To solve these problems, we use the Central Limit Theorem, which states that for a normally distributed population, the sampling distribution of the sample mean ($\bar{x}$) is also normally distributed with:

• Mean: $\mu_{\bar{x}} = \mu = 575$
• Standard error: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

where:

• $\sigma = 40$ (population standard deviation)
• $n$ is the sample size.

Part (a): Find $P(\bar{x} < 555)$ when $n = 16$

1. Compute the standard error:

   $\sigma_{\bar{x}} = \frac{40}{\sqrt{16}} = \frac{40}{4} = 10$

2. Compute the $z$-score for $\bar{x} = 555$:

   $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{555 - 575}{10} = \frac{-20}{10} = -2$

3. Find the probability using the standard normal table:

   $P(Z < -2) = 0.0228$

Thus, the probability is 0.0228.

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Part (b): Find $P(\bar{x} \geq 593)$ when $n = 25$

1. Compute the standard error:

   $\sigma_{\bar{x}} = \frac{40}{\sqrt{25}} = \frac{40}{5} = 8$

2. Compute the $z$-score for $\bar{x} = 593$:

   $z = \frac{593 - 575}{8} = \frac{18}{8} = 2.25$

3. Find the probability:

   $P(Z \geq 2.25) = 1 - P(Z \leq 2.25)$

   From the standard normal table:

   $P(Z \leq 2.25) = 0.9878$

   So,

   $P(Z \geq 2.25) = 1 - 0.9878 = 0.0122$

Thus, the probability is 0.0122.

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Final Answers:

(a) $P(\bar{x} < 555) = 0.0228$
(b) $P(\bar{x} \geq 593) = 0.0122$

Would you like a detailed explanation of how to read the standard normal table or any clarifications?

Related Questions:

1. How does the Central Limit Theorem affect the shape of the sampling distribution?
2. What happens to the probability if the sample size increases?
3. How do you calculate probabilities for a non-normal population?
4. What is the probability of obtaining a sample mean between two given values?
5. How would the probability change if the population standard deviation were