rounded to 1.1
b. μ = 5.8 games and σ = 1.1 games
c. No; since P(x=4) = 0.1919 > 0.05, winning in 4 games is not an unusual event.
19. a. The given table is a probability distribution since 0P(x)1 for each x and ΣP(x)=1.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0.824 0 0 0 = 0.435, rounded to 0.4
1 0.083 0.083 1 0.083 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.039 0.078 4 0.156 = 1.821 – (0.435)^{2}
3 0.014 0.042 9 0.126 = 1.635
4 0.012 0.048 16 0.192 σ = 1.279, rounded to 1.3
5 0.008 0.040 25 0.200
6 0.008 0.048 36 0.288
7 0.004 0.028 49 0.196
8 0.004 0.032 64 0.256
9 0.004 0.036 81 0.324
1.000 0.435 1.821
b. μ = 0.4 bumper stickers and σ = 1.3 bumper stickers
c. The range rule of thumb suggests that “usual” values are those within two standard
deviations of the mean.
minimum usual value = μ – 2σ = 0.4 – 2(1.3) = -2.2, truncated to 0
maximum usual value = μ + 2σ = 0.4 + 2(1.3) = 3.0
The range of values for usual numbers of bumper stickers is from 0 to 3.0.
d. No; since 0 < 1 < 3.0, it is not unusual to have more than one bumper sticker.
21. There are eight equally likely possible outcomes: GGG GGB GBG BGG GBB BGB BBG BBB.
The following table describes the situation, where x is the number of girls per family of 3.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0.125 0 0 0 = 1.500, rounded to 1.5 girls
1 0.375 0.375 1 0.375 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.375 0.750 4 1.500 = 3.000 – (1.500)^{2}
3 0.125 0.375 9 1.125 = 0.75
1.000 1.500 3.000 σ = 0.8660, rounded to 0.9 girls
No. Since P(x=3) = 0.125 > 0.05, it is not unusual for a family of 3 children to have all girls.
23. The following table describes the situation.
x P(x) x∙P(x) x^{2} x^{2}∙P(x) μ = Σ[x∙P(x)]
0 0.1 0 0 0 = 4.5
1 0.1 0.1 1 0.1 σ^{2} = Σ[x^{2}∙P(x)] – μ^{2}
2 0.1 0.2 4 0.4 = 28.5 – (4.5)^{2}
3 0.1 0.3 9 0.9 = 8.25
4 0.1 0.4 16 1.6 σ = 2.8723, rounded to 2.9
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
1.0 4.5 28.5
The probability histogram for this distribution would be flat, with each bar having the same
height.
25. a. Since each of the 3 positions could be filled (with replacement) by any of the 10 digits
0,1,2,3,4,5,6,7,8,9, there are 10∙10∙10 = 1000 possible different selections.
b. Let W = winning (i.e., matching the selection drawn).
Since only one of the possible 1000 selections is a inner, P(W) = 1/1000 = 0.001
c. The net profit is the payoff minus the original bet, in this case $250.00 − $0.50 = $249.50.
d. The following table describes the situation.
x P(x) x∙P(x) E = Σ[x∙P(x)] = -0.2500 [i.e., a loss of 25¢]
-0.50 0.999 -0.4995
249.50 0.001 0.2495
1.000 -0.2500
The expected value is -25¢.
e. Since both games have the same expected value, neither bet is better than the other.
27. a. The following table describes the situation.
x P(x) x∙P(x) E = Σ[x∙P(x)] = -15/38 = -.3947, rounded to -39.4¢
-5 33/38 -165/38
30 5/38 150/38
38/38 -15/38
b. Since -26 > -39.4, wagering $5 on the number 13 is the better bet.
29. a. From the 30-year-old male’s perspective, the two possible outcome values are
-$161, if he lives
100,000 – 161 = $99,839, if he dies
b. The following table describes the situation.
x P(x) x∙P(x) E = Σ[x∙P(x)] = -21.0000, rounded to -$21.0
-161 0.9986 -160.7746
99,839 0.0014 139.7746
1.0000 -21.0000
c. Yes; the insurance company can expect to make an average of $21.0 per policy.
31. For every $1000, Bond A gives a profit of (0.06)($1000) = $60 with probability 0.99.
The following table describes the situation.
x P(x) x∙P(x) E = Σ[x∙P(x)] = 49.40, rounded to 49.4 [i.e., a profit of $49.4]
-1000 0.01 -10.00
60 0.99 59.40.
