Homework SourseA study of the attendance at the University of Alabamas bask
A study of the attendance at the University of Alabama\'s basketball games revealed that the distribution of attendance is normally distributed with a mean of 9,650 and a standard deviation of 1,900.
What is the probability a particular game has an attendance of 13,500 or more? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
What percent of the games have an attendance between 8,000 and 11,500? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
Ten percent of the games have an attendance of how many or less?
| A study of the attendance at the University of Alabama\'s basketball games revealed that the distribution of attendance is normally distributed with a mean of 9,650 and a standard deviation of 1,900. |
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 13500
u = mean = 9650
s = standard deviation = 1900
Thus,
z = (x - u) / s = 2.03
Thus, using a table/technology, the right tailed area of this is
P(z > 2.03 ) = 0.02117827 [answer]
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B)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 8000
x2 = upper bound = 11500
u = mean = 9650
s = standard deviation = 1900
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.87
z2 = upper z score = (x2 - u) / s = 0.97
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.192150202
P(z < z2) = 0.833976754
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.641826552 [ANSWER]
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C)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.1
Then, using table or technology,
z = -1.281551566
As x = u + z * s,
where
u = mean = 9650
z = the critical z score = -1.28
s = standard deviation = 1900
Then
x = critical value = 7218 [ANSWER]
