Question

Sample size 115 pop mean 48 standard dev 12

Probablililoty x will be between 45 and 50

Answer 1

To find the probability that a sample mean \( \bar{x} \) falls between 45 and 50 based on the sample size, population mean, and population standard deviation, we can use the central limit theorem (CLT).

Given:
- Population mean \( \mu = 48 \)
- Population standard deviation \( \sigma = 12 \)
- Sample size \( n = 115 \)

First, we need to find the standard error (SE) of the sample mean:

\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{115}} \approx 1.12
\]

Next, we can standardize the values 45 and 50 using the z-score formula:

\[
z = \frac{\bar{x} - \mu}{SE}
\]

1. For \( \bar{x} = 45 \):

\[
z_{45} = \frac{45 - 48}{1.12} \approx \frac{-3}{1.12} \approx -2.68
\]

2. For \( \bar{x} = 50 \):

\[
z_{50} = \frac{50 - 48}{1.12} \approx \frac{2}{1.12} \approx 1.79
\]

Now, we can look up these z-scores in the standard normal distribution table or use a calculator to find the corresponding probabilities.

- For \( z = -2.68 \), the probability (or area to the left) is approximately 0.0037 (or 0.37%).
- For \( z = 1.79 \), the probability (or area to the left) is approximately 0.9633 (or 96.33%).

Finally, to find the probability that \( \bar{x} \) is between 45 and 50, we subtract the probability at \( z = -2.68 \) from the probability at \( z = 1.79 \):

\[
P(45 < \bar{x} < 50) = P(z < 1.79) - P(z < -2.68) \approx 0.9633 - 0.0037 \approx 0.9596
\]

Thus, the probability that the sample mean \( \bar{x} \) is between 45 and 50 is approximately **0.9596** or **95.96%**.