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Concept explainers
a.
Find the number of problems that is expected to be resolved today.
Find the standard deviation.
a.
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Answer to Problem 23E
The number of problems that is expected to be resolved today is 10.5.
The standard deviation is 1.7748.
Explanation of Solution
Here, n=15; π=0.70.
The expected number of problems to be resolved today is calculated as follows:
μ=nπ=15×0.7=10.5
Therefore, the expected number of problems to be resolved today is 10.5.
The standard deviation is calculated as follows:
σ=√nπ(1−π)=√15×0.7×(1−0.7)=√3.15=1.7748
Therefore, the standard deviation is 1.7748.
b.
Compute the
b.
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Answer to Problem 23E
The probability that 10 of the problems can be resolved today is 0.2061.
Explanation of Solution
The formula to find the binomial probability is as follows:
P(X)=Cnxπx(1−π)(n−x)where, C is the combination.n is the number of trials.X is the random variable.π is the probability of success.
The probability that 10 of the problems can be resolved today is calculated as follows:
P(x=10)=C1510(0.70)10(1−0.70)(15−10)=15!10!(15−10)!×0.7010×0.305=3003×0.0282×0.0024=0.2061
Therefore, the probability that 10 of the problems can be resolved today is 0.2061.
c.
Compute the probability that 10 or 11 of the problems can be resolved today.
c.
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Answer to Problem 23E
The probability that 10 or 11 of the problems can be resolved today is 0.4247.
Explanation of Solution
The probability that 10 or 11 of the problems can be resolved today is calculated as follows:
P(x=10)+P(x=11)=(C1510(0.70)10(1−0.70)(15−10))+(C1511(0.70)11(1−0.70)(15−11))=(15!10!(15−10)!×0.7010×0.305)+(15!11!(15−11)!×0.7011×0.304)=(3,003×0.0282×0.0024)+(1,365×0.0198×0.0081)=0.2061+0.2186=0.4247
Therefore, the probability that 10 or 11 of the problems can be resolved today is 0.4247.
d.
Compute the probability that more than 10 of the problems can be resolved today.
d.
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Answer to Problem 23E
The probability that more than 10 of the problems can be resolved today is 0.5154.
Explanation of Solution
The probability that more than 10 of the problems can be resolved today is calculated as follows:
P(x>10)=P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)={(C1511(0.70)11(1−0.70)(15−11))+(C1512(0.70)12(1−0.70)(15−12))+(C1513(0.70)13(1−0.70)(15−13))+(C1514(0.70)14(1−0.70)(15−14))+(C1515(0.70)15(1−0.70)(15−15))}=[(15!11!(15−11)!×0.7011×0.304)+(15!12!(15−12)!×0.7012×0.303)+(15!13!(15−13)!×0.7013×0.302)+(15!14!(15−14)!×0.7014×0.301)+(15!15!(15−15)!×0.7015×0.300)]=0.2186+0.1700+0.0916+0.0305+0.0047=0.5154
Therefore, the probability that more than 10 of the problems can be resolved today is 0.5154.
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Chapter 6 Solutions
EBK STATISTICAL TECHNIQUES IN BUSINESS
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