Answer to Question #98918 in Statistics and Probability for Preet

Question

The time it takes for Norman to travel from his house to the 10:00 AM MATH 1F92 lecture is
approximately normal with a mean of 25 minutes and standard deviation of 3 minutes.
(Note: Labelled diagrams and proper notation are required for all parts.)
a) One morning, Norman leaves his house at 9:40 AM. The professor is going to give hints about
the test during the first 7 minutes of lecture. What is the probability that he will miss all the hints?
b) On the morning of the test, Norman wants to arrive by 9:30 AM. What time must Norman leave
his home to ensure a 97% probability that he will arrive before 9:30 AM?

Answer 1

a) He missed all the hints, that means he reached the class after 10:07

That means total time taken by him in reaching class is more than or equal to 27 minutes as he started at 9:40 and reached at or after 10:07

So, we need to find $P(X \ge 27)$

First we convert it into Z score, then using normal distribution table, we will find the required probability.

So,

$Z= (x-\mu)/\sigma$

Where the mean is 25 minutes and the standard deviation is 3 minutes.

Putting up the value, we will get

$Z=(27-25)/3$

$=2/3=0.667$

From the normal probability distribution table, it can be seen that $P(Z \le 0.667)=0.748$

Hence the $P(Z \ge 0.667)=1-0.748$

$=0.252$

b) Given that $P(Z \le z)=0.97$

From the normal distribution table we can find the value of z to be 1.881

So let's suppose the time taken by him to reach office is x minutes.

Then,

$(x-25)/3=1.881$

$x=(3*1.881)+25$

$x=30.643$

So, he must leave his home by 9:00 to reach the lecture by 9:30 with 97% probability.