Suppose that the flying hours for aircraft operated by regional airlines in the U. S. during 2006 are normally distributed with mean 2500 and standard deviation 400. Find the number of flying hours such that about 97.5% of the aircraft has flying hours exceeding this number. A) 1,700
B) 2,100
C) 2.500
D) 2,900
E) 3,300

Step-by-step explanation:

Given that :

Mean (μ) = 2500

Standard deviation (σ) = 400

P = 97.5% = 0.975

Zscore corresponding to p(z > x) = 0.975

Z = - 1.96 ( Z probability calculator)

Using the relation :

Zscore = (score (x) - μ) / σ

-1.96 = (x - 2500) / 400

x - 2500 = 400 * - 1.96

x - 2500 = - 784

x = - 784 + 2500

x = 1716

Hence, number of flying hours required is 1716

if im not mistaking its

step-by-step explanation: