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Dual Enrollment Statistics
Chapter 5 Review
Do all work on a SEPARATE SHEET!!!!
1. A marketing survey compiled data on the number of personal computers in households. If X = the number of
computers in a randomly-selected household, and we omit the rare cases of more than 5 computers, then X has
the following distribution:
X
P(X)
0
0.24
1
0.37
2
0.20
3
0.11
4
0.05
5
0.03
What is the probability that a randomly chosen household has at least two personal computers?
A. 0.19
B.0.20
C.0.29
D.0.39
E.0.61
2. Determine whether the random variable described satisfies the conditions for a binomial setting. Support your
conclusion. A high school principal goes to 10 different classrooms and randomly selects one student from each
class. X = the number of female students in his group of 10 students.
3. A manufacturer produces a large number of toasters. From past experience, the manufacturer knows that
approximately 2% are defective. In a quality control procedure, we randomly select 20 toasters for testing.
(a) Determine the probability that exactly one of the toasters is defective.
(b) Find the probability that at most two of the toasters are defective.
(c) Let X = the number of defective toasters in the sample of 20. Find the mean and standard deviation of X.
(d) What is the range “normal values” for the distribution?
(e) If 3 toasters were found to be defective would that be unusual? Explain.
(f) In order for a set of 20 toasters to be rejected, at least 2 must be defective. Find the probability that a set will
not be rejected
4. Determine whether the random variable described satisfies the conditions for a binomial setting. Support your
conclusion. Joey buys a Virginia lottery ticket every week. X is the number of times in a year that he wins a
prize.
5. A company that rents DVDs from vending machines in grocery stores has developed the following probability
distribution for the random variable X = the number of DVDs a customer rents per visit to a machine.
X
P(X)
1
0.5
2
0.3
3
0.1
4
0.1
(a) Find and interpret the mean (expected value) of X. (by hand)
(b) Find and interpret the standard deviation of X. (by hand)
Ch 5 Rev
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6. Picard Partners is planning a major investment. The amount of profit X is uncertain but a probabilistic estimate
gives the following distribution (in millions of dollars):
X
P(X)
1
0.2
2
0.5
4
0.2
10
0.1
(a) Find an interpret the mean (expected value) of X. (with calculator)
(b) Find and interpret the standard deviation of X. (with calculator)
7. Suppose there are three balls in a box. On one of the balls is the number 1, on another is the number 2, and on the
third is the number 3. You select two balls at random and without replacement from the box and note the two
numbers observed. Let X be the sum of the numbers on two balls selected. Construct a table for the
probability distribution for X, then find the mean and standard deviation.
8. Is this a probability distribution? Why or why not?
X
P(X)
0
0.24
1
0.37
2
0.20
3
0.11
4
0.05
5
0.03
9. Is this a probability distribution? Why or why not?
X
P(X)
0
-0.24
1
0.44
2
0.60
3
0.12
4
0.05
5
0.03
10. Is this a probability distribution? Why or why not?
X
P(X)
0
0.24
1
0.04
2
0.60
3
0.02
4
0.05
5
0.04
11. In New Jersey’s Pick 4 lottery game, you pay 50¢ to select a sequence of 4 digits, such as 1332. If you select
the same sequence of four digits that are drawn, you win and collect $2788.
a. How many different selections are possible?
b. What is the probability of winning?
c. If you win, what is your net profit?
d. Find the expected value.
e. If you bet 50¢ in Illinois’ Pick 4 game the expected value is -25¢. Which bet is better: A
50¢ bet in the Illinois Pick 4 game or a 50¢ bet in New Jersey’s Pick 4 game? Explain.
12. What are the two requirements for a probability distribution?
13. What are the four requirements for a binomial distribution?
14. A fair coin is flipped 20 times.
(a) Determine the probability that the coin comes up tails exactly 15 times.
(b) Let X = the number of tails in the 20 flips. Find the mean and standard deviation of X.
Ch 5 Rev
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Revised 3/5/2015
Dual Enrollment Statistics
Answer Section
Chapter 5 Review
1. D
2. This is not binomial , because the probability of success (selecting a female student) is probably different in
each trial, unless every classroom has exactly the same proportion of females, which is unlikely.
3. (a) 0.272
(b) 0.993
(c) 𝜇 = 0.4 toasters, 𝜎 = 0.6 toaster
(d) -.9 to 1.7 toasters
(e) Yes, it would be unusual. 3 is larger than 1.66 which is the maximum “normal value” ; therefore, it is
considered unusual. Also, the probability of getting 3 defective toasters p(3) is .00647 which is less than 0.05.
(f ) 0.940
4. This is a binomial setting: Two outcomes (Joey wins or loses), Independent trials (whether he wins this week
does not influence whether he wins next week), Number of trials is fixed at 52 (weeks), and the probability of
Success—however minuscule—does not change.
; The average number of DVD’s per customer visit,
5. (a)
over the long run, is 1.8.
(b)
; the long-run average
distance between how many DVD’s are rented per visit and the mean is about 1.0 DVD’s.
6. (a)
the average profit would be 3 million dollars.
; If this kind of investment was made many times,
, 𝜎 = $2.5 million
(b)
7.
X
Prob
3
4
5
= 4
= 0.8
8. yes, because the sum of the probabilities is 1 and all probabilities are between 0 and 1 inclusive ( 0 < p < 1)
9. No, because all probabilities are not between 0 and 1 inclusive ( 0 < p < 1). One of the probabilities is -0.24.
10. Yes, because all probabilities are between 0 and 1 inclusive ( 0 < p < 1). Even though the sum of the
probabilities is 0.99 instead of 1, it is still allowable due to potential rounding error.
11. a. 10,000
b. 0.0001
c. $2787.50
d. -22.12¢ or -22¢
e. The 50¢ bet in New Jersey’s Pick 4 game is better, because it has a larger expected value (-22¢ is better than
-25¢).
12. 1. The probabilities must be between 0 and 1 inclusive (0 < p < 1)
2. The sum of the probabilities must = 1. (1.001 or .999 is also acceptable)
13. 1. There is a fixed number of trials.
2. The trials must be independent.
3. Each trial must have only 2 outcomes (success or failure)
4. The probability of a success remains the same in all trials.
14. (a) 0.0148
(b) 𝜇 = 10 tails, 𝜎 = 2.2 tails
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