Question

Problem 4 The average air speed of an F-22 and F-15E fighter jets are 2778.3 km/h and 3087 km/h, respectively. The F-22 and F-15E jets have standard deviations of 400.75 km/h and 495.45 km/h, respectively. The air force is planning on inspecting 20 of eadch variant of aircraft. Determine P(R F15-E- X F22) > 250 Note we are not interested in the probability that P(R F15-E-K F22)< -250

Answer 1

4) The sample mean is approximately normally distributed according to Central Limit Theorem (CLT).

Thus,

XF15-E-XF22 N(3087-2778.3,495.454+400.75*) N(308.7,495.452+ 400.752)

The probability,

250 308.7 P XF15-E-XF22> 250 PZ> V495.452 400.752 P (XF15-E XF22> 250 P (Z> -0.092116403) P (XF15-E-XF22> 250 (0.092116403) P (XF15-E-Xf22> 250 = 0.5367

We are required to solve only one question. Please post the remaining questions as another post.