To solve this problem, we will apply the Central Limit Theorem (CLT), which states that the sampling distribution of the sample mean will approximate a normal distribution as the sample size increases, even if the population distribution is skewed. Given that we have a sample size of 35, which is relatively large, we can proceed with this approach.

Step-by-Step Solution

1. Define the given values:

   • Population mean, $\mu = 79$
   • Population standard deviation, $\sigma = 8.2$
   • Sample size, $n = 35$
   • We want to find the probability that the sample mean $\bar{X}$ is between 76 and 82.

2. Calculate the standard error of the mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{8.2}{\sqrt{35}} \approx 1.3855$

3. Convert the bounds 76 and 82 to standard $z$-scores:

   For $X = 76$: $z = \frac{76 - 79}{1.3855} \approx -2.1656$

   For $X = 82$: $z = \frac{82 - 79}{1.3855} \approx 2.1656$

4. Find the probabilities associated with these $z$-scores: Using a standard normal distribution table or calculator:

   • $P(Z \leq -2.1656) \approx 0.0152$
   • $P(Z \leq 2.1656) \approx 0.9848$

5. Calculate the probability that the sample mean is between 76 and 82: $P(76 \leq \bar{X} \leq 82) = P(Z \leq 2.1656) - P(Z \leq -2.1656)$ $P(76 \leq \bar{X} \leq 82) = 0.9848 - 0.0152 = 0.9696$

Answer

The probability that the average final exam grade of this sample is between 76 and 82 is approximately 0.9696 (rounded to 4 decimal places).

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Would you like further explanation on any of these steps or additional examples?

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1. How does the Central Limit Theorem apply to sample sizes smaller than 30?
2. How would the probability change if the sample size increased to 50?
3. How is the standard error of the mean calculated, and why does it decrease with larger samples?
4. What would be the effect on the probability if the standard deviation were larger?
5. How can you use the $z$-score table to find probabilities for other intervals?

Tip: When working with sample means, always remember that increasing the sample size reduces the standard error, making the sample mean more likely to be close to the population mean.