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 [tex]\mu[/tex] and standard deviation [tex]\sigma\sqrt{n}[/tex]
Sample size, n = 37
Standard Deviation =
[tex]=\sigma\times \sqrt{n} = 3.5\times \sqrt{37}=21.28[/tex]
P(taxi and takeoff time will be less than 320 minutes)
[tex]P( \sum x < 320) = P( z < \displaystyle\frac{320 - 37(8.4)}{21.28}) = P(z < 0.4323)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(\sum x < 320) =0.6672 = 66.72\%[/tex]
0.6672 is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
