1. The problem involves a normal distribution with mean $\mu = 62\%$ and standard deviation $\sigma = 11\%$. We are given probabilities or percentages related to this distribution. 2. To solve problems involving normal distributions, we use the standard normal variable $Z = \frac{X - \mu}{\sigma}$, which converts any normal variable $X$ to a standard normal variable with mean 0 and standard deviation 1. 3. The values 0.1587, 0.1376, 79.10%, and 0.1629 likely represent probabilities or cumulative probabilities related to $Z$-scores. 4. For example, 0.1587 corresponds approximately to the probability $P(Z < -1)$ or $P(Z > 1)$ since $\Phi(-1) \approx 0.1587$. 5. Similarly, 0.1376 corresponds to a $Z$-score near -1.1, and 0.1629 corresponds to a $Z$-score near -0.99. 6. The 79.10% likely represents a cumulative probability $P(Z < z)$ or $P(X < x)$. 7. To find the $X$ value corresponding to a cumulative probability $p$, use the inverse standard normal function $z = \Phi^{-1}(p)$, then convert back: $$X = \mu + z \sigma$$ 8. For example, if $p = 0.1587$, then $z = -1$, so $$X = 62 + (-1)(11) = 62 - 11 = 51\%$$ 9. This means about 15.87% of the data lies below 51%. 10. Similarly, for $p = 0.1376$, find $z \approx -1.1$, so $$X = 62 + (-1.1)(11) = 62 - 12.1 = 49.9\%$$ 11. For $p = 0.1629$, $z \approx -0.99$, so $$X = 62 + (-0.99)(11) = 62 - 10.89 = 51.11\%$$ 12. For $p = 0.7910$ (79.10%), $z \approx 0.81$, so $$X = 62 + 0.81 \times 11 = 62 + 8.91 = 70.91\%$$ 13. These calculations show how to interpret probabilities and percentages in terms of the original variable $X$ using the normal distribution. Final answers: - $P(X < 51\%) \approx 0.1587$ - $P(X < 49.9\%) \approx 0.1376$ - $P(X < 51.11\%) \approx 0.1629$ - $P(X < 70.91\%) \approx 0.7910$