1. The problem involves a normal distribution with mean $\mu = 15.7$ and standard deviation $\sigma = 0.7$. We want to find the probability that a value is less than $15.2$ (the cutoff point). 2. The formula to standardize a value $x$ in a normal distribution is the z-score formula: $$z = \frac{x - \mu}{\sigma}$$ This converts the value to a standard normal distribution with mean 0 and standard deviation 1. 3. Calculate the z-score for $x = 15.2$: $$z = \frac{15.2 - 15.7}{0.7} = \frac{-0.5}{0.7} = -\frac{5}{7} \approx -0.7143$$ 4. The z-score tells us how many standard deviations $15.2$ is below the mean. 5. To find the probability that a value is less than $15.2$, we look up the cumulative probability for $z = -0.7143$ in the standard normal distribution table or use a calculator. 6. Using a standard normal table or calculator, $P(Z < -0.7143) \approx 0.2375$. 7. Therefore, the probability that a value from this distribution is less than $15.2$ is approximately $0.2375$ or 23.75%.