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 \( \sigma = 61.9 \).
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Calculate the z-score for \( X = 97 \): \[ z = \frac{X - \mu}{\sigma} = \frac{97 - 65.4}{61.9} = \frac{31.6}{61.9} \approx 0.510 \]
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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): \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{61.9}{\sqrt{24}} \approx \frac{61.9}{4.899} \approx 12.63 \]
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Calculate the z-score for the sample mean \( M < 97 \): \[ z = \frac{M - \mu_M}{\sigma_M} = \frac{97 - 65.4}{12.63} = \frac{31.6}{12.63} \approx 2.50 \]
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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: \[ P(M < 97) \approx 0.9938 \]