Questions: A population of values has a normal distribution with μ=75.1 and σ=39.7. You intend to draw a random sample of size n=173. Find P60, which is the score separating the bottom 60% scores from the top 40% scores. P60(for single values) = 98.8 Find P60, which is the mean separating the bottom 60% means from the top 40% means. P60(for sample means) = 78.9 Enter your answers as numbers accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Solution Steps

To find the 60th percentile, we first calculate the z-score corresponding to this percentile in a standard normal distribution. The z-score is given by the formula:

\[ z = \frac{X - \mu}{\sigma} \]

For the 60th percentile, we have:

\[ z = \frac{0.6 - 0}{1} = 0.6 \]

Using the z-score calculated above, we can find the value that separates the bottom 60% of scores from the top 40% in the population. The formula to calculate \( P_{60} \) for single values is:

\[ P_{60} = \mu + z \cdot \sigma \]

Substituting the values:

\[ P_{60} = 75.1 + 0.6 \cdot 39.7 = 98.9 \]

Next, we calculate the standard error of the mean (SEM) for the sample size \( n = 173 \):

\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{39.7}{\sqrt{173}} \approx 3.034 \]

Now, we can find the mean that separates the bottom 60% of sample means from the top 40%. The formula for \( P_{60} \) for sample means is:

\[ P_{60} = \mu + z \cdot \text{SEM} \]

Substituting the values:

\[ P_{60} = 75.1 + 0.6 \cdot 3.034 \approx 76.9 \]

Final Answer