Question

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) =

Answer 1

To find the probability that a single randomly selected value \( X \) from a normally distributed population is less than 97, we can use the z-score formula. The z-score is calculated using the following formula:

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

where:
- \( X \) is the value we are interested in (in this case, 97),
- \( \mu \) is the mean of the distribution (65.4),
- \( \sigma \) is the standard deviation of the distribution (61.9).

Let's calculate the z-score for \( X = 97 \):

\[
z = \frac{97 - 65.4}{61.9} = \frac{31.6}{61.9} \approx 0.5101
\]

Next, we need to find the probability \( P(X < 97) \), which corresponds to the cumulative distribution function (CDF) for the z-score we just calculated.

We can use the standard normal distribution table or a calculator to find the probability associated with \( z \approx 0.5101 \).

Using a standard normal distribution table or calculator:

\[
P(Z < 0.5101) \approx 0.6950
\]

Thus, the probability that a single randomly selected value is less than 97 is approximately:

\[
P(X < 97) \approx 0.6950
\]

So, the final answer is:
\[
P(X < 97) \approx 0.6950
\]