In: Math
The mean operating cost of a 737 airplane is $2071 per day. Suppose you take a sample of 49 of these 737 airplanes and find a mean operating cost of $2050 with a sample standard deviation of $106.
A.) what is the probability that a 737 will have an operating cost that is greater than the sample mean you have found? (show work)
B.) what is the probability that a plane would have an operating cost that is between $2050 and 2088.60 per day? (show work)
Solution: Given that μ = 2050, σ = 106, n = 49, x = 2071
A. P(X > 2071) = P((X-μ)/(σ/sqrt(n) >
(2071-2050)/(106/sqrt(49)))
= P(Z > 1.3868)
= 1 − P(Z < 1.3868)
= 1 − 0.9177
= 0.0823
B. P(2050 < X < 2088.60) :
= P((2071-2050)/(106/sqrt(49)) < Z <
(2088.6-2050)/(106/sqrt(49)))
= P(1.3868 < Z < 2.5491)
= P(Z < 2.5491) − P(Z < 1.3868)
= 0.9946 - 0.9177
= 0.0769
P(Z < 2.5491) can be found by using the following standard normal table:
P(Z < 1.3868) can be found by using the following standard normal table: