# Central limit theorem question

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• Oct 18th 2008, 08:52 PM
lllll
Central limit theorem question
For a given exam the standard deviation for all seniors in a particular high school is 2.5. If we take a sample of 100 seniors, find the probability that the difference between the sample mean and the mean of all seniors does not exceed 0.5.

so far I have:

$P\left( \frac{\overline{Y}-X}{2.5 / \sqrt{100}} \leq \frac{0.5}{0.25} \right) \longrightarrow P\left( \frac{\overline{Y}-X}{0.25} \leq 2 \right)$

if I let $P(Z\leq 2) = 1- 0.228 = 0.9972$

but the answer in the back of the book is 0.9544 which would correspond to $P(Z\leq1.69)$. I'm guessing it has something to do with the population mean, but I can't figure it out.
• Oct 18th 2008, 09:17 PM
mr fantastic
Quote:

Originally Posted by lllll
For a given exam the standard deviation for all seniors in a particular high school is 2.5. If we take a sample of 100 seniors, find the probability that the difference between the sample mean and the mean of all seniors does not exceed 0.5.

so far I have:

$P\left( \frac{\overline{Y}-X}{2.5 / \sqrt{100}} \leq \frac{0.5}{0.25} \right) \longrightarrow P\left( \frac{\overline{Y}-X}{0.25} \leq 2 \right)$

if I let $P(Z\leq 2) = 1- 0.228 = 0.9972$

but the answer in the back of the book is 0.9544 which would correspond to $P(Z\leq1.69)$. I'm guessing it has something to do with the population mean, but I can't figure it out.

You know that $\bar{X}$ ~ Normal $\left(\mu, ~ \frac{2.5}{\sqrt{100}} = 0.25\right)$ where $\mu$ is the population mean.

You need to calculate $\Pr(\mu - 0.5 < \bar{X} < \mu + 0.5)$.

$z = \frac{(\mu + 0.5) - \mu}{0.25} = 2$ and $z = \frac{(\mu - 0.5) - \mu}{0.25} = -2$.

So $\Pr(\mu - 0.5 < \bar{X} < \mu + 0.5) = \Pr(-2 < Z < 2) = 0.9545$, correct to four decimal places.