Suppose that the time required to complete a 1040R tax form is normal distributed with a mean of 100 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in less than 119 minutes? Round your answer to at least four decimal places.
 


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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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.

Suppose that the time required to complete a 1040R tax form is normal distributed with a mean of 100 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in less than 119 minutes? Round your answer to at least four decimal places.

Find the proportion of 1040R tax forms completed in less than 119 minutes using the normal distribution. Given a mean of 100 minutes and a standard deviation of 15 minutes, calculate the Z-score and determine the percentage. Suitable for students in Grades 10-12.

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