Every year, more than 100,000 testers take the 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 and calculate the mean and standard deviation of each sample.

Calculate the mean and standard deviation of the sampling distribution of \(\bar{x}\). Round to one decimal place.

A. Mean = 151, Standard Deviation = 1.5
B. Mean = 151, Standard Deviation = 1.8
C. Mean = 151, Standard Deviation = 1.2
D. Mean = 151, Standard Deviation = 1.0

To calculate the mean and standard deviation of the sampling distribution of x-bar for the LSAT scores with a mean of 151 and a standard deviation of 9, we use the formulas for the sampling distribution of the sample mean.

1. The mean of the sampling distribution of x-bar is equal to the population mean. Therefore, the mean of the sampling distribution is 151.

2. The standard deviation of the sampling distribution of x-bar is equal to the population standard deviation divided by the square root of the sample size. Given that the sample size is 36, we calculate the standard deviation of the sampling distribution as:

Standard deviation = 9 / √36 = 9 / 6 = 1.5

Considering the provided options:

a) Mean = 151, Standard Deviation = 1.5 - This option matches the calculated values.

b) Mean = 151, Standard Deviation = 1.8 - This standard deviation does not match the calculated value.

c) Mean = 151, Standard Deviation = 1.2 - This standard deviation does not match the calculated value.

d) Mean = 151, Standard Deviation = 1.0 - This standard deviation does not match the calculated value.

Therefore, the correct mean and standard deviation of the sampling distribution of x-bar are:

Mean = 151

Standard Deviation = 1.5

Hence, option a) Mean = 151, Standard Deviation = 1.5 is the correct answer.