The waiting time until a customer is served at a fast food restaurant during lunch hours...
The waiting time until a customer is served at a fast food
restaurant during lunch hours has a skewed distribution with a mean
of 2.4 minutes and a standard deviation of 0.4 minute. Suppose that
a random sample of 44 waiting times will be taken. Compute the
probability that the mean waiting time for the sample will be
longer than 2.5 minutes.
Answer: (Round to 4 decimal places.)
Homework Answers
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/
2σ^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 ) = 2.4
standard Deviation ( sd )= 0.4/ Sqrt ( 44 ) =0.0603
sample size (n) = 44
the probability that the mean waiting time for the sample will
be longer than 2.5 minutes
P(X > 2.5) = (2.5-2.4)/0.4/ Sqrt ( 44 )
= 0.1/0.06= 1.6583
= P ( Z >1.6583) From Standard Normal Table
= 0.0486
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