Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean

Question

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 38 hours. hours and a standard deviation of 5.4 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 9 batteries.

a.) What can you say about the shape of the distribution of the sample mean? Sample Mean:_____?

b.) What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.) Standard Error: ______?

c.) What proportion of the samples will have a mean useful life of more than 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability:_____?

d.) What proportion of the sample will have a mean useful life greater than 36.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability:_______?

e.) What proportion of the sample will have a mean useful life between 36.5 and 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probabilityt:_______?

a.) What can you say about the shape of the distribution of the sample mean? Sample Mean:_____?

b.) What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.) Standard Error: ______?

c.) What proportion of the samples will have a mean useful life of more than 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability:_____?

d.) What proportion of the sample will have a mean useful life greater than 36.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability:_______?

e.) What proportion of the sample will have a mean useful life between 36.5 and 39 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probabilityt:_______?

Explanation / Answer

a) sample mean =38

b)Standard Error:=std deviaiton/(n)1/2=1.8

c)P(X>39)=1-P(Z<(39-38)/1.8)=1-P(Z<0.5556)=1-0.7107=0.2893

d)P(X>36.5)=1-P(Z<(36.5-38)/1.8)=1-P(Z<-0.833)=1-0.2023=0.7977

e)P(36.5<X<39)=P(-0.8333<Z<0.5556)=0.7107-0.2023=0.5084

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.