During a busy time of day, the wait times (in minutes) at an airport security ch

Question

During a busy time of day, the wait times (in minutes) at an airport security checkpoint have a mean of 50.3 minutes and a standard deviation of 13.55 minutes. Use this information to answer the followinq questions.

The standard deviation of the sample means, x¯, for all groups of 43 passengers is (round your answer to exactly four decimal places).

Suppose you take a simple random sample of 43 passengers at this time of the day. The probability that the average wait time (for the passengers in your sample) will be less than 45 minutes is

Suppose you take a simple random sample of 43 passengers at this time of the day. The probability that the average wait time (for the passengers in your sample) will be between 45 minutes and 60 minutes is

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 50.3
standard Deviation ( sd )= 13.55/ Sqrt ( 43 ) =2.0664
sample size (n) = 43

a.
The standard deviation of the sample means, x¯, for all groups of 43 passengers is 2.0664

b.
LESS THAN
P(X < 45) = (45-50.3)/13.55/ Sqrt ( 43 )
= -5.3/2.0664= -2.5649
= P ( Z <-2.5649) From Standard NOrmal Table
= 0.00516

c.
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 45) = (45-50.3)/13.55/ Sqrt ( 43 )
= -5.3/2.06636
= -2.5649
= P ( Z <-2.5649) From Standard Normal Table
= 0.00516
P(X < 60) = (60-50.3)/13.55/ Sqrt ( 43 )
= 9.7/2.06636 = 4.69425
= P ( Z <4.69425) From Standard Normal Table
= 1
P(45 < X < 60) = 1-0.00516 = 0.99484

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