Question

commute to work: megan drives to work each morning. her commute time is normally distributed with mean 30 minutes and standard deviation 4 minutes. her workday begins at 9:00 a.m. at what time should she leave for work so that the probability she is on time is 90%?
megan should leave for work at
:
a.m.

Explanation:

Step1: Find the z-score for 90% probability

We need to find the z - score corresponding to a cumulative probability of 0.90. Looking at the standard normal distribution table (or using a calculator with inverse - normal function), the z - score \(z\) such that \(P(Z\leq z)=0.90\) is approximately \(z = 1.28\) (from standard normal tables, the value of \(z\) for which the area to the left is 0.90 is about 1.28).

Step2: Use the z - score formula to find the required commute time

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value we want to find (the maximum commute time she can have and still be on time with 90% probability), \(\mu\) is the mean (\(\mu = 30\) minutes), and \(\sigma\) is the standard deviation (\(\sigma = 4\) minutes).

We can rearrange the formula to solve for \(x\): \(x=\mu+z\sigma\)

Substitute \(\mu = 30\), \(z = 1.28\), and \(\sigma = 4\) into the formula:

\(x=30 + 1.28\times4\)

First, calculate \(1.28\times4=5.12\)

Then, \(x=30 + 5.12=35.12\approx35\) minutes (we can round to the nearest minute for practical purposes).

Step3: Calculate the departure time

Her work starts at 9:00 A.M. She needs to account for a maximum commute time of approximately 35 minutes. To find the departure time, we subtract 35 minutes from 9:00 A.M.

9:00 A.M. is 9 hours and 0 minutes. Subtracting 35 minutes: 9 hours and 0 minutes - 0 hours and 35 minutes.

We know that 1 hour = 60 minutes, so we borrow 1 hour (60 minutes) from the 9 hours. So we have 8 hours and 60 minutes - 0 hours and 35 minutes = 8 hours and 25 minutes, which is 8:25 A.M.

Answer:

8:25