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Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 515 and a standard deviation of 107. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect.

If 18 of the men are randomly selected, find the probability that their mean score is at least 557.9.

P( > 557.9) =

Answer 1

Thinking:

N < 30, so I choose to use the Student's t Distribution (instead of the Normal Distribution) to answer the question. Note that the question revolves around the Sampling Distribution of the SampleMean, xBar.

Parameters:

N = 18: Sample Size

df = 17 = N - 1: Parameter Degrees of Freedom

Hypothesized Mean-Score SampleMean = 557.9

Test Score DistributionMean = 515

s = 107: Test Score Distribution Std Deviation

s_xBar = s / √N = 107 / √18 = 25.220: Mean-Score Sampling Distribution Standard Error (from the Central Limit Theorem)

t-Score Test Statistic:

t = (SampleMean - DistributionMean) / s_xBar

t = (557.9 - 515) / 25.220 = 1.701

From Student's t Distribution Statistical Table (17 degrees of freedom):

t (p = 0.10) = 1.333

t (p = 0.05) = 1.740

Interpolating to Approximate Answer:

Slope of line = m = (0.05 - 0.10) / (1.740 - 1.333)

p = 0.10 + (1.701 - 1.333) * m = 0.0548

p(xBar > 557.9) = 0.0548 = Probability the mean score (for N = 18) will exceed 557.9

Online Calculator Answer (a more accurate answer):

p(xBar > 557.9) = 0.0536 = Probability the mean score (for N = 18) will exceed 557.9

https://www.danielsoper.com/statcalc/calculator.aspx?id=10

Hoping the above will help you,

Gary

Answer 2

use transformation z = (sample mean - population mean) / standard error

where sample mean (xbar) = 557.9, population mean = 515, standard error = SD/square root (sample size) = 107/sqrt(15) = 27.63

so P(xbar > 557.9) = P(z > (557.9 - 515)/27.63) = P(z > 1.55) = 1 - P(z < 1.55) = 1 - 0.9394 = 0.0606