To solve this problem, we use the Central Limit Theorem (CLT), which tells us that for large sample sizes, the distribution of the sample mean will be approximately normal, even if the population distribution is not normal. Here, the sample size is 64, which is sufficiently large.

Given information:

• Population mean ($\mu$) = 35 minutes
• Population standard deviation ($\sigma$) = 12 minutes
• Sample size ($n$) = 64

We are asked to find the probability that the sample mean ($\bar{x}$) is between 30 and 40 minutes.

Step 1: Calculate the standard error of the mean (SE)

The standard error (SE) is given by:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{64}} = \frac{12}{8} = 1.5$

Step 2: Convert the sample means to z-scores

The z-score is calculated using the formula:

$z = \frac{\bar{x} - \mu}{SE}$

For the sample mean of 30 minutes:

$z_{30} = \frac{30 - 35}{1.5} = \frac{-5}{1.5} = -3.33$

For the sample mean of 40 minutes:

$z_{40} = \frac{40 - 35}{1.5} = \frac{5}{1.5} = 3.33$

Step 3: Find the probabilities corresponding to these z-scores

Using a standard normal distribution table or calculator, we find the cumulative probabilities:

• $P(z < -3.33)$ ≈ 0.0004
• $P(z < 3.33)$ ≈ 0.9996

Step 4: Find the probability between the two z-scores

The probability that the sample mean is between 30 and 40 minutes is the difference between these two cumulative probabilities:

$P(30 < \bar{x} < 40) = P(z < 3.33) - P(z < -3.33) = 0.9996 - 0.0004 = 0.9992$

Thus, the probability that the sample mean is between 30 and 40 minutes is 0.9992.

Would you like any details or further clarification on this solution?

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Related Questions:

1. How does the Central Limit Theorem apply to this problem?
2. What is the significance of a large sample size in using the CLT?
3. How would the result change if the sample size were smaller?
4. How do you interpret the z-scores in a normal distribution?
5. What if the standard deviation had been different, say 10 minutes?

Tip: When using the Central Limit Theorem, always check that your sample size is sufficiently large (usually $n > 30$) to justify the normal approximation.