Suppose that the average number of miles driven by a CSULA students is 15,000 and the...

Suppose that the average number of miles driven by a CSULA students is 15,000 and the standard error is 5000 miles. Find the interval of miles within which 99% of the drivers fall. Please show all your work

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Expert Solution

solution:

middle 99% of score is

P(-z < Z < z) = 0.99

P(Z < z) - P(Z < -z) = 0.99

2 P(Z < z) - 1 = 0.99

2 P(Z < z) = 1 + 0.99 = 1.99

P(Z < z) = 1.99/ 2 = 0.995

P(Z <2.58 ) = 0.995

z ± 2.58

Using z-score formula

= +z * +

= 2.58*5000+15000

= 27900

=- z * +

= -2.58*5000+15000

= 2100

orchestra answered 1 year ago

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