Chapter 3
Random Variables
3-1. a. = 1.0
b. / x / F(x)0 / 0.3
1 / 0.5
2 / 0.7
3 / 0.8
4 / 0.9
5 / 1.0
c. P(X > 2) = 1 - F(2) = 0.3
3-2. a. = 1.0
b. / x / F(x)0 / 0.01
1 / 0.10
2 / 0.40
3 / 0.60
4 / 0.80
5 / 0.90
6 / 1.00
c. = F(4) = 0.80
d. = 1 - F(1) = 0.90
3-3. a. = 1.0
b. / x / F(x)0 / 0.10
10 / 0.30
20 / 0.65
30 / 0.85
40 / 0.95
50 / 1.00
c. P(X > 20) = 1 - F(20) = 0.35
3-4. a. = 0.2 + 0.2 + 0.3 = 0.7
b. / x / F(x)0 / 0.1
1 / 0.2
2 / 0.4
3 / 0.6
4 / 0.9
5 / 1.0
c. = 1.0
3-5. first die outcome:
1 / 2 / 3 / 4 / 5 / 61 / · / · / · / · / · / ·
2 / · / · / · / · / · / ·
3 / · / · / · / · / · / ·
Second / 4 / · / · / · / · / · / ·
die out- / 5 / · / · / · / · / · / ·
come: / 6 / · / · / · / · / · / ·
X = sum of the two dice. Note from the sample space that one outcome leads to a sum of 2, two equally-likely outcomes lead to a sum of 3, three lead to a sum of 4, four to a sum of 5, five to a sum of 6, and six (the highest number of outcomes) lead to a sum of 7. Thus, the most likely sum is x = 7. Afterwards, the probabilities decline: five outcomes lead to a sum of 8, four to a sum of 9, three to a sum of 10, two to a sum of 11, and one outcome (two “sixes”) leads to a sum of 12. We thus have the following probability distribution and cumulative distribution function:
X / P(x) / F(x)2 / 1/36 / 1/36
3 / 2/36 / 3/36
4 / 3/36 / 6/36
5 / 4/36 / 10/36
6 / 5/36 / 15/36
7 / 6/36 / 21/36
8 / 5/36 / 26/36
9 / 4/36 / 30/36
10 / 3/36 / 33/36
11 / 2/36 / 35/36
12 / 1/36 / 1.0
3-6. a. = 1.0
b. / x / F(x)0 / 0.1
1 / 0.3
2 / 0.7
3 / 0.8
4 / 0.9
5 / 1.00
c. = F(4) - F(0) = 0.9 - 0.1 = 0.8
d. P(extra costs) = P(X > 3) = 1 - F(3) = 0.2
e. P(no orders for 5 days) = (0.1)5 = 0.00001
f. P(extra costs for 2 days) = (0.2)2 = 0.04
3-7. a. = P(4) + P(5) + P(6) + P(7) = 0.55
b. / x / F(x)2 / 0.20
3 / 0.40
4 / 0.70
5 / 0.80
6 / 0.90
7 / 0.95
8 / 1.00
c. = F(6) = 0.9
d. = F(6) - F(3) = 0.9 - 0.4 = 0.5
3-8. a. = 1.0
b. = P(2) + P(3) = 0.6
c. = 0.7
d. / x / F(x)0 / 0.1
1 / 0.3
2 / 0.6
3 / 0.9
4 / 1.0
3-9. a. = 1.0
b. = 0.20 + 0.15 + 0.10 + 0.05 = 0.50
c. / x / F(x)9 / 0.05
10 / 0.20
11 / 0.50
12 / 0.70
13 / 0.85
14 / 0.95
15 / 1.00
3-10. a. = 1.0
b. 0.85, found by summing the probabilities for 9 - 12:
P(9) + P(10) + P(11) + P(12) = 0.25 + 0.40 + 0.15 + 0.05 = 0.85
c. 0.80 found by: 1 - P(11) - P(12) = 1 - 0.15 – 0.05 = 0.80
d. 0.36 found by: P(price >9) = P(10) + P(11) + P(12) = 0.40 + 0.15 + 0.05 = 0.60
for the price to be above 9 for two days in a row: (0.60)2 = 0.36
3-11. For the random variable in Problem 3-1:
E(X) = 1.8 E(X 2) = 6 V(X) = 2.76 SD(X) = 1.661
Statistics of XMean / 1.8
Variance / 2.76
Std. Devn. / 1.66132
