Originally Posted by

**meks08999** Here is the problem:

The distribution of heights of a certain breed of terrier dogs has a mean height of 72 cm and a standard deviation of 10 cm, whereas the distribution of heights of a certain breed of poodles has a mean height of 28 cm with a stanndard deviation of 5 cm. Assuming that the sample means can be measured to any degree of accuracy, find the probability that the sample mean for a random sample of heights of 64 terriers exceeds the sample mean for a random sample of heights of 100 poodles by at most 44.2 cm.

The wording is slightly confusing.

Steps to take:

1. Use CLT to find distribution of mean height for the terriers.

2. Use CLT for poodles then

Not sure what comes nexts.

so CLT is Z = (X(bar) - mean (mu))/((s.d.)/n)

so it is (X(bar) - 72)/(10/sqrt(64)) for the terriers

and (X(bar) - 28)/(5/sqrt(100)) for the poodles.

But are both of those supposed to have an X(bar) in them still?

Then what do I do with those two CLT results to see get the probability that the sample mean of the terriers exceeds the poodles by 44.2 cm?

Thanks