A population of values has a normal distribution with [tex]\mu = 224.4[/tex] and [tex]\sigma = 98.2[/tex]. You intend to draw a random sample of size [tex]n = 38[/tex]. Find [tex]P_{92}[/tex], which is the score separating the bottom 92% of scores from the top 8%.

To find P92, you need to find the z-score that corresponds to the bottom 92% scores in a standard normal distribution. The score separating the bottom 92% scores is approximately 86.562.

Explanation:

To find P92, we need to find the z-score that corresponds to the bottom 92% scores in a standard normal distribution. The z-score can be calculated using the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

First, we need to find the z-score corresponding to the bottom 92%, which is 0.08 (1 - 0.92 = 0.08). Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to 0.08 is approximately -1.41.

Next, we can use the formula z = (x - μ) / σ to solve for x. Rearranging the formula, we get x = μ + z * σ. Plugging in the values μ = 224.4, σ = 98.2, and z = -1.41, we can calculate the score separating the bottom 92% scores:

x = 224.4 + (-1.41) * 98.2 = 86.562. Therefore, the score separating the bottom 92% scores is approximately 86.562.

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