1.00 49.40
For every $1000, Bond B gives a profit of (0.08)($1000) = $80 with probability 0.95.
The following table describes the situation.
x P(x) x∙P(x) E = Σ[x∙P(x)] = 26.00, rounded to 26.0 [i.e., a profit of $26.0]
-1000 0.05 -50.00
80 0.95 76.40.
1.00 26.00
Bond A is the better bond since it has the higher expected value – i.e., 49.4 > 26.0. Since
both bonds have positive expectations, either one would be a reasonable selection. Although
Bond A has a higher expectation, a person willing to assume more risk in hope of a higher
payoff might opt for Bond B.
33. Since each die has 6 faces, there will be 6∙6 = 36 possible outcomes. If each of the 12 sums
1,2,3,…,12 is to be equally likely, each sum must occur exactly 3 times. As a starting point,
suppose one of the dice is normal. If one die contains the usual digits 1,2,3,4,5,6, the other die
must have three 0’s to pair with the 1 to get three sums of 1.
must have three 6’s to pair with the 6 to get three sums of 12.
1-0 2-0 3-0 4-0 5-0 6-0
1-0 2-0 3-0 4-0 5-0 6-0
1-0 2-0 3-0 4-0 5-0 6-0
1-6 2-6 3-6 4-6 5-6 6-6
1-6 2-6 3-6 4-6 5-6 6-6
1-6 2-6 3-6 4-6 5-6 6-6
The 36 possible outcomes generated by such dice areGiven in the box at the right. Inspection indicates that
each of the sums 1,2,3,…,12 appears exactly 3 times so
that P(x) = 3/36 = 1/12 for x = 1,2,3,…,12. A solution,
therefore, is to mark one die normally 1,2,3,4,5,6 and
mark the other die 0,0,0,6,6,6.
5-3 Binomial Probability Distributions
1. Getting two people with blue eyes followed by three people with eyes that are not blue is only
one way of getting exactly two people with blue eyes in a sample of five randomly selected
people. All the ways of getting exactly two people as described must be considered.
3. Yes. Since 30/1236 = 0.024 0.05, the 5% rule can be applied to determine that the repeated
selections can be treated as if they were independent.
5. Yes. All four requirements are met.
7. No. Requirement (3) is not met. There are more than two possible outcomes for the ages of
the parents.
9. No. Requirements (2) and (4) are not met. Since 20/100 = 0.20 > 0.05, and the selections are
done without replacement, the trials cannot be considered independent. The value p of
obtaining a success changes from trial to trial as each selection without replacement changes
the population from which the next selection is made.
11. Yes. All four requirements are met. Since 500/2,800,000 = 0.00018 0.05, the selections
can considered to be independent – even though the they are made without replacement.
13. a. P(WWC) = P(W)∙P(W)∙P(C)
= (4/5)(4/5)(1/5)
= 16/125 = 0.128
b. There are three possible arrangements: WWC, WCW, CWW
P(WWC) = P(W)∙P(W)∙P(C) = (4/5)(4/5)(1/5) = 16/125
P(WCW) = P(W)∙P(C)∙P(W) = (4/5)(1/5)(4/5) = 16/125
P(CWW) = P(C)∙P(W)∙P(W) = (1/5)(4/5)(4/5) = 16/125
c. P(exactly one correct) = P(WWC or WCW or CWW)
= P(WWC) + P(WCW) + P(CWW)
= 16/125 + 16/125 + 16/125
= 48/125 = 0.384
15. From Table A-1 in the .30 column and the 2-1 row, P(x=1) = 0.420.
17. From Table A-1 in the .99 column and the 15-11 row, P(x=11) = 0+.
19. From Table A-1 in the .05 column and the 10-2 row, P(x=2) = 0.075.
21. P(x) = p^{x}q^{n-x}
P(x=10) = [12!/(2!10!)](0.75)^{10}(0.25)^{2}
= [66](0.0563)(0.0625)
= 0.232
NOTE: The intermediate values 66, 0.0563, and 0.0625 are given in Exercise 21 to help those with an incorrect answer to identify the portion of the problem in which the mistake was made. In the future, only the value n!/[(n-x)!x!] will be given separately. In practice, all calculations can be done in one step on a calculator. You may choose (or be asked) to write down intermediate values for your own (or the instructor’s) benefit, but…
• Never round off in the middle of a problem.
• Do not write the values down on paper and then re-enter them in the calculator.
• Use the memory to let the calculator remember with complete accuracy any intermediate values
that will be used in subsequent calculations.
In addition, always make certain that n!/[(n-x)!x!] is a whole number and that the final answer is
between 0 and 1.
23. P(x) = p^{x}q^{n-x}
P(x=4) = [20!/(16!4!)](0.15)^{4}(0.85)^{16}
= [4845](0.15)^{4}(0.85)^{16}
= 0.182
25. P(x1) = 1 – P(x=0)
= 1 – 0.050328
= 0.949672, rounded to 0.0950
Yes, it is reasonable to expect that at least one group O donor will be obtained.
27. P(x=5) = 0.018453, rounded to 0.018
Yes. Since 0.018 0.05, getting all 5 donors from group O would be considered unusual.
29. Let x = the number of consumers who recognize the Mrs. Fields brand name.
binomial problem: n = 10 and p = 0.90, use Table A-1
a. P(x=9) = 0.387
b. P(x≠9) = 1 – P(x=9)
= 1 – 0.387
= 0.613
31. Let x = the number of people with brown eyes.
binomial problem: n = 14 and p = 0.40, use Table A-1
P(x12) = P(x=12) + P(x=13) + P(x=14)
= 0.001 + 0^{+} + 0^{+}
= 0.001
Yes. Since 0.001 0.05, getting at least 12 persons with brown eyes would be unusual.
32. Let x = the number of delinquencies.
binomial problem: n = 12 and p = 0.01, use Table A-1
P(x1) = 1 – P(x=0)
= 1 – 0.886
= 0.114
Yes. Since 0.114 > 0.05, it would not be unusual for at least one of the people to become
delinquent. The bank should make plans for dealing with a delinquency.
33. Let x = the number of offspring peas with green pods.
binomial problem: n = 10 and p = 0.75, use the binomial formula
P(x) = p^{x}q^{n-x}
P(x9) = P(x=9) + P(x=10)
= [10!/(1!9!)](0.75)^{9}(0.25)^{1} + [10!/(0!10!)](0.75)^{10}(0.25)^{0}
= [10](0.75)^{9}(0.25)^{1} + [1](0.75)^{10}(0.25)^{0}
= 0.1877 + 0.0563
= 0.2440, rounded to 0.244
No. Since 0.244 > 0.05, getting at least 9 peas with green pods is not unusual.
35. Let x = the number of special program students who graduated.
binomial problem: n = 10 and p = 0.94, use the binomial formula
P(x) = p^{x}q^{n-x}
a. P(x9) = P(x=9) + P(x=10)
= [10!/(1!9!)](0.94)^{9}(0.06)^{1} + [10!/(0!10!)](0.94)^{10}(0.06)^{0}
= [10](0.94)^{9}(0.06)^{1} + [1](0.94)^{10}(0.06)^{0}
= 0.3438 + 0.5386
= 0.8824, rounded to 0.882
b. P(x=8) = [10!/(2!8!)](0.94)^{8}(0.06)^{2}
= [45](0.94)^{8}(0.06)^{2}
= 0.0988
P(x8) = P(x=8) + P(x=9) + P(x=10)
= 0.0988 + 0.3438 + 0.5386
= 0.9812
P(x7) = 1 – P(x8)
= 1 – 0.9812
= 0.0188
Yes. Since P(x7) = 0.0188 0.05, getting only 7 that graduated would be unusual.
NOTE: Remember that in situations involving multiple ordered outcomes, the unusualness of
a particular outcome is generally determined by the probability of getting that outcome or a
more extreme outcome.
37. Let x = the number of households tuned to the NBC game if the stated share is correct.
binomial problem: n = 20 and p = 0.22, use the binomial formula
P(x) = p^{x}q^{n-x}
a. P(x=0) = [20!/(20!0!)](0.22)^{0}(0.78)^{20}
= [1](0.22)^{0}(0.78)^{20}
= 0.00695
b. P(x1) = 1 – P(x=0)
= 1 – 0.00695
= 0.99305, rounded to 0.993
c. P(x=1) = [20!/(19!1!)](0.22)^{1}(0.78)^{19 }
^{ }= [20](0.22)^{1}(0.78)^{19}
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