Thread: Central Limit Theorem

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

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

Use the fact that:

1. The sample mean follows a normal distribution.

2. The difference of two normal random variables is also a normal random variable. (Use Google to get the required pdf).

Then find Pr(difference > 0) in the usual way.

Originally Posted by mr fantastic

2. The difference of two normal random variables is also a normal random variable. (Use Google to get the required pdf).

Google should not be needed the OP should know that the difference of two normal RVs is normal with mean equal to the difference of the means and variance equal to the sum of the variances.

CB

Ok I got it. I ended up with ((44.2)-(72-28))/(sqrt((100/64)+(25/100))) and I got a z value of .15 and using a table I got a probability of .5596. Thanks for the help.