In: Statistics and Probability
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
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