A population of values has a normal distribution with μ=65.4 and o=61.9. You intend to draw a random sample of size n=24.

Find the probability that a single randomly selected value is less than 97.
P(X < 97) =

Find the probability that a sample of size n=24 is randomly selected with a mean less than 97.
P(M < 97) =

1 answer

To find the probabilities you requested, we will use the properties of the normal distribution.

Part 1: Probability that a single randomly selected value is less than 97

We need to calculate \( P(X < 97) \) where \( X \) is a normally distributed random variable with mean \( \mu = 65.4 \) and standard deviation \( \sigma = 61.9 \).

  1. Calculate the z-score for \( X = 97 \): \[ z = \frac{X - \mu}{\sigma} = \frac{97 - 65.4}{61.9} = \frac{31.6}{61.9} \approx 0.510 \]

  2. Look up the z-score in the standard normal distribution table, or use a calculator or software to find \( P(Z < 0.510) \).

Using a z-table or calculator, we find: \[ P(Z < 0.510) \approx 0.695 \]

Thus, \[ P(X < 97) \approx 0.695 \]

Part 2: Probability that the sample mean is less than 97

For the sample mean \( M \) of a sample of size \( n = 24 \), we need to use the central limit theorem which tells us that the sampling distribution of the sample mean will be normally distributed with:

  • Mean: \( \mu_M = \mu = 65.4 \)
  • Standard deviation (standard error): \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{61.9}{\sqrt{24}} \approx \frac{61.9}{4.899} \approx 12.63 \]
  1. Calculate the z-score for the sample mean \( M < 97 \): \[ z = \frac{M - \mu_M}{\sigma_M} = \frac{97 - 65.4}{12.63} = \frac{31.6}{12.63} \approx 2.50 \]

  2. Look up the z-score in the standard normal distribution table: Using the z-table or a calculator: \[ P(Z < 2.50) \approx 0.9938 \]

Thus, \[ P(M < 97) \approx 0.9938 \]

Summary

  • Probability that a single randomly selected value is less than 97: \[ P(X < 97) \approx 0.695 \]
  • Probability that a sample of size n=24 has a mean less than 97: \[ P(M < 97) \approx 0.9938 \]