5. A telephone company has determined that during non-holidays the number of pho

Question

5. A telephone company has determined that during non-holidays the number of phone calls that pass through the main branch office each hour follows the normal distribution with mean u = 80000 and standard deviation o = 35000. Suppose that a random sample of 60 nonholiday hours is selected and the sample mean X of the incoming phone calls is computed. a. Describe the distribution of X. State the distribution type and the parameter values. b. Find the probability that the sample mean X of the incoming phone calls for these 60 hours is larger than 91970. c. Is it more likely that the sample average X will be greater than 75000 hours, or that one hour's incoming calls will be?

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)
a.
mean of the sampling distribution ( x ) = 80000
standard Deviation ( sd )= 35000/ Sqrt ( 60 ) =4518.4806
sample size (n) = 60
b.
P(X > 91970) = (91970-80000)/35000/ Sqrt ( 60 )
= 11970/4518.481= 2.6491
= P ( Z >2.6491) From Standard Normal Table
= 0.004
c.
P(X > 75000) = (75000-80000)/35000/ Sqrt ( 60 )
= -5000/4518.481= -1.1066
= P ( Z >-1.1066) From Standard Normal Table
= 0.8658

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