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Chapter 5 Review (w/key)
1) a. With four flips of a balanced coin, there are ____ likely possibilities. Name
them all below:
TTTT
TTTH TTHT THTT HTTT
HHTT HTHT HTTH THTH TTHH THHT
HHHT HHTH HTHH THHH
HHHH
b. Is this a probability distribution? Why?
Yes; All probabilities are between 0 and 1, and the probabilities of the
sample space adds up to 1.
c. Fill in the columns with the appropriate headings.
# of Heads=L1 Probability=L2
L3=L1*L2
L1-squared
0
1/16
0
0
1
4/16
.25
1
2
6/16
.75
4
3
4/16
.75
9
4
1/
.25
16
L5=L4*L2
0
.25
1.5
2.25
1
d. Find the mean. Sum L3 = 2
e. Find the variance. 5-(2^2)=1
f. Find the standard deviation. Square root of e = 1.
g. You are going to flip the coin four times, and want to know the probability of
getting two heads. Is this a binomial distribution? Why or why not?
Yes, because there are 2 outcomes (heads/tails), a set number of trials (4),
Flips are independent, and the probability of getting a head is set for every
trial at .5.
2) Check whether the following probability distribution given by
x3
P ( x) 
for x = 1, 2, 3
15
Is a probability distribution of some random variable.
X = L1
1
2
3
Probability=L2
4/15
5/15
3/15
L3=L1*L2
L1-squared
L5=L4*L2
No, if you sum up the probabilities you get 12/15 which is not equal to 1;
therefore, not a probability distribution.
3) If the probability is 0.70 that any one registered voter (randomly selected form
official roles) will vote in a given election, what is the probability that two of five
registered voters will vote in the election?
Find P(X=2): Check 2NIP; its binomial. Use the binomial formula to prove that the
probability is .1323 (binompdf(5, .7, 2)).
4) The probability is 0.30 that a person shopping at a certain supermarket will take
advantage of its special promotion of ice cream.
a. Find the probabilities that among six persons shopping at this market there
will be 0, 1, 2, 3, 4, 5, or 6 who will take advantage of this promotion.
X=L1
Probability=L2
L3=L1*L2
L1-squared
L5=L4*L2
0
.1176
1
.3025
2
.3241
3
.18522
4
.0595
5
.0102
6
.0007
Binompdf(6,.3,0)=.1176, Binompdf(6,.3,1)=.3025, etc….
b. Find the mean number of persons, among six shopping at the market, who
will take advantage of the special promotion.
Mean = np = 6 (.3) = 1.8.
5) If the probability is 0.05 that there will be a serious accident at a certain
intersection on a weekday, what is the probability that there will be a serious
accident at that intersection on a least three of 20 workdays?
P(X is greater than or equal to 3) = 1-binomialcdf(20, .05, 2)=.0755
6) It is known that 2 percent of the books bound at a certain bindery have defective
bindings. Use the Poisson approximation to find the probability that 5 of 400
books bound by this bindery will have defective bindings.
Mean = 5/400 = .008. Using the Poisson probability formula, P(x=5) = 2.71 x 10^-13
Check with calculator: poissonpdf(.008, 5)
7) Records show that the probability is 0.00006 that a car tire will go flat while being
driven through a certain tunnel. Use the Poisson approximation to find the
probability that at least 2 of 10,000 cars passing through that tunnel will get flat
tires.
First, calculate the mean: .00006*10,000 = .6
P(X is greater than or equal to 2) = 1-P(X is less than or equal to 1) = 1Poissoncdf(.6, 1) = .1219
8) If the probabilities are 0.06, 0.21, 0.24, 0.18, 0.14, 0.10, 0.04, 0.02, 0.01 than at
airline office at a certain airport will receive 0, 1, 2, 3, 4, 5, 6, 7, or 8 complaints
per day about its luggage handling,
a. How many such complaints can it expect per day?
X
0
1
2
3
4
5
6
7
P(x)
.06
.21
.24
.18
.14
.1
.04
.02
Mean of a distribution = expected value = use formula 5.1. E(X) = 2.75.
8
.01
b. Find the mean of this probability distribution.
Same answer as a! Expected value = mean of a probability
distribution = 2.75.
c. Find the standard deviation. Use formula 5.4 to find it. Variance =
3.0275, so Standard Deviation = 1.74.
9) The probabilities are 0.22, 0.36, 0.28, and 0.14 that an investor will be able to sell
a piece of property at a profit of $2,500, at a profit of $1,500, at a profit of $500,
or at a loss of $500. What is the investor’s expected profit?
X
2500
1500
500
-500
P(X)
.22
.36
.28
.14
E(X) = Mean of a probability distribution = $1160.