Question

A distribution of values is normal with a mean of 152.4 and a standard deviation of 47.9.

Find the probability that a randomly selected value is less than 272.2.
P(X < 272.2) =


Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

To find the probability that a randomly selected value \( X \) is less than 272.2 in a normal distribution with mean \( \mu = 152.4 \) and standard deviation \( \sigma = 47.9 \), we first need to calculate the z-score.

The z-score is calculated using the formula:

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

Substituting the given values:

\[
z = \frac{272.2 - 152.4}{47.9} = \frac{119.8}{47.9} \approx 2.497
\]

Next, we need to find the probability corresponding to this z-score. We can use a standard normal distribution table or a calculator to find \( P(Z < 2.497) \).

Using a z-table or calculator, we find:

\[
P(Z < 2.497) \approx 0.9933
\]

Thus, the probability that a randomly selected value is less than 272.2 is:

\[
P(X < 272.2) \approx 0.9933
\]

Finally, rounding to four decimal places, the answer is:

\[
\boxed{0.9933}
\]