Question

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?

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma\sqrt{n}$

Sample size, n = 37

Standard Deviation =

$=\sigma\times \sqrt{n} = 3.5\times \sqrt{37}=21.28$

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

$P( \sum x < 320) = P( z < \displaystyle\frac{320 - 37(8.4)}{21.28}) = P(z < 0.4323)$

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

$P(\sum x < 320) =0.6672 = 66.72\%$

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