statistics question - mean and stardard deviation?

Question

Originally, rivet breaking strengths are normally distributed with mean of 2500lbs, and standard deviation of 150lbs.

Suppose 100 rivets of the type are selected.

-Calculate the probability that 10 or more of the rivets will have a tensile strength exceeding 2725 pounds.

-Calculate the probability that the average tensile strength of all the rivets tested will exceed 2540 pounds.

Thanks for your help!!

Answer 1

I can tell that nobody wants to just do this problem. There's nothing interesting about it. It's a typical problem that a stat student does to show understanding. Get out your textbook and study up how to solve it.

Answer 2

rivet breaking strengths are normally distributed with mean of 2500lbs, and standard deviation of 150lbs.

Suppose 100 rivets of the type are selected

A. the probability that 10 or more of the rivets will have a tensile strength exceeding 2725 pounds

We need to find P(x>2725) = 1-P(0 = 1-0.0668
= 0.933193.

B. the probability that the average tensile strength of all the rivets tested will exceed 2540 pounds.
We need to find P(meanx>2540)=P(x>2540*sigma*sigma/mean)
=P(x>0.0610)
[This x is a standard normal variable]
=0.475680.

Thats your answer.

Right now I am busy, but I will try to come back to your question.

An apology for errors

All the best.