Let's solve each part of the question step-by-step.

Given Data

• Population mean ($\mu$) = 188.1
• Population standard deviation ($\sigma$) = 97.2
• Sample size ($n$) = 21

Part (a)

We need to find the probability that a single randomly selected value is greater than 162.6.

1. Calculate the Z-score for a single value $X$: $Z = \frac{X - \mu}{\sigma}$ Plugging in the values: $Z = \frac{162.6 - 188.1}{97.2} = \frac{-25.5}{97.2} \approx -0.2623$

2. Find $P(X > 162.6)$: This is equivalent to finding $P(Z > -0.2623)$.

   Using the standard normal distribution table, we find $P(Z < -0.2623) \approx 0.3966$.

   Therefore, $P(X > 162.6) = 1 - 0.3966 = 0.6034$.

   So, the answer to part (a) is approximately: $P(X > 162.6) \approx 0.6034$

Part (b)

Now we need to find the probability that a sample mean ($M$) of size $n = 21$ is greater than 162.6.

1. Calculate the standard error of the mean ($\sigma_M$): $\sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{97.2}{\sqrt{21}} \approx 21.22$

2. Calculate the Z-score for the sample mean $M$: $Z = \frac{M - \mu}{\sigma_M} = \frac{162.6 - 188.1}{21.22} \approx -1.2024$

3. Find $P(M > 162.6)$: This is equivalent to finding $P(Z > -1.2024)$.

   Using the standard normal distribution table, we find $P(Z < -1.2024) \approx 0.1143$.

   Therefore, $P(M > 162.6) = 1 - 0.1143 = 0.8857$.

   So, the answer to part (b) is approximately: $P(M > 162.6) \approx 0.8857$

Summary of Answers

• (a) $P(X > 162.6) \approx 0.6034$
• (b) $P(M > 162.6) \approx 0.8857$

Would you like further details on any step or clarification on how to interpret these probabilities?

Additional Related Questions

1. What would happen to $P(X > 162.6)$ if the mean $\mu$ were lower?
2. How does the sample size affect the probability in part (b)?
3. What is the probability that a single randomly selected value is less than 162.6?
4. How would the answers change if the standard deviation were smaller?
5. What would be the Z-score if the sample mean $M$ were exactly equal to 188.1?

Tip:

When calculating probabilities involving sample means, always adjust the standard deviation by dividing by the square root of the sample size, as this accounts for the Central Limit Theorem.