3-12. For Problem 3-2:
E(X) = 3.19 E(X 2) = 12.39 V(X) = 2.2139 SD(X) = 1.4879
3-13. For Problem 3-3:
E(X) = 21.5 E(X 2) = 625 V(X) = 162.75
3-14. For Problem 3-4:
E(X) = 2.8 E(X 2) = 10 V(X) = 2.16 SD(X) = 1.47
Statistics of XMean / 2.8
Variance / 2.16
Std. Devn. / 1.46969
3-15. E(sum of two dice) = 7. Derived as follows:
x / P(x) / xP(x)2 / 1/36 / 2/36
3 / 2/36 / 6/36
4 / 3/36 / 12/36
5 / 4/36 / 20/36
6 / 5/36 / 30/36
7 / 6/36 / 42/36
8 / 5/36 / 40/36
9 / 4/36 / 36/36
10 / 3/36 / 30/36
11 / 2/36 / 22/36
12 / 1/36 / 12/36
252/36 / = 7
3-16. For Problem 3-6:
E(X) = 2.2
P(X > E(X)) = P(X > 2.2) = P(3) + P(4) + P(5) = 0.3
3-17. For Problem 3-7:
E(X) = 4.05 E(X 2) = 19.15 V(X) = 2.7475 SD(X) = 1.6576
3-18. By Chebyshev’s theorem:
P(|X - m| < ks) 1-1/k2
For k = 4, Probability 1 - 1/42 = 1 - 1/16 = 0.9375
3-19. Three standard deviations. Because:
8/9 = 1 - 1/32, so k = 3 in Chebyshev’s theorem.
3-20. E(X) = 8.3 V(X) = 2.3
E(Y) = 8.4 V(Y) = 6.4
Y is a riskier stock because V(Y) > V(X) while E(X) is close to E(Y).
3-21. P(X = 0) is 0.2.
a. The most likely outcome is $2,000 (prob. = 0.3)
b. Is the venture likely to be successful? Yes: P(X > 0) = 0.6
c. The long-term average is E(X): E(X) = +800
d. A good measure of the risk is the standard deviation.
E(X 2) = 2,800,000 V(X) = 2,160,000 SD(X) = 1,469.69
Statistics of XMean / 800
Variance / 2160000
Std. Devn. / 1469.69
3-22. The expected value of a random variable is the long-term average. Hence the airline must raise its fare by the expected value of a claim per passenger. The mean claim is:
E(X) = 0 x P(no claim) + 600 x P(claim) = 0 + 600(0.005) = $3
3-23. We use Equation (3-6) for a linear function of a random variable:
E(aX + b) = aE(X) + b
Hence: E(monthly cost of the operation)
= 25,000 + 5,000E(X) = 25,000 + 5,000(4.05) = $45,250
3-24. Refer to Problem 3-4. Operation cost: C(X) = 300
x / P(x) / C(x) / C(x)P(x)0 / 0.1 / 0 / 0
1 / 0.1 / 300 / 30
2 / 0.2 / 424.26 / 84.85
3 / 0.2 / 519.62 / 103.92
4 / 0.3 / 600 / 180
5 / 0.1 / 670.82 / 67.08
465.85 / = E(cost)
3-25. In Problem 3-2, penalty = X 2.
Find E(penalty). From Problem 3-11: E(X 2) = $12.39
3-26.
Descriptive Statistics of a Random Variable / Problem 3-26x / P(x) / F(x)
-1 / 0.45 / 0.45 / Statistics of X
0 / 0.07 / 0.52 / Mean / 0.03
1 / 0.48 / 1 / Variance / 0.9291
Std. Devn. / 0.9639
Skewness / -0.0599
(Relative) Kurtosis / -1.921
3-27. The variance is a measure of the spread or uncertainty of the random variable. The variance is the expected squared deviation of the value of the random variable from its mean. One use of the variance is as a measure of the risk of an investment.
