We are dealing with a problem related to the distribution of sample means. The population has a mean $\mu = 200$ and a standard deviation $\sigma = 50$. Since a simple random sample of size $n = 100$ is selected, we can apply the Central Limit Theorem (CLT), which tells us that the sampling distribution of the sample mean $\bar{X}$ will be approximately normally distributed with:

• Mean: $\mu_{\bar{X}} = \mu = 200$
• Standard error (SE) of the mean: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = 5$

a) What is the probability that the sample mean will be below 193?

We need to find the probability $P(\bar{X} < 193)$. To do this, we first calculate the z-score:

$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{193 - 200}{5} = \frac{-7}{5} = -1.4$

Using a standard normal distribution table or calculator, the probability corresponding to a z-score of $-1.4$ is approximately 0.0808.

Thus, the probability that the sample mean will be below 193 is:

$P(\bar{X} < 193) \approx 0.0808$

b) What is the probability that the sample mean will be below 207?

We need to find $P(\bar{X} < 207)$. Again, we first calculate the z-score:

$z = \frac{207 - 200}{5} = \frac{7}{5} = 1.4$

The probability corresponding to a z-score of 1.4 is approximately 0.9192.

Thus, the probability that the sample mean will be below 207 is:

$P(\bar{X} < 207) \approx 0.9192$

c) What is the probability that the sample mean will be within 7 of the population mean?

We want to find $P(193 < \bar{X} < 207)$. This is equivalent to finding the probability that the sample mean is within 7 units of the population mean. To do this, we calculate the z-scores for both 193 and 207, which we've already done:

• For 193: $z = -1.4$
• For 207: $z = 1.4$

Now, we calculate the probability between these two z-scores:

$P(193 < \bar{X} < 207) = P(-1.4 < z < 1.4)$

From the standard normal table, the probability corresponding to $z = -1.4$ is 0.0808, and for $z = 1.4$, it's 0.9192. The probability between these two values is:

$P(193 < \bar{X} < 207) = 0.9192 - 0.0808 = 0.8384$

Thus, the probability that the sample mean will be within 7 units of the population mean is:

$P(193 < \bar{X} < 207) \approx 0.8384$

Summary of Results:

a) $P(\bar{X} < 193) \approx 0.0808$
b) $P(\bar{X} < 207) \approx 0.9192$
c) $P(193 < \bar{X} < 207) \approx 0.8384$

Do you want more details or have any questions about this?

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Here are 5 related questions to expand your understanding:

1. What happens to the probability if the sample size is increased to 400?
2. How would the probabilities change if the population standard deviation was 100 instead of 50?
3. What is the probability that the sample mean will exceed 215?
4. How does the Central Limit Theorem help in approximating the distribution of sample means?
5. How does the standard error of the mean affect the precision of estimates for the population mean?

Tip: When sample sizes are large, the distribution of the sample mean approaches normality, regardless of the shape of the population distribution, thanks to the Central Limit Theorem.