Question

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 501 psi. (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,900 and 10,200

Answer 1

`z=(9900-10000)/((501)/(√(40)))= -1.262`

`z=(10200-10000)/((501)/(√(40)))= 2.525`

And we can find the probability with this difference and using the normal standard distribution table and we got:

`P(-1.262 < z< 2.525) =P(Z<2.525) -P(Z<-1.262) = 0.994-0.103= 0.891`

Explanation:

For this problem we have the following info:

`\mu = 10000` represent the true mean

`\sigma = 501` represent the deviation

`n = 40` representthe sample size selected

We want to find the following probability:

`P(9900 <\bar X <10200)`

And for this case since the sample size is large enough we can use the central limit theorem and then we can use the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

Andreplacing we got:

`z=(9900-10000)/((501)/(√(40)))= -1.262`

`z=(10200-10000)/((501)/(√(40)))= 2.525`

And we can find the probability with this difference and using the normal standard distribution table and we got:

`P(-1.262 < z< 2.525) =P(Z<2.525) -P(Z<-1.262) = 0.994-0.103= 0.891`