In a recent test the mean time to log into the hertz website

In a recent test, the mean time to log into the hertz website through a smartphone was 7.524 seconds. Suppose that the download time is normally distributed, with a standard deviation of 1.7 seconds. What is the probabilty that a download time is:

P(X < 2)

P(1.5 < X < 2.5)

P(X > 1.8)

P(A < X) = 0.99. What is A?

Solution

We use excel function to find teh probability

P(X<2) = 0.0006. The excel function used is =NORMDIST(2,7.524,1.7,1)

P(1.5<X<2.5) = 0.0014 The excel function used is =NORMDIST(2.5,7.524,1.7,1)-NORMDIST(1.5,7.524,1.7,1)

P(X>1.8) = 0.99962. The excel function used is =1-NORMDIST(1.8,7.524,1.7,1)

P(A<X)=0.99 . That is P(X>A) = 0.99 . A = 3.569. The excel function used is =NORMINV(1-0.99,7.524,1.7) We take 1-0.99 because norminv return cumulative normal distribution and it is X>A. 0.99 is area on the right of A