To solve this problem, we are asked to find the probability that the sample mean ($\bar{X}$) is less than 95, given the following information:

• Population mean ($\mu$) = 101
• Population standard deviation ($\sigma$) = 10
• Sample size ($n$) = 25
• We need to find $P(\bar{X} < 95)$

Step 1: Find the distribution of the sample mean $\bar{X}$

The sample mean $\bar{X}$ is normally distributed with:

• Mean: $\mu_{\bar{X}} = \mu = 101$
• Standard deviation (standard error): $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$

Thus, $\bar{X}$ is normally distributed with a mean of 101 and a standard deviation (standard error) of 2.

Step 2: Standardize the value

To find the probability $P(\bar{X} < 95)$, we need to standardize 95 to a z-score using the formula: $z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}$ Substitute the values: $z = \frac{95 - 101}{2} = \frac{-6}{2} = -3$

Step 3: Find the probability using the z-score

Now, we need to find the probability corresponding to $z = -3$. Using standard normal distribution tables or a calculator, we find:

$P(Z < -3) = 0.00135$

Conclusion:

The probability that the sample mean $\bar{X}$ is less than 95 is approximately 0.00135, or 0.135%.

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Do you have any questions or would you like more details on this solution?

Here are 5 related questions:

1. How does changing the sample size $n$ affect the standard error?
2. What is the probability that $\bar{X}$ is greater than 105 in this same distribution?
3. How would the probability change if $\mu = 95$ and $\sigma = 5$?
4. Can you explain the relationship between sample size and the accuracy of sample mean estimates?
5. How would you calculate the probability if $n = 100$ instead of 25?

Tip: The larger the sample size $n$, the smaller the standard error, which means the sample mean becomes a more precise estimate of the population mean.