EBK STATISTICAL TECHNIQUES IN BUSINESS
EBK STATISTICAL TECHNIQUES IN BUSINESS
17th Edition
ISBN: 9781259924163
Author: Lind
Publisher: MCGRAW HILL BOOK COMPANY
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Question
Chapter 6, Problem 52CE

a.

To determine

Create a probability distribution for the number of students in the sample who watch soap operas.

a.

Expert Solution
Check Mark

Answer to Problem 52CE

The probability distribution for the number of students in the sample who watch soap operas is as follows:

Number of studentsProbability
00.0025
10.0207
20.0763
30.1665
40.2384
50.2340
60.1596
70.0746
80.0229
90.0042
100.0003

Explanation of Solution

The formula to find the binomial probability is as follows:

P(X)=Cnxπx(1π)(nx)where, C is the combination.n is the number of trials.X is the random variable.π is the probability of success.

Let x denotes the students who watch soap operas.

Here, n=10; π=0.45

For the number of students x=0, the probability is calculated as follows:

P(0)=C100(0.45)0(10.45)(100)=10!0!(100)!(0.45)0(0.55)10=1×1×0.0025=0.0025

Similarly the probability value for the remaining x values is calculated.

Thus, the probability distribution for the number of students in the sample who watch soap operas is obtained as follows:

Number of studentsProbability
0C100(0.45)0(10.45)(100)=0.0025
1C101(0.45)1(10.45)(101)=0.0207
2C102(0.45)2(10.45)(102)=0.0763
3C103(0.45)3(10.45)(103)=0.1665
4C104(0.45)4(10.45)(104)=0.2384
5C105(0.45)5(10.45)(105)=0.2340
6C106(0.45)6(10.45)(106)=0.1596
7C107(0.45)7(10.45)(107)=0.0746
8C108(0.45)8(10.45)(108)=0.0229
9C109(0.45)9(10.45)(109)=0.0042
10C1010(0.45)10(10.45)(1010)=0.0003

b.

To determine

Compute the mean and standard deviation of this distribution.

b.

Expert Solution
Check Mark

Answer to Problem 52CE

The mean is 4.5.

The standard deviation is 1.5732.

Explanation of Solution

The mean and standard deviation of the binomial distribution are as follows:

μ=nπσ=nπ(1π)

The mean of the distribution is calculated as follows:

μ=10×0.45=4.5

Therefore, the mean of the distribution is 4.5.

The standard deviation of the distribution is calculated as follows:

σ=10×0.45×(10.45)=2.475=1.5732

Therefore, the standard deviation of the distribution is 1.5732.

c.

To determine

Compute the probability of finding exactly four students who watch soap operas.

c.

Expert Solution
Check Mark

Answer to Problem 52CE

The probability of finding exactly four students who watch soap operas is 0.2384.

Explanation of Solution

The probability of finding exactly four students who watch soap operas is calculated as follows:

P(x=4)=C104(0.45)4(10.45)(104)=10!4!(104)!×0.454×0.556=210×0.0410×0.0277=0.2384

Therefore, the probability of finding exactly four students who watch soap operas is 0.2384.

d.

To determine

Compute the probability that less than half of the students selected watch soap operas.

d.

Expert Solution
Check Mark

Answer to Problem 52CE

The probability that less than half of the students selected watch soap operas is 0.5044.

Explanation of Solution

The probability that less than half of the students selected watch soap operas is calculated as follows:

P(x<5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)={[C100(0.45)0(10.45)(100)]+[C101(0.45)1(10.45)(101)]+[C102(0.45)2(10.45)(102)]+[C103(0.45)3(10.45)(103)]+[C104(0.45)4(10.45)(104)]}={[10!0!(100)!×0.450×0.5510]+[10!1!(101)!×0.451×0.559]+[10!2!(102)!×0.452×0.558]+[10!3!(103)!×0.453×0.557]+[10!4!(104)!×0.454×0.556]}=(0.0025+0.0207+0.0763+0.1665+0.2384)=0.5044

Therefore, the probability that less than half of the students selected watch soap operas is 0.5044.

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Chapter 6 Solutions

EBK STATISTICAL TECHNIQUES IN BUSINESS

Ch. 6 - Ninety-five percent of the employees at the J. M....Ch. 6 - In a binomial situation, n = 4 and = .25....Ch. 6 - In a binomial situation, n = 5 and = .40....Ch. 6 - Prob. 11ECh. 6 - Assume a binomial distribution where n = 5 and =...Ch. 6 - An American Society of Investors survey found 30%...Ch. 6 - Prob. 14ECh. 6 - Prob. 15ECh. 6 - FILE A telemarketer makes six phone calls per hour...Ch. 6 - FILE A recent survey by the American Accounting...Ch. 6 - Prob. 18ECh. 6 - Prob. 4SRCh. 6 - Prob. 19ECh. 6 - In a binomial distribution, n = 12 and = .60....Ch. 6 - FILE In a recent study, 90% of the homes in the...Ch. 6 - FILE A manufacturer of window frames knows from...Ch. 6 - Prob. 23ECh. 6 - Prob. 24ECh. 6 - Prob. 5SRCh. 6 - Prob. 25ECh. 6 - A population consists of 15 items, 10 of which are...Ch. 6 - Prob. 27ECh. 6 - Prob. 28ECh. 6 - Prob. 29ECh. 6 - Prob. 30ECh. 6 - Prob. 6SRCh. 6 - Prob. 31ECh. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Automobiles arrive at the Elkhart exit of the...Ch. 6 - It is estimated that 0.5% of the callers to the...Ch. 6 - Prob. 36ECh. 6 - Prob. 37CECh. 6 - For each of the following indicate whether the...Ch. 6 - Prob. 39CECh. 6 - Prob. 40CECh. 6 - Prob. 41CECh. 6 - The payouts for the Powerball lottery and their...Ch. 6 - In a recent study, 35% of people surveyed...Ch. 6 - Prob. 44CECh. 6 - An auditor for Health Maintenance Services of...Ch. 6 - Prob. 46CECh. 6 - Prob. 47CECh. 6 - The Bank of Hawaii reports that 7% of its credit...Ch. 6 - Prob. 49CECh. 6 - Prob. 50CECh. 6 - Prob. 51CECh. 6 - Prob. 52CECh. 6 - Prob. 53CECh. 6 - Prob. 54CECh. 6 - Prob. 55CECh. 6 - Prob. 56CECh. 6 - Prob. 57CECh. 6 - Prob. 58CECh. 6 - Prob. 59CECh. 6 - Prob. 60CECh. 6 - Prob. 61CECh. 6 - Prob. 62CECh. 6 - Prob. 63CECh. 6 - Prob. 64CECh. 6 - The National Aeronautics and Space Administration...Ch. 6 - Prob. 66CECh. 6 - Prob. 67CECh. 6 - Prob. 68CECh. 6 - A recent CBS News survey reported that 67% of...Ch. 6 - Prob. 70DACh. 6 - Prob. 71DA
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