The taxi and takeoff time for and takeoff times is approximately normal. You may

Question

The taxi and takeoff time for 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 jets is a random variable x with a mean of 8.5 minutes and a of taxi (a) What is the probability that for 36 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places) (b) What is the probability that for 36 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.) (c) What is the probability that for 36 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.) Need Help?

Explanation / Answer

Mean taxi and takoff time = 8.5 minutes

Standard deiation of taxi and takeoff time = 2.8 minutes

Sample size = 36

(a Mean of Total taxi and takeoff time of 36 jets T= 36 * 8.5 = 306 minutes

Variance of Total taxi and takeoff time of 36 jets = 36 * 2.82 = 282.24 mintes

Standard deviation of Total taxi and takeoff time of 36 jets T= 16.8 minutes

Pr(Total time < 320 mins; 306 mins ; 16.8 mins) = ?

Z = (320 - 306)/ 16.8 = 0.8333

so Pr( T < 320 ; 306; 16.8) = Pr(Z < 0.8333) = 0.7977

(b) Pr(Total time > 275 mins; 306 mins ; 16.8 mins) = 1 - Pr(Total time < 275 mins; 306 mins ; 16.8 mins)

Z = (275 - 306)/ 16.8 = -1.8452

so Pr( T > 275 ; 306; 16.8) = 1 - Pr(Z < -1.8452) = 1 - 0.0325 = 0.9675

(c) Pr (275 minutes < X < 320 minutes ; 306 mins ; 16.8 mins) = Pr(Total time < 320 mins; 306 mins ; 16.8 mins) - Pr(Total time < 275 mins; 306 mins ; 16.8 mins)

= 0.7977 - (1 - 0.9675)

= 0.7652

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