The distribution of the number of viewers for the American I

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 26 million with a standard deviation of 8 million. What is the probability next weeks show will: A. Have between 30 and 37 million viewers? B. Have at least 21 million viewers? C. Exceed 49 million viewers?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    30      
x2 = upper bound =    37      
u = mean =    26      
          
s = standard deviation =    8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.5      
z2 = upper z score = (x2 - u) / s =    1.375      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.691462461      
P(z < z2) =    0.915434278      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.223971816   [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    21      
u = mean =    26      
          
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) / s =    -0.625      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.625   ) =    0.734014471 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    49      
u = mean =    26      
          
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) / s =    2.875      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.875   ) =    0.002020137 [ANSWER]