The distribution of the number of viewers for the American Idol television show
ID: 3155023 • Letter: T
Question
The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 27 million with a standard deviation of 10 million.
Have between 29 and 37 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
Have at least 21 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
Exceed 43 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
What is the probability next week's show will:
Explanation / Answer
A)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 29
x2 = upper bound = 37
u = mean = 27
s = standard deviation = 10
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.2
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.5793
P(z < z2) = 0.8413
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.262 [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 = 27
s = standard deviation = 10
Thus,
z = (x - u) / s = -0.6
Thus, using a table/technology, the right tailed area of this is
P(z > -0.6 ) = 0.7257 [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 = 43
u = mean = 27
s = standard deviation = 10
Thus,
z = (x - u) / s = 1.6
Thus, using a table/technology, the right tailed area of this is
P(z > 1.6 ) = 0.0548 [ANSWER]
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