Question

Jeannie is an experienced business traveler, often traveling back and forth from San Francisco to the East Coast several times per month. To catch her flights from San Francisco she leaves her office one hour before her flight leaves. Her travel time from her office to the departing gate at the San Francisco airport, including the time to park and go through security screening, is normally distributed with a mean of 46 minutes and a standard deviation of 5 minutes. What is the probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour? Round your final answer to 4 decimal places.

Use the same information about Jeannie in the above problem. Jeannie is known to be somewhat lax about getting to the airport in time to catch her flight. Suppose that she decides to leave her office so that she has a 96% chance of catching her flight; consequently there is a 4% chance that she will miss her flight. How many minutes before the flight leaves should she leave her office? Round your final answer to the nearest minute, then input that whole number below. DO NOT include units. Just input the numerical answer or the system will mark your answer wrong even if you input the correct value.

Answer 1

Answer:

The probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour is 0.0026.

Jeannie should leave her office 55 minutes early.

Step-by-step explanation:

Let X = time it requires Jeannie to catch her plane.

The random variable X is normally distributed, $N(46, 5)$.

Compute the probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour as follows:

$P(X>60)=P(\frac{X-\mu}{\sigma} >\frac{60-46}{5} )\\=P(Z>2.8)\\=1-P(Z<2.8)\\=1-0.9974\\=0.0026$

The probability that Jeannie will miss her flight because her total time for catching her plane exceeds one hour is 0.0026.

Now, it is provided that Jeannie has 96% chance of catching her flight if she reaches under x minutes.

That is, P (X < x) = 0.96.

Compute the value of x as follows:

$P(X<x)=0.96\\P(\frac{X-\mu}{\sigma}<\frac{x-46}{4})=0.96\\ P(Z<z)=0.96$

**Use the z-table to compute the value of z.

The value of z is 1.751.

Compute the value of x as follows:

$z=\frac{x-46}{5}\\ 1.751=\frac{x-46}{5}\\x=46+(1.751\times5)\\=54.755\\\approx 55$

Thus, Jeannie should leave her office 55 minutes early.