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:
\[ \boxed{0.9933} \]