the breaking strength of a rivets a metal pin has a mean value of 10.000 psi a s

Question

the breaking strength of a rivets a metal pin has a mean value of 10.000 psi a standard deviation of 500 psi.A random sample of 40 rivets is selected from a production line y=1000 ,n=40

A) what is the approximate distribution of the sample mean y?

my answer is -----> normal distribution

B) what is the standard deviation of y?

my answer is -----> Q/sqrt 40 = 500/sqrt40 =79.056

C) what is the mean of y?

my answer is -----> 10000 psi

D) What theorem you used to justify your answer to A) ,B) and C) ?

my answer is -----> Central limit theorem

F) what is the probability that the sample mean breaking strength for the 40 rivets is less than 10,079.1psi

I NEED ANSWER FOR THIS QUESTION PLSE

Explanation / Answer

P = 0.5629, by applying Normal Distribution..

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