Homework SourseThe 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.
Have between 30 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 49 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
| 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. |
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 = 1.38
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.691462461
P(z < z2) = 0.916206678
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.224744216 [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 = -0.63
Thus, using a table/technology, the right tailed area of this is
P(z > -0.63 ) = 0.735652708 [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 = 2.88
Thus, using a table/technology, the right tailed area of this is
P(z > 2.88 ) = 0.001988376 [ANSWER]
