Question

Question 19 (4 points) The numbers of minutes that travelers arrives at the airport before their scheduled departure are approximately normally distributed with mean 80.5 and standard deviation 9.9. What proportion of travelers arrive at the airport more than 65.0 minutes before their departure? Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage. Your Answer: Answer Question 2014 points) Natural Gas bills in a certainty have 576.75. Assume the bills are normally

Answer 1

Solution:

Given

µ = 80.5

σ = 9.9

We have to find P(X>65) = 1 – P(X<65)

Z = (X - µ)/σ

Z = (65 - 80.5)/9.9

Z = -1.56566

P(Z<-1.56566) = P(X<65) = 0.058715

(by using z-table)

P(X>65) = 1 – P(X<65)

P(X>65) = 1 – 0.058715

P(X>65) = 0.941285

Required probability = 0.9413