Time in min. it takes Jessica to bicycle to school

[okep]

Homework Statement

The time in minutes (min.) it takes Jessica to bicycle to school is normally distributed with mean 15 and variance 4. Jessica has to be at school at 8:00 am. What time shoud she leave her house so she will be late only 4% of the time?
A. 8:00 B) 11.5 min. before 8:00 C) 22 min. before 8:00 D) 18.5 min. before 8:00.
Answer: D) 18.5 min. before 8:00.

Homework Equations

Z = 1.96 for 95% CI
phat = 0.04; qhat = 0.96
+/- Z[phat*qhat/sqrt(n)]
15 +/- Z[phat*qhat/sqrt(n)]

The Attempt at a Solution

15 +/- Z[phat*qhat/sqrt(n)]
15 +/- 1.96[0.04*0.96/sqrt(60)]
= 0.025 min <--incorrect

The answer should be D) 18.5 min. But where did i go wrong? How do i solve this problem? Thanks.

 

[mighty2000]

Your attempt at a solution is close, but there are a few errors. Here is the correct approach:

We want to find the time that Jessica should leave her house so that she is late only 4% of the time. This means that we are looking for the time that corresponds to the 96th percentile of the normal distribution, since we know that the time taken to bike to school is normally distributed with mean 15 and variance 4.

To find the 96th percentile, we can use the standard normal table or a calculator. Using the standard normal table, we find that the z-score corresponding to the 96th percentile is approximately 1.75. This means that 96% of the time, the time taken to bike to school will be less than 15 + 1.75 = 16.75 minutes.

Now, we can use the formula for the 95% confidence interval to solve for the time that corresponds to the 96th percentile:

15 +/- Z[phat*qhat/sqrt(n)] = 15 +/- 1.96[0.04*0.96/sqrt(n)] = 15 +/- 0.025

Since we are looking for the time that corresponds to the 96th percentile, we can set this equal to 16.75 and solve for n:

15 +/- 0.025 = 16.75
0.025 = 16.75 - 15
0.025 = 1.75
n = (1.75/0.025)^2 = 4900

So, Jessica should leave her house 4900 minutes before 8:00 am, which is equivalent to 18.5 minutes before 8:00 am. Therefore, the correct answer is D) 18.5 min. before 8:00.