(1 point) Once checked in at the ticket kiosk, the amount of time it takes a air
ID: 3314646 • Letter: #
Question
(1 point) Once checked in at the ticket kiosk, the amount of time it takes a airline passenger at the Calgary International Airport to clear security is a random variable that can be modeled by the Exponential distribution with a = 36.7 minutes. A statistician is randomly select n = 33 airline passengers and record how long, in minutes, it takes each to clear security once each has checked in at the ticket kiosk. (a) Complete the statement below. Use at least two decimals in each numeric answer. The distribution of X is exactly Normal $ with a mean = 36.7 minutes and a standard deviation Ox= 6.3886 minutes. (b) What is the probability that the average time for this sample of n 33 airline passengers to clear security is between 25 minutes and 51 minutes? Use at least four decimals in your answer. (c) 94% of the time, the mean time it takes n = 33 airline passengers to clear security after ticket kiosk check-in is at most how many minutes? Enter your answer to at least two decimals. i minutesExplanation / Answer
Solution:
a) Sample mean follows normal distribution Mean of sampling means:
x = = 36.7
Standard deviation of sample mean or standard error:
SE = x = /n = 36.7/33= 6.389
b) Since =36.7 and =36.7 we have:
P( 25<X<51 ) = P( 2536.7/36.7<X/<5136.7/36.7)
= P ( 0.32<Z<0.39 )
= 0.2772
c) P(X < x0) = 0.94
Normal distribution =36.7, =36.7, n = 33
P(X < x0) = 0.94
=> P(X//33 < X//n) = 0.94
=> P(X//n) = 0.95
Use z-table to get invNorm(0.94) = 1.88
=> x0 = 36.7 + 36.7/33 *(1.88) = 48.710662 48.711
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