The engineering staff for the Bentim Manufacturing Company is considering sampling 30 bolts produced by a supplier for the company to determine the average diameter.The population is 200 bolts, the population standard deviation is known to be 0.015 inches,and the output measures are normally distributed.
a) Determine the standard error of the sampling distribution
b)The staff is troubled by bolts that do not fit the parts Bentrim is producing.Therefore,they wish to develop a method of determining when the bolts are too narrow or broad to fit the parts they are producing.If the mean diameter of bolt population is 0.75 inches,determine the largest average diameter a sample of 30 bolts should produce.
c)The staff has concluded that if the average they calculate from their sample is larger that the 95th percentile of the average bolts width,they will return the shipment of bolts and seek another supplier of bolts.What will be the minimum average width that will cause Bentrim to return the shipment?

2. (a)

$SE = \frac{\sigma}{\sqrt{n}}$
$SE$ is the standard error, which is the standard deviation of the sample mean. (Because samples are taken randomly, the sample mean is also a random variable, so it has a standard deviation)
$\sigma$ is the standard deviation of your population
$n$ is the number of data points in your sample

(b)
EDIT: corrected 2nd term
Sample means from a normal distribution are normally distributed. Using the information above, your sample mean follows a normal distribution $N \left( 0.75,\left(\frac{0.015}{\sqrt{30}}\right)^2 \right)$

Try and use this information to answer parts (b) and (c)

3. OMG THANK YOU VERY MUCH!!!!!!!!!!!!!!!!!!!!!!!

4. Lol i cant solve it(((

5. ok, heres some help with (b)

(b) "determine the largest average diameter a sample of 30 bolts should produce."

the question doesn't tell us how to interpret "largest value that should be produced". Ill interpret it as asking us to find the 99th percentile of the distribution (as 99% of sample means should be below this value).

If X is our sample mean. We know that $X \sim N(0.015,SE^2)$ .This is slightly different from what i told you earlier, i forgot the square.

We want to find the value K such that
$P(X

This is pretty hard Unless you have a computer. We can make it easier by standardising the variable. Standardising in the usual way: $\frac{X-0.015}{SE} \sim N(0,1)$

Aside:
You can lookup in your statistical tables that for any standard normal variable Z
$P(Z<2.3263) = 0.99$

Now, consider:
$P(\frac{X-0.015}{SE}

We know $\frac{X-0.015}{SE}$ is a standard normal variable, So
$P(\frac{X-0.015}{SE}<2.3263) = 0.99$

$P(X < (SE*2.3263) +0.015) = 0.99$

So the highest value of X you would expect is
$(SE*2.3263) +0.015$

part (c) is almost the same except you want the 95th percentile. The 95th percentile of the standard normal distribution is 1.6449

6. A deadline suggests that this question is part of work that counts towards your final grade. MHF policy is to not knowingly help with such work. It's meant to be the student's own work, not the work of others.