The distribution of the number of viewers for the American Idol television show
ID: 3378437 • Letter: T
Question
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?
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?
Explanation / Answer
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]
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