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Math Help - Central limit theorem question

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    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.
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    Quote Originally Posted by lllll View Post
    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.
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