The amount of time that a customer spends waiting at an airport
check-in counter (X) is a random variable with
mean 8.2 minutes and standard deviation 2.5 minutes. If a random
sample of n = 30 customers were observed, then
what is the probability that the sample average waiting time
exceeds 9 minutes? Show your work.
Solution :
Given that ,
mean = = 8.2
standard deviation = =2.5
n = 30
= 8.2
= / n =2.5 / 30= 0.4564
P( >9 ) = 1 - P( <9 )
= 1 - P[( - ) / < (9-8.2) /0.4564]
= 1 - P(z < 1.75)
Using z table
= 1 - 0.9599
= 0.0401
probability= 0.0401
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