3-28. The variance is a squared quantity, and thus in applications it has less meaning than its square root, the standard deviation. The standard deviation is in the original units of the problem.
3-29. Refer to Problem 3-23:
Cost = 25,000 + 5,000X
By Equation (3-10): V(aX + b) = a2V(X)
From Problem 3-16, we know that V(X) = 2.7475, so:
V(Cost) = (5,000)2(2.7475) = 68,687,500
SD(Cost) = 8,287.79
3-30. [Using template: Random Variable.xls]
x / P(x)7 / 0.05
8 / 0.1
9 / 0.25
10 / 0.4
11 / 0.15
12 / 0.05
Statistics of X
Mean / 9.65
Variance / 1.3275
Std. Devn. / 1.15217
3-31. 3.11 found by:
P(x) v(x) P(x)v(x)
0.1 2.0 0.20
0.1 2.5 0.25
0.3 3.0 0.90
0.2 3.25 0.65
0.2 3.40 0.68
0.05 3.60 0.18
0.05 5.00 0.25
3.11
3-32. F is not binomial: N/n is not greater than 10 (the trials are not independent and p is not constant).
3-33. X is binomial if sales calls are independent of each other.
3-34. E(X) = np = 100(.10) = 10
SD(X) = = = 3.00
By Chebyshev’s theorem: P(|X - m| < ks) 1 - 1/k2
1 - 1/k2 = 0.75, so k = 2
m - 2s < X < m + 2s
10 - 2(3) < X < 10 + 2(3)
4 < X < 16 are the desired bounds.
3-35. X is not binomial because members of the same family are related and not independent of each other.
3-36. Bernoulli trials are a sequence of identical, independent trials each resulting in a success or a failure with constant success probability, p. A binomial random variable counts the number of successes in a sequence of n Bernoulli trials.
3-37.
a) distribution for n=5, p=0.6 is slightly skewed. Holding p constant, as n is increased from 5 to 20, the distribution becomes more symmetric.
b) Holding n constant, the closer p is to 0.5 the more symmetrical the distribution.
3-38. Using the Binomial Template, we get the following answers:
a) 0.1993
b) 0.7651
c) 0.4955
d) 0.5382
e) 30
f) 0.27859
3-39. Using the Binomial Template, we get the following answers:
a) 0.8889
b) 11
c) 0.55
3-40. Using the Binomial Template, we get the following answers:
a) 0.5769
b) I) 27
ii) 37
iii) 0.9576
3-41.
a) 0.9981
b) 0.889, 0.935
c) 4, 5
d) increase the reliability of each engine.
3-42.
a) Poisson
b) negative binomial
c) hypergeometric
d) Poisson
e) exponential
f) Poisson
g) binomial
h) negative binomial
i) binomial
j) Poisson
k) exponential
l) uniform
m) hypergeometric
n) binomial
o) Poisson
p) exponential
q) uniform
3-43.
a)
Negative Binomial Distributions / p / Mean / Variance / Stdev.
3 / 0.4800 / 6.25 / 6.77083 / 2.602082
b) 61.80%
c) 11
x / P(Exactly x) / P(At most x) / P(At least x)3 / 0.1106 / 0.1106 / 1.0000
4 / 0.1725 / 0.2831 / 0.8894
5 / 0.1794 / 0.4625 / 0.7169
6 / 0.1555 / 0.6180 / 0.5375
7 / 0.1213 / 0.7393 / 0.3820
8 / 0.0883 / 0.8276 / 0.2607
9 / 0.0612 / 0.8889 / 0.1724
10 / 0.0409 / 0.9298 / 0.1111
11 / 0.0266 / 0.9564 / 0.0702
d) p= 0.7272 Setting D9 to 0.95 by changing cell C3
s / p3 / 0.7272
x / P(Exactly x) / P(At most x) / P(At least x)
3 / 0.3846 / 0.3846 / 1.0000
4 / 0.3147 / 0.6993 / 0.6154
5 / 0.1717 / 0.8710 / 0.3007
6 / 0.0781 / 0.9491 / 0.1290
Negative Binomial Distribution
s / p
4 / 0.2400
x / P(Exactly x) / P(At most x) / P(At least x)
4 / 0.0033 / 0.0033 / 1.0000
5 / 0.0101 / 0.0134 / 0.9967
6 / 0.0192 / 0.0326 / 0.9866
7 / 0.0291 / 0.0617 / 0.9674
8 / 0.0387 / 0.1004 / 0.9383
9 / 0.0471 / 0.1475 / 0.8996
10 / 0.0537 / 0.2012 / 0.8525
11 / 0.0583 / 0.2596 / 0.7988
12 / 0.0609 / 0.3205 / 0.7404
13 / 0.0617 / 0.3822 / 0.6795
14 / 0.0610 / 0.4432 / 0.6178
15 / 0.0590 / 0.5022 / 0.5568
16 / 0.0561 / 0.5583 / 0.4978
17 / 0.0524 / 0.6107 / 0.4417
18 / 0.0484 / 0.6591 / 0.3893
19 / 0.0441 / 0.7032 / 0.3409
20 / 0.0398 / 0.7431 / 0.2968
a) 0.5022
b) 19
c) 0.2987
s / p4 / 0.2987
x / P(Exactly x) / P(At most x) / P(At least x)
4 / 0.0080 / 0.0080 / 1.0000
5 / 0.0223 / 0.0303 / 0.9920
6 / 0.0391 / 0.0694 / 0.9697
7 / 0.0549 / 0.1243 / 0.9306
8 / 0.0674 / 0.1917 / 0.8757
9 / 0.0756 / 0.2673 / 0.8083
10 / 0.0795 / 0.3468 / 0.7327
11 / 0.0797 / 0.4265 / 0.6532
12 / 0.0768 / 0.5034 / 0.5735
13 / 0.0719 / 0.5752 / 0.4966
14 / 0.0655 / 0.6407 / 0.4248
15 / 0.0585 / 0.6992 / 0.3593
16 / 0.0513 / 0.7505 / 0.3008
d)
4 / 0.2000 / 0.3518
0.2500 / 0.5387
0.3000 / 0.7031
0.3500 / 0.8273
0.4000 / 0.9095
0.4500 / 0.9576
0.5000 / 0.9824
0.5500 / 0.9937
0.6000 / 0.9981
3-45.
a) mean = 2.857, var = 5.306
Geometric Distributionp / Mean / Variance / Stdev.
0.35 / 2.857143 / 5.306122 / 2.303502
b) 82.15%
c) 7
X / P(Exactly x) / P(At most x) / P(At least x)1 / 0.3500 / 0.3500 / 1.0000
2 / 0.2275 / 0.5775 / 0.6500
3 / 0.1479 / 0.7254 / 0.4225
4 / 0.0961 / 0.8215 / 0.2746
5 / 0.0625 / 0.8840 / 0.1785
6 / 0.0406 / 0.9246 / 0.1160
7 / 0.0264 / 0.9510 / 0.0754
d) p = 0.5269 Setting cell D10 to 0.95
p0.526858267
x / P(Exactly x) / P(At most x) / P(At least x)
1 / 0.5269 / 0.5269 / 1.0000
2 / 0.2493 / 0.7761 / 0.4731
3 / 0.1179 / 0.8941 / 0.2239
4 / 0.0558 / 0.9499 / 0.1059
5 / 0.0264 / 0.9763 / 0.0501
a) 0.7863
Hypergeometric Distributionn / S / N
3 / 2 / 27
x / P(exactly x) / P(at most x) / P(al least x)
0 / 0.7863 / 0.7863 / 1.0000
1 / 0.2051 / 0.9915 / 0.2137
2 / 0.0085 / 1.0000 / 0.0085
b) add 32 good pins to be assured of at least 90% acceptance