The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.4 minutes and a standard deviation of 3.5 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway. What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

0.6672 is the required probability.            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.4 minutes

Standard Deviation, σ = 3.5 minutes

We are given that the distribution of distribution of taxi and takeoff times is a bell shaped distribution that is a normal distribution.

According to central limit theorem the sum measurement of n is normal with mean and standard deviation

Sample size, n = 37

Standard Deviation =

P(taxi and takeoff time will be less than 320 minutes)

Calculation the value from standard normal z table, we have,  

0.6672 is the probability  that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

it is probably the graph on the bottom right.

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.