Random Variable/Prob. Dist. Worksheet Answers
1.
The following data show 𝑥 = the number of fish caught in a 6 – hour period on a
given lake. The percentage data are the percentages of fisherman who catch 𝑥 fish in
a 6 – hour period from shore.
𝑥
%
𝑃(𝑥)
𝑥 ∙ 𝑃(𝑥)
𝑥 − 𝑥̅
(𝑥 − 𝑥̅ )2
(𝑥 − ̅̅̅
𝑥)2 ∙ 𝑃(𝑥)
0
44%
0.44
0
−0.82
0.6724
0.297
1
36%
0.36
0.36
0.18
0.0324
0.012
2
15%
0.15
0.30
1.18
1.3924
0.209
3
4%
0.04
0.12
2.18
4.7524
0.190
≥4
1%
0.01
0.04
3.18
10.1124
0.101
∑ = 0.82
∑ = 0.809
a) Convert the percentages into probabilities P(x) and create a histogram
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
b) Find the probability that a fisherman selected at random fishing from shore will
catch one or more fish in a 6 – hour period.
𝑃(1 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑓𝑖𝑠ℎ) = 0.36 + 0.15 + 0.04 + 0.01 = 0.56
c) Compute μ, the expected value of the number of fish caught per fisherman in a 6 –
hour period ( round ≥ 4 to just 4 )
𝜇 = 0.82
d) Compute σ, the standard deviation of the number of fish caught per fisherman in
a 6 – hour period ( round ≥ 4 to just 4 )
𝜎 = 0.809
2. The college hiking club is having a fundraiser to buy new equipment for the fall and
winter outings. The club is selling fortune cookies at a price of $1 per cookie. Each
cookie contains a piece of paper with a number written on it. A random drawing will
determine which number is the winner of a dinner for two at a local Chinese
restaurant valued at $35. The fortune cookies were donated by the restaurant so we
can ignore the cost of the cookies. The club sold 719 cookies.
a) If Lisa buys 15 cookies, what is her probability of winning the free dinner ?
𝑃(𝑤𝑖𝑛𝑛𝑖𝑛𝑔) =
15
= 0.021
719
b) Lisa’s expected earnings can be found by multiplying the value of the dinner by
the probability that she will win. What are Lisa’s expected earnings.
Expected earning 0.021 × $35 = $0.74
3. Jim is a 60 – year old Anglo male in reasonably good health. He wants to take out a
$50,000 term life insurance policy until he is 65. The policy will expire on his 65th
birthday. The probability of death in a given year is given by the Vital Statistics
Section of the Statistical Abstract of the United States ( 116th Ed. )
𝑥 = age
60
61
62
63
64
𝑃(𝑑𝑒𝑎𝑡ℎ 𝑎𝑡 𝑡ℎ𝑖𝑠 𝑎𝑔𝑒)
0.01191
0.01292
0.01396
0.01503
0.01613
a) What is the probability that Jim will die in his 60th year ?
𝑃(𝑑𝑖𝑒 𝑖𝑛 60𝑡ℎ 𝑦𝑟) = 0.01191
b) Using this probability and the $50,000 death benefit, what is the expected cost to
Big Rock Insurance ?
Expected cost = $50000 × 0.01191 = $595.50
c) Repeat parts a) and b) for years 61 thru 64.
P(death in year 61) = 0.01292
Expected cost = $646.00
P(death in year 62) = 0.01396
Expected cost = $698.00
P(death in year 63) = 0.01503
Expected cost = $751.50
P(death in year 64) = 0.01613
Expected cost = $806.50
d) What is the total expected cost ?
Total expected cost = $595.50+$646.00+$698.00+$751.50+$806.50
Total expected cost = $3497.50
e) If Big Rock Insurance wants to make a $700 profit above the expected total cost
paid out for Jim’s death, how much should it charge Jim for the policy ?
$3497.50 + $700 = $4197.50
4.
Ron and Gary are entered in a local golf tournament. Both have played the local
course many times. Their scores are random variables with the following means and
standard deviations,
Ron, 𝑥1 ∶ 𝜇1 = 115; 𝜎1 = 12
Gary, 𝑥2 ∶ 𝜇2 = 100; 𝜎2 = 8
a) The difference between their scores is 𝑊 = 𝑥1 − 𝑥2 . Compute the mean, variance,
and standard deviation for the random variable 𝑊.
𝑎 = 1, 𝑏 = −1
𝜇 = 𝑎𝜇1 + 𝑏𝜇2 = 1(115) + (−1)(100) = 115 − 100 = 15
2
𝜎𝑊
= 𝑎2 𝜎12 + 𝑏 2 𝜎22 = (1)2 (12)2 + (−1)2 (8)2 = 144 + 64 = 208
2
𝜎𝑊 = √𝜎𝑊
= √208 = 14.42
b) The average of their scores is = 0.5𝑥1 + 0.5𝑥2 . Compute the mean, variance, and
standard deviation for the random variable 𝑊.
𝑎 = 0.5 , 𝑏 = 0.5
𝜇 = 𝑎𝜇1 + 𝑏𝜇2 = 0.5(115) + (−0.5)(100) = 57.5 + 50 = 107.5
2
𝜎𝑊
= 𝑎2 𝜎12 + 𝑏 2 𝜎22 = (0.5)2 (12)2 + (0.5)2 (8)2 = 36 + 16 = 52
2
𝜎𝑊 = √𝜎𝑊
= √52 = 7.2
c) The tournament rules have a special handicap system for each player. For Ron,
the handicap formula is 𝐿 = 0.8𝑥1 − 2. Compute the mean, variance, and
standard deviation for the random variable 𝐿.
𝑎 = −2, 𝑏 = 0.8
𝜇𝐿 = 𝑎 + 𝑏𝜇 = −2 + 0.8(115) = −2 + 92 = 90
𝜎𝐿2 = 𝑏 2 𝜎 2 = (0.8)2 (12)2 = (0.64)(144) = 92.16
𝜎𝐿 = √𝜎𝐿2 = √92.16 = 9.6
d) For Gary, the handicap formula is 𝐿 = 0.95𝑥2 − 5. Compute the mean, variance,
and standard deviation for the random variable 𝐿.
𝑎 = −5 , 𝑏 = 0.95
𝜇𝐿 = 𝑎 + 𝑏𝜇 = −5 + 0.95(100) = −5 + 95 = 90
𝜎𝐿2 = 𝑏 2 𝜎 2 = (0.95)2 (8)2 = (0.9025)(64) = 57.76
𝜎𝐿 = √𝜎𝐿2 = √57.76 = 7.6