# sample means...need help

• Jul 3rd 2008, 02:10 PM
peanutbutter
sample means...need help
Distances between students' homes and school observe normal distribution with mean of 4.76 miles and standard deviation of 1.74 miles. Suppose all examples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

Z=Mean-mean of sample means/ std of sample means, so i have

Z=6.78-4.76/(1.74/sqrt[12])=4.02

I know the next step is to look for a percentage in the table that corresponds to this zscore, but I can't find 4.02 in the table? Can someone talk me through what I *should* be looking for? step by step?
• Jul 3rd 2008, 03:43 PM
mr fantastic
Quote:

Originally Posted by peanutbutter
Distances between students' homes and school observe normal distribution with mean of 4.76 miles and standard deviation of 1.74 miles. Suppose all examples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

Z=Mean-mean of sample means/ std of sample means, so i have

Z=6.78-4.76/(1.74/sqrt[12])=4.02

I know the next step is to look for a percentage in the table that corresponds to this zscore, but I can't find 4.02 in the table? Can someone talk me through what I *should* be looking for? step by step?

The sample mean is a random variable and follows a normal distribution.

In fact, if you let $\bar{X}$ represent the random variable mean of the sample, then

$\bar{X}$ ~ Normal $\left( \mu = 4.76, ~ \sigma = \frac{1.74}{\sqrt{12}} \approx 0.5\right)$.

You first need to find $\Pr \left(\bar{X} > 6.78\right)$:

$Z = \frac{6.78 - 4.76}{0.5} = 4.04$.

This z-value is really, really big (more than 4 standard deviations from the mean) so it's no surprise that you can't find it in your tables. $\Pr(Z > 4.04) = 0$, correct to four decimal places.

Then you multiply the answer by 100 to convert into a percentage: 0%.