To solve for the probability that a normally distributed random variable $X$ is between 31 and 97, we use the standard normal distribution.

Step 1: Convert the values of $X = 31$ and $X = 97$ to standard normal $Z$-scores using the formula:

$Z = \frac{X - \mu}{\sigma}$

where:

• $X$ is the value of the random variable,
• $\mu = 69$ is the mean,
• $\sigma = 16$ is the standard deviation.

For $X = 31$:

$Z_1 = \frac{31 - 69}{16} = \frac{-38}{16} = -2.375$

For $X = 97$:

$Z_2 = \frac{97 - 69}{16} = \frac{28}{16} = 1.75$

Step 2: Look up the cumulative probabilities for $Z_1 = -2.375$ and $Z_2 = 1.75$ using the standard normal distribution table (or a calculator).

• The cumulative probability for $Z_1 = -2.375$ is approximately $0.0088$.
• The cumulative probability for $Z_2 = 1.75$ is approximately $0.9599$.

Step 3: Find the probability that $X$ is between 31 and 97 by subtracting the cumulative probabilities:

$P(31 < X < 97) = P(Z_2) - P(Z_1) = 0.9599 - 0.0088 = 0.9511$

Final Answer:

The probability that $X$ is between 31 and 97 is approximately $0.951$ (rounded to the nearest thousandth).

Would you like further details or explanations? Here are 5 related questions to extend this problem:

1. How do you calculate the probability for values outside of the range 31 to 97?
2. What would the probability be if the interval was between 40 and 90?
3. How does changing the standard deviation affect the probability calculation?
4. What is the significance of the Z-scores in the context of this problem?
5. How do you interpret the result in terms of real-world applications of normal distributions?

Tip: When calculating probabilities for normally distributed variables, always check if it's easier to standardize the distribution using Z-scores, which makes calculations simpler.