Step 1 :The problem is asking for the value that separates the bottom 40% of sample means from the top 60% in a normally distributed population. This is also known as the 40th percentile of the sample means.
Step 2 :In a normal distribution, the sample mean is equal to the population mean, which is given as \( \mu = 126.4 \).
Step 3 :The standard deviation of the sample means (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size. So, \( \sigma = \frac{50.2}{\sqrt{242}} = 3.22697821959679 \).
Step 4 :We can use the z-score formula to find the value corresponding to the 40th percentile. The z-score is the number of standard deviations a particular value is from the mean. The z-score corresponding to the 40th percentile is \( z = -0.2533471031357997 \).
Step 5 :Once we have the z-score, we can use the formula for the sample mean to find the value that separates the bottom 40% of sample means from the top 60%. So, \( P_{40} = \mu + z \cdot \sigma = 125.58245441618284 \).
Step 6 :The value that separates the bottom 40% of sample means from the top 60% in a normally distributed population is approximately 125.6.
Step 7 :Final Answer: \( P_{40} \text { (for sample means) }= \boxed{125.6} \)