Time taken by a randomly selected applicant for a mortgage to fill out a certain

Question

Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 4 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day Is at most 11 min? (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question. Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.1. (Round your answers to four decimal places.) If the distribution is normal, what Is the probability that the sample mean hardness for a random sample of 10 pins is at least 51? What is the (approximate) probability that the sample mean hardness for a random sample of 38 pins is at least 51? You may need to use the appropriate table in the Appendix of table to answer this question.

Explanation / Answer

Using central limit theorem the sampling distribution of sampling mean is also normal with mean is 10 min and standard deviation is 4/sq rt 5

=1.79

P(X is at most 11)

=P(X<=11)

For X bar=11, z=(x bar-mu)/sigma=(11-10)/1.79=0.56

Thus, P(X bar<=11)

=P(z<=0.56)

=0.7123

Using central limit theorem the sampling distribution of sampling mean is also normal with mean is 50 and standard deviation is 1.1/sq rt 10

=0.35

(a)X bar=51, z=(51-50)/0.35=2.85

P(X bar>=51)=P(X bar>=2.85)

=1-P(X bar<=2.85)

=1-0.9978

=0.0022

Using central limit theorem the sampling distribution of sampling mean is also normal with mean is 50 and standard deviation is 1.1/sq rt 38

=0.18

(b)For X bar=51, z=(51-50)/0.18=5.56

P(X bar>=51)=P(X bar>=5.56)

=1-P(X bar<=5.56)

=1-0.9999

=0.0001

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