Question

Let Z ∼ N(0, 1).

Find a constant c for which P(Z ≥ c) = 0.1587. Round the answer to two decimal places.

Find a constant c for which P(c ≤ Z ≤ 0) = 0.4772. Round the answer to two decimal places.

Find a constant c for which P(−c ≤ Z ≤ c) = 0.8664. Round the answer to two decimal places.

Find a constant c for which P(0 ≤ Z ≤ c) = 0.2967. Round the answer to two decimal places.

Find a constant c for which P(|Z| ≥ c) = 0.1470. Round the answer to two decimal places.

Answer 1

-) Answer: To find c when z N(o, 1) p(2> c) = 0.1587 since p(271) = 0.1587 comparing (1) and (2), | c= 1.00 e) 2 >N(0,1) HA001587 -coo i too plcszso) 0.47721 Here cis must be negative. Pegative . p(270)-0.5. P(cz Z sol = 0.4772 => 1 - P(Z >C) - P(2 0) = 0.4772 1- 0.5 - PC2>) 0.4772. > 0.5 - P(?>c) = 0.4772 ş. 0.5 -0.4772 =pcZ>e) p(2>C) 0.0228 since. P(272) = 0,0228 comparing © & (). 0 . CE -2.001 0,4772 2 NCOV VII -2 -60 0 To

Pi-CS ZSC) = 0.8664 >> 1-2p(Z>c) = 0.8664_ > 1-0, 8664 - 2P(2>C). => 0.1336 = 2 P(2@7c) 32 _p(2>c) 0.9336/2 0) P(Z>c) - 0.0668 since, P(Z >1.50) - 0.0668 0.8664 42 2->N(0,1) comparing.(1) f(2). I ca logo --6150 is to PC05250) = 0.2967 - P(Z > 0) - P(Z >C) = 0.2967_ >> 0.5 - p(230) - 0.2967_ p(270)-0.5 p(Z>c) = 0.5-0.2967 → P(2>C) = 0.2033 - (1) since PC 2 0.83) 0.2033 comparing ) c=0.83 2-> NO) Afo. 2967 --06 o 0.83 too

vl P(1217,C) = 0.1470 > I-p(121<C) = 0.1470 l-0.1470 = P(12]<c) => P(Z <C) = 0.853 >> => - PC-c<Zcc) - 0.853 1-2pcZ>c) = 0.853 2 pcZ>c) = 0.147 PLZ>C) 200735 since, P(Z > 1,45) = 0.0735 comparing f®. co 1.45 - 2->N6,1) -00-C c too outside of region. with probability 0.1470