The exercise in the image involves a scenario where the completion time for IRS Form 1040 follows a normal distribution. Let's break down and solve the questions step by step:

1. Single Taxpayer Case

The question asks to find the probability that a randomly selected taxpayer took more than 13.28 hours to complete the form. We're given:

• Mean ($\mu$) = 12.76 hours
• Standard deviation ($\sigma$) = 1.9 hours
• The time we're interested in is 13.28 hours.

Steps:

1. Z-Score Calculation: The Z-score formula is:

   $Z = \frac{X - \mu}{\sigma}$

   Plugging in the values:

   $Z = \frac{13.28 - 12.76}{1.9} \approx 0.2737$

   This Z-score is around 0.2737.

2. Finding Probability: To find the probability that a taxpayer took more than 13.28 hours, we use the cumulative distribution function (CDF). From the Z-table or a calculator, the probability associated with $Z = 0.2737$ is approximately 0.608. Since we're looking for the area to the right (i.e., more than 13.28 hours), the probability is:

   $P(X > 13.28) = 1 - 0.608 = 0.392$

   So, the probability is 0.392 or 39.2%.

2. 65 Taxpayers Case

The second part asks for the probability that, on average, 65 taxpayers took more than 13.28 hours.

• Mean ($\mu$) = 12.76 hours
• Standard deviation for the sample mean = $\frac{\sigma}{\sqrt{n}}$, where $n = 65$.

Thus, the new standard deviation is:

$\sigma_{\text{new}} = \frac{1.9}{\sqrt{65}} \approx 0.2357$

Steps:

1. Z-Score Calculation:

Using the same formula for Z-score, but with the new standard deviation:

$Z = \frac{13.28 - 12.76}{0.2357} \approx 2.203$

2. Finding Probability:

Using the Z-table, the probability associated with $Z = 2.203$ is approximately 0.986. Since we want the probability of taking more than 13.28 hours, we subtract this from 1:

$P(X > 13.28) = 1 - 0.986 = 0.014$

Thus, the probability is approximately 0.014 or 1.4%.

Would you like more details on any step or a further explanation?

5 Questions to Explore Further:

1. How does the sample size affect the standard deviation in this type of problem?
2. What is the significance of using Z-scores in normal distribution problems?
3. How would the probability change if the standard deviation were larger?
4. Can you explain how the central limit theorem applies in the second part of this problem?
5. How can we use software like Excel or calculators to quickly find Z-scores and probabilities?

Tip:

Remember, as the sample size increases, the distribution of the sample mean becomes more concentrated around the population mean, which is why the standard deviation decreases.