Question

Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.) (a) P(0 ≤ Z ≤ 2.13) (b) P(0 ≤ Z ≤ 1) (c) P(−2.20 ≤ Z ≤ 0) (d) P(−2.20 ≤ Z ≤ 2.20) (e) P(Z ≤ 1.93) (f) P(−1.15 ≤ Z) (g) P(−1.20 ≤ Z ≤ 2.00) (h) P(1.93 ≤ Z ≤ 2.50) (i) P(1.20 ≤ Z) (j) P(|Z| ≤ 2.50)

Answer 1

T ZWN0o, non Z @ PCO<2<2.13) - A et =(A+B)-B . a 7 B 2.13 =P(zzo) - P(Z 72.13) 0.5 -0.016586 = 0.48346 6 P osz=1) = P(270) - P(Z >1) = 0.5- 0. 15866 = 0. 34134

OP(-2.2 czco) - B A+B+CCI = 1 - A-C - P(Z < -2.2) - P(Z > - P(Z > 2 2)- P(Z >O) due to symmetry -22 C = 1 0 oligo3 -05 = 0.4861 @ p(-2.2 - - A-C Z32.2) =B A = - P(aL-2.2) - P(2722) due to symmetry - 2 2 2.2 2 = 1- P(Z > 2.2) - P(Z 72.2) =1-0.013903 - 0013903 - 0.97 2194 P(Z < 1.93) = 1 - P(Z >1.93) due to symmetry = -0.026803 - 0973197 1.93 6 P(-1.1557) - ro P(Z7-1.15) - 1- P(Z <-1.15) 1- [P(Z > 1.15))

1 - 0.12507 087493 - 1.20 9 P(-1. 202Z < 2.00)=B 1-A-C = 1 - P(Z <-1.2) - P(Z72) due to symmetry - P(Z > 1.2) - P(Z > 2) = 1 -0.11507 - 0.022750 = 0.86218 6 Pll.gos z <2.5) P(Z > 1.90) - P(Z > 25 = 0.028717- 0.00 6 2097 = 0.0225013 1.go 2.5 1 P(1.2002) = P(Z Z 1.20) = 0.11507 1:20

1) POIZI<2.50) = P(Z > 2.5) +PCZL-25) = P(Z > 2.5) + P(Z > 2 .5 due to symmetry = 0.00 62097 +0.0062091 -0.0124194 - 2.5 2 .5