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

Calculate the mean and standard deviation of the sampling distribution of [tex]\bar{x}[/tex].

Answer :

Final answer:

The mean of the sampling distribution would be 151 (the same as the LSAT scores), whereas the standard deviation of the sampling distribution would be 1.5 given by the formulae σ/√n.

Explanation:

In this context, we are specifically dealing with the concept of a sampling distribution in statistics.

The mean of the sampling distribution of the sample mean (also known as the expectation of the sample mean) is equal to the mean of the population from which the samples are drawn. So in this case, the mean of the sampling distribution will be the same as the mean of the LSAT scores, which is given as 151.

The standard deviation of the sampling distribution of the sample mean is given by σ/√n, where σ is the standard deviation of the population and n is the sample size. In this case, the standard deviation is 9 and the sample size is 36, so the standard deviation of the sampling distribution would be 9/√36 = 1.5.

Learn more about Sampling Distribution here:

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