Question

Every year, more than 100,000 testers take the Law School Admission Test (LSAT). One year, the scores had a mean and standard deviation of approximately 151 and 9 points, respectively. Suppose that in the scoring process, test officials audit random samples of 36 tests, which involves calculating the sample mean score x/bar. Calculate the mean and standard deviation of the sampling distribution of x/bar

Answer 1

Answer:

look at the screenshot

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Step-by-step explanation:

Answer 2

Concept

This problem is related to the mean and the standard deviation of the sampling distribution of X/bar which states that if a random variable x is normally distributed with mean µ and standard deviation σ, then the distribution of the sample mean, x/bar, is normally distributed with mean

µ x/bar = µ and

standard deviation σ x/bar = σ/√n

Given

We have given every year, more than 100,000 testers take the Law School Admission Test (LSAT). One year, the scores had a mean and standard deviation of approximately 151 and 9 points, respectively. Suppose that in the scoring process, test officials audit random samples of 36 tests, which involves calculating the sample mean score x/bar.

To Find

We have to calculate the mean and standard deviation of the sampling distribution of x/bar

Solution

Here,

One year, the score had a Mean $\mu=151$

And standard deviation $\sigma=9$

Since random samples of size $n = 36$

So, $\bar{X}$, is normally distributed with mean

$\mu_{\bar{X}}=\mu=151$

and

standard deviation  $\sigma_{\bar{X}}= \frac{\sigma}{\sqrt{n}}$

Put the values we get,

$\sigma_{\bar{X}}= \frac{9}{\sqrt{36}}$

$\sigma_{\bar{X}}= 1.5$

As a result, the mean and standard deviation of the sampling distribution of x/bar are $151$ and $1.5$ respectively.

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