The mean ACT test score reported by the college’s Institutional Research Depart was 20.8, with a standard deviation of 4.8. If all possible random samples of size 169 are taken from this population, determine the following : (ROUND ALL FINAL ANSWERS TO THE NEAREST TENTHOUSANDTH) a) name of the Sampling Distribution b) mean and standard error of the sampling distribution of the mean (use the correct name and symbol for each) c) percent of sample means for a sample of 169 college students that are greater than 21.8. d) probability that sample means for samples of size 169 fall between 19.3 and 22. e) Below which sample mean can we expect to find the lowest 25% of all the sample means?

Accepted Solution

Answer:a) Sampling distribution of sample means; b) The mean, μ, is 20.8 and the standard error, [tex]\sigma_{\bar{x}}[/tex], is 0.3692; c) 0.0034; d) 0.9994; e) 20.55626.Step-by-step explanation:For part a,The name of a distribution of sample means is the sampling distribution of sample means.  The mean of a distribution of sample means is the same as the population mean, μ, which is 20.8.  The standard error of a sampling distribution of sample means is represented by [tex]\sigma_{\bar{x}}[/tex], and is given by σ/√n:4.8/√169 = 4.8/13 = 0.3692.For part c,We use the formula for a z score, [tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]z = (21.8-20.8)/0.3692 = 1/0.3692 = 2.71Using a z table, we see that the area under the curve to the left of this is 0.9966.  However, we want the area under the curve to the right, so we subtract from 1:1-0.9966 = 0.0034For part d,We want P(19.3 ≤ X ≤ 22):z = (19.3-20.8)/0.3692 = -1.5/0.3692 = -4.06z = (22-20.8)/0.3692 = 1.2/0.3692 = 3.25The area under the curve to the right of z = -4.06 is 0.00.  The area under the curve to the right of z = 3.25 is 0.9994.  This makes the area between them0.9994 - 0.00 = 0.9994.For part e,We look up 25%, or 0.25, in the z table.  The closest value to this is 0.2514, which corresponds to a z score of -0.67:-0.67 = (X-20.8)/0.3692Multiply both sides by 0.3692:0.3692(-0.67) = ((X-20.8)/0.3692)(0.3692)-0.247364 = X-20.8Add 20.8 to each side:-0.247364+20.8 = X-20.8+20.820.5526 = X