To solve this problem, we will use the sampling distribution of the sample mean.

Step 1: Identify Given Values

• Population Mean ($\mu$): 158 minutes
• Population Standard Deviation ($\sigma$): 7 minutes
• Sample Size ($n$): 99

Step 2: Calculate the Standard Error of the Mean (SEM)

The standard error of the mean is given by: $\text{SEM} = \frac{\sigma}{\sqrt{n}}$ Substitute the values: $\text{SEM} = \frac{7}{\sqrt{99}} \approx 0.704$

Step 3: Calculate Z-Scores for the Range

We want to find the probability that the sample mean is between 157.71 minutes and 160.31 minutes. To do this, we calculate the corresponding $z$-scores.

1. Lower Bound (157.71 minutes): $z = \frac{157.71 - 158}{0.704} \approx -0.41$

2. Upper Bound (160.31 minutes): $z = \frac{160.31 - 158}{0.704} \approx 3.28$

Step 4: Find the Probability Using Z-Scores

Using a standard normal distribution table or calculator:

• P(Z < -0.41) ≈ 0.3409
• P(Z < 3.28) ≈ 0.9995

The probability that the sample mean is between 157.71 and 160.31 minutes is: $P(-0.41 < Z < 3.28) = 0.9995 - 0.3409 = 0.6586$

Conclusion

The probability that the mean commuting time for a sample of 99 employees is between 157.71 and 160.31 minutes is approximately 0.6594.

So, the correct answer is: $\boxed{0.6594}$

Would you like further details or explanations on any step?

Relative Questions

1. How would the probability change if the sample size were smaller?
2. What would the probability be if the range was between 155 and 161 minutes?
3. How would a change in the population standard deviation affect this probability?
4. What if the commuting times were not normally distributed—how would that impact this calculation?
5. How is the central limit theorem applied in this context?

Tip: When working with sample means, remember that the standard deviation decreases by a factor of $\sqrt{n}$ due to the sample size. This makes larger samples less variable, improving the precision of the sample mean.