Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 110 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in 0 less than 85 minutes? Round your answer to at least four decimal placese
Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 110 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in 0 less than 85 minutes? Round your answer to at least four decimal placese
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![**Problem Statement:**
Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 110 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in less than 85 minutes? Round your answer to at least four decimal places.
**Solution:**
First, calculate the z-score, which standardizes the value in question:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \( X = 85 \) (the value we are comparing)
- \( \mu = 110 \) (the mean)
- \( \sigma = 15 \) (the standard deviation)
\[ z = \frac{85 - 110}{15} \]
\[ z = \frac{-25}{15} \]
\[ z = -1.6667 \]
Next, use the standard normal distribution table or a calculator to find the proportion corresponding to the z-score of -1.6667.
The proportion or probability \( P(Z < -1.6667) \) gives the required proportion of tax forms completed in less than 85 minutes.
After consulting a z-table or using a calculator, you should find:
\[ P(Z < -1.6667) \approx 0.0478 \]
Therefore, approximately 4.78% of the 1040R tax forms will be completed in less than 85 minutes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b2e5bc1-517a-41fd-bb7e-bd7ffddb0842%2Fd310447a-8a4c-4ffa-99d2-7ba599ab2e0c%2Fvbi287d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 110 minutes and a standard deviation of 15 minutes. What proportion of 1040R tax forms will be completed in less than 85 minutes? Round your answer to at least four decimal places.
**Solution:**
First, calculate the z-score, which standardizes the value in question:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \( X = 85 \) (the value we are comparing)
- \( \mu = 110 \) (the mean)
- \( \sigma = 15 \) (the standard deviation)
\[ z = \frac{85 - 110}{15} \]
\[ z = \frac{-25}{15} \]
\[ z = -1.6667 \]
Next, use the standard normal distribution table or a calculator to find the proportion corresponding to the z-score of -1.6667.
The proportion or probability \( P(Z < -1.6667) \) gives the required proportion of tax forms completed in less than 85 minutes.
After consulting a z-table or using a calculator, you should find:
\[ P(Z < -1.6667) \approx 0.0478 \]
Therefore, approximately 4.78% of the 1040R tax forms will be completed in less than 85 minutes.
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