Answer to Question #98917 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)

$P(X>27)=P(Z>\frac{27-25}{3})=P(Z>0.67)=1-P(Z<0.67)=\\=0.2514.$

b)

$P(Z<z)=0.97\to z=1.88\to 1.88=\frac{x-25}{3}\to x=31\;min.$

Norman must leave his home at 9:30-31=8:59.