Question

1. Use Table A to find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to each question. a. z<-1.83 b.z > 2.18 c. Z>-0.51 d. z <0.33 e. 1.42<z <2.68 f.-2 <2<1.53

Answer 1

as ZX-1.83 PCZL-1.83) = 0.0336 (from 2-table) Area 0.0336 D -1.83 0 b> P(Z >2-18) = 1-p(222.18) = 1-0-9854... Corom z-tables -0.0146 P(Z > 2.18) - T0-0146 Area 0.0146 2.18 es Z} -0.51 Plzs-0.51)-1-P(Z <-0.51) =) - 0.3050 - 0.6950 ...cfrom za table) P(2}-0.51)-0.6950 Area 0.6950 -0.510 dy a<0.33 P(Z<0.33)=10.6293 (from z-table) Area 0.6293 O 0.33

Area 0.0761 es 1.42<2<2.68 P(1.42<Z22.68 - P(Z <2-68)-P(2<1.42) = 0.9963 0.9222...from z table). =0.0741 P(1.42<<<2.68) -10.0741 1.42 2.68 Area 0.9142 -2 $>-2<x< 1.53 P (-2<2<1.53) -P(221-53)-P(ZZ-2) 0.9370-0.0228 ... (from 2-table) 0.9142 P(-2<2<1:53)-10.9142 1.53 as l=30 o=4 XN Normal cel=30,034) a ļx 240 P(X<40)=P(* 40-3 L 39) 4 - P(Z < 2.5) -0.9938 PCXC40) = 10.9938 (from 2-table) 9 b) P(x + 21) = \-PCX < 20) - |- P(X-4, 2) - 239) =1-P(Z <-2.25) (from a-table) -0.9878 PCX321) = 0.9878 E1-0.0122

ch P(30< x <35) - p(30-4x4235-4 4 = P(30-30 4 2 2 35-30) -P (0 2 <125) - P(7< 