CRA CDs Inc wants the mean lengths of the cuts on a CD to be

CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 146 seconds (2 minutes and 26 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 11 seconds. Suppose we select a sample of 19 cuts from various CDs sold by CRA CDs Inc.

What percent of the sample means will be greater than 150 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

What percent of the sample means will be greater than 139 seconds? (Round your answer to 2 decimal places.)

What percent of the sample means will be greater than 139 but less than 150 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Solution

a)

By central limit theorem, it is NORMAL.

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b)

standard error = s/sqrt(n) = 11/sqrt(19)

= 2.523573073 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    150      
u = mean =    146      
n = sample size =    19      
s = standard deviation =    11      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.585054161      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.585054161   ) =    0.056477015 or 5.6477015 % [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    139      
u = mean =    146      
n = sample size =    19      
s = standard deviation =    11      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.773844782      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -2.773844782   ) =    0.997230095 or 99.7230095% [ANSWER]

E)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    139      
x2 = upper bound =    150      
u = mean =    146      
n = sample size =    19      
s = standard deviation =    11      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.773844782      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.585054161      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.002769905      
P(z < z2) =    0.943522985      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.94075308 or    94.075308% [ANSWER]