Question

During the 2019 regular Major League baseball season there were 6755 home runs hit where the distance traveled was tracked. The distances of these home runs had a normal distribution with mean=400.21 feet and standard deviation 26.28 feet. Suppose we randomly select 2 home runs, and look at their distances traveled, D1 and D2. a) What is the probability that D1 > 428? b) What is the probability that D1 < 390? c) What is the expected value of D1 + D2? d) What is the standard deviation of D1 + D2? e) What is the variance of the random variable D2 f) What is the expected value of D1 - D2? g) What is the standard deviation of D1 - D2? h) What is the probability that D1 + D2 > 856? i) What is the variance of 10D1?

Answer 1

- 18:1. Let tawo random variables x,y follows a normal distribution with mean " Mz and standard deviation 5, and q respectively. ie. x wN4,802) and Y ~N (42, 82²) then x+y ~ N(Hitz, 5,²+52) and x-y ~ N(M, Hz, s? +62) 2. E(X) = mean , E(X+Y) = EK) I EU) = H, I M2 3. Var (ax) = a Var(x) and Var (x) = (standard deviation) ? 4. If x~ N(4,62) then z=X-4 ~ N(0,1) Also Plac Zab) = $(6) - $(9) - where $(2) = P(Z <2) = 5² €4/2 du 1 and $ez) = 1–0(2) Given. Mean (4) = 400.21 feet, Standard deviation (5)=26-28 feet so D, ~ N(4,02) and Dz ~ N(+1, 62) Z = X © PID, >428) = P(2.>1.06) = 428-400 2 - - P(Z <106) ~ 1.06 -1- 01.06) = 1-0.8554 = 0.1446 And. P (D, <390) = P (2, <-0.39) -0.39 = 0.33) = 1- 40.39) 1-0-6517 = 0.3483 And O ED, + Dz) = E(B) + E(D2) = ut 4=2u = 2x 460.21 = 800.42 so expected value of DTDz is 800.42 feet 26.28 02- 2- 336 - 400.21 - 26.28 Z-92

6 Standard deviation of D, TD2 Since Dithe Wutu, s²+0) i eN(24,202) so Var (0,+Dz) = 20 Standard deviation (D+ DJ = Variance - 1.4140 26.28 2 37.16 Aus © Var (D2) = r² = (26.283 ~ 690.64 And © ED.-D ) = f (OJ – ECD:] = 4-M = 0 ® Since Di-D ~ N(Mall, o²+02) ie. N(6,202) so Var (21-D2) = 202 Standard deviation = variance =sir ~ 37.16 Aug (6 Let x = Dit Dz ~Nutu, o2+02 so X ~ N(24,202) ce. X has mean(st= 2x 400.21 = 800.42 and standard deviation (otr o = 37.16 Now P(D,+D=> 856) = P(Z > 1.49) z = 856-800.42 37.16 =1.49 214 - 1-P(Z <1:49) - - $(149) = -0.9319 = 0.0681 Aug į) Var (10D,) = (10) x Var (D) = 100 x 62 = 100x 690.64 - 69064 And