Question 1182185
Here's how to solve this problem:

**a. μM = 73**

The mean of the sampling distribution of the mean (μM) is equal to the population mean (μ).  Therefore, μM = 73.

**b. σM = 2**

The standard deviation of the sampling distribution of the mean (σM), also known as the standard error, is calculated as:

σM = σ / √n

Where:

* σ = population standard deviation = 20
* n = sample size = 100

σM = 20 / √100 = 20 / 10 = 2

**c. The shape of the comparison distribution is approximately normal.**

Even though the population distribution is rectangular, the Central Limit Theorem states that the sampling distribution of the mean will approach a normal distribution as the sample size increases.  With a sample size of 100, the sampling distribution will be approximately normal.

**d. Lower limit for the 99% confidence interval:**

For a 99% confidence interval, the alpha (α) level is 1 - 0.99 = 0.01.  Alpha/2 is 0.01 / 2 = 0.005. The critical z-score (zα/2) for a 99% confidence interval is approximately 2.576.

The formula for the confidence interval is:

CI = M ± zα/2 * σM

Where:

* M = sample mean = 75

Lower Limit = M - zα/2 * σM
Lower Limit = 75 - (2.576 * 2)
Lower Limit ≈ 75 - 5.152
Lower Limit ≈ 69.85

**e. Upper limit for the 99% confidence interval:**

Upper Limit = M + zα/2 * σM
Upper Limit = 75 + (2.576 * 2)
Upper Limit ≈ 75 + 5.152
Upper Limit ≈ 80.15

**f. Lower limit for the 95% confidence interval:**

For a 95% confidence interval, α = 1 - 0.95 = 0.05. Alpha/2 is 0.05 / 2 = 0.025. The critical z-score (zα/2) for a 95% confidence interval is approximately 1.96.

Lower Limit = M - zα/2 * σM
Lower Limit = 75 - (1.96 * 2)
Lower Limit ≈ 75 - 3.92
Lower Limit ≈ 71.08

**g. Upper limit for the 95% confidence interval:**

Upper Limit = M + zα/2 * σM
Upper Limit = 75 + (1.96 * 2)
Upper Limit ≈ 75 + 3.92
Upper Limit ≈ 78.92