To find the proportion of 1040R tax forms completed in less than 119 minutes, we will use the standard normal distribution (Z-distribution) formula.

The given problem specifies:

• Mean $\mu = 100$ minutes,
• Standard deviation $\sigma = 15$ minutes,
• Time $X = 119$ minutes.

The formula to convert a value $X$ from a normal distribution to the standard normal distribution (Z-distribution) is:

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

Substituting the values:

$Z = \frac{119 - 100}{15} = \frac{19}{15} \approx 1.2667$

Next, we find the proportion of the population that has a Z-score less than 1.2667. Using a Z-table or a standard normal distribution calculator:

$P(Z < 1.2667) \approx 0.8970$

So, approximately 89.70% of the 1040R tax forms will be completed in less than 119 minutes.

Final Answer:

The proportion is 0.8970 or 89.70%.

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Do you want more details or have any questions?

Here are 5 related questions for further exploration:

1. What proportion of 1040R tax forms will be completed in more than 119 minutes?
2. What is the probability that a form will be completed in exactly 100 minutes?
3. How much time will the top 5% of tax forms take to complete?
4. If someone took 90 minutes to complete the form, what is their Z-score?
5. What time corresponds to the 50th percentile for completing the form?

Tip: For a standard normal distribution, 68% of the data lies within 1 standard deviation from the mean.