Smallville Radio songs on the CDs it plays has a mean length of 135 seconds. Thi

Question

Smallville Radio songs on the CDs it plays has a mean length of 135 seconds. This allows the DJs to have plenty of time for commercials. Assume the distribution of the length of the songs follows a normal distribution with a population standard deviation of 8 seconds. If we select 16 songs from the various CDs: What can we say about the shape of the distribution of sample means? What is the standard error of the mean? What percent of the sample means will be greater than 140 seconds? What percent of the sample means will be greater than 128 seconds? What percent of the sample means will be greater than 128 seconds but less than 140 seconds?

Explanation / Answer

If we select 16 songs from the various CDs: What can we say about the shape of the distribution of sample means? What is the standard error of the mean?

By central limit theorem, it is also normally distributed, with the same mean,

u(X) = 135

and a reduced standard error,

sigm(X) = sigma/sqrt(n) = 8/sqrt(16) = 2 [ANSWER]

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What percent of the sample means will be greater than 140 seconds?

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    140      
u = mean =    135      
n = sample size =    16      
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.5   ) =    0.006209665 = 0.6209665% [ANSWER]

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What percent of the sample means will be greater than 128 seconds?

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    128      
u = mean =    135      
n = sample size =    16      
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -3.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -3.5   ) =    0.999767371 = 99.9767371% [ANSWER]

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What percent of the sample means will be greater than 128 seconds but less than 140 seconds?

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    128      
x2 = upper bound =    140      
u = mean =    135      
n = sample size =    16      
s = standard deviation =    8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -3.5      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.000232629      
P(z < z2) =    0.993790335      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.993557706 = 99.3557706% [ANSWER]      

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