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CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 138 seconds (2

ID: 3159998 • Letter: C

Question

CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 138 seconds (2 minutes and 18 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 7 seconds. Suppose we select a sample of 17 cuts from various CDs sold by CRA CDs Inc.

a. What is the standard error of the mean? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Standard error of the mean _______ seconds

b. What percent of the sample means will be greater than 140 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Percent ____%

c. What percent of the sample means will be greater than 133 seconds? (Round your answer to 2 decimal places.)

Percent ____%

d. What percent of the sample means will be greater than 133 but less than 140 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Percent ____%

Explanation / Answer

a.
Mean ( u ) =138
Standard Deviation ( sd )=7
Number ( n ) = 17
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  

Standard error = sd/Sqrt(n) = 7/ Sqrt ( 17 ) = 1.698

b.
P(X > 140) = (140-138)/7/ Sqrt ( 17 )
= 2/1.698= 1.178
= P ( Z >1.178) From Standard Normal Table
= 0.1194                  

11.94% sample means will be greater than 140 seconds

c.
P(X > 133) = (133-138)/7/ Sqrt ( 17 )
= -5/1.698= -2.9451
= P ( Z >-2.9451) From Standard Normal Table
= 0.9984                  

99.84% percent of the sample means will be greater than 133 seconds

d.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 133) = (133-138)/7/ Sqrt ( 17 )
= -5/1.6977
= -2.9451
= P ( Z <-2.9451) From Standard Normal Table
= 0.00161
P(X < 140) = (140-138)/7/ Sqrt ( 17 )
= 2/1.6977 = 1.178
= P ( Z <1.178) From Standard Normal Table
= 0.88061
P(133 < X < 140) = 0.88061-0.00161 = 0.879                  

87.9% percent of the sample means will be greater than 133 but less than 140 seconds