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Assume that the mean score on a certain aptitude test across the nation is 100, and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is between 99.5 and 100.5.
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Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
To find the probability of the mean aptitude test score falling between 99.5 and 100.5 for a randomly selected group of 150 8th graders, we will use the Central Limit Theorem and the Z-score.
Step 1: Determine the population mean and the standard deviation of the population mean (also known as the standard error).
Given that the mean score on the aptitude test across the nation is 100 and the standard deviation is 20, we can calculate the standard error as follows:
Standard Error (SE) = Standard Deviation (σ) / Square Root of Sample Size (n)
SE = 20 / √150
SE ≈ 1.632993
Step 2: Calculate the Z-scores for the lower and upper bounds of the desired range.
Z-score = (X - μ) / SE
Where:
X = Score
μ = Population Mean
SE = Standard Error
For the lower bound:
Z1 = (99.5 - 100) / 1.632993
Z1 ≈ -0.30615
For the upper bound:
Z2 = (100.5 - 100) / 1.632993
Z2 ≈ 0.30615
Step 3: Look up the corresponding area under the standard normal distribution curve for the Z-scores found in Step 2. This will give us the probability.
Using a Z-table or a statistical calculator, we can find that the area to the left of Z1 is approximately 0.381, and the area to the left of Z2 is also approximately 0.619.
Step 4: Calculate the probability between the two bounds.
To find the probability that the mean aptitude test score falls between 99.5 and 100.5, we subtract the lower probability from the upper probability.
Probability = Area(Z2) - Area(Z1)
= 0.619 - 0.381
= 0.238 or 23.8%
Therefore, the probability that the mean aptitude test score for a randomly selected group of 150 8th graders falls between 99.5 and 100.5 is approximately 23.8%.