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

151μxˉ​≈151mu, start subscript, x, with, \bar, on top, end subscript, approximately equals, 151 pointsσxˉ≈1.5σxˉ​≈1.5sigma, start subscript, x, with, \bar, on top, end subscript, approximately equals, 1, point, 5 pointsShow Calculator

Explanation:

The mean of the sampling distribution:The mean of the sampling distribution of a sample mean xˉxˉx, with, \bar, on top is equal to the population mean:μxˉ=μμxˉ​=μmu, start subscript, x, with, \bar, on top, end subscript, equals, muThe population mean is reported as μ≈151μ≈151mu, approximately equals, 151 points.So μxˉ=μ≈151μxˉ​=μ≈151mu, start subscript, x, with, \bar, on top, end subscript, equals, mu, approximately equals, 151Hint #22 / 3The standard deviation of the sampling distribution:Since there are more than 100,000100,000100, comma, 000 testers in the population, a sample of n=36n=36n, equals, 36 testers is less than 10%10%10, percent of population, and we can assume independence between the testers in a sample. So the standard deviation of the sampling distribution of xˉxˉx, with, \bar, on top can be found using this formula:σxˉ=σnσxˉ​=n​σ​sigma, start subscript, x, with, \bar, on top, end subscript, equals, start fraction, sigma, divided by, square root of, n, end square root, end fractionThe standard deviation of the sampling distribution isσxˉ=σn≈936≈96≈1.5σxˉ​=n​σ​≈36​9​≈69​≈

Answer 2

Explanation:

151 points and 1.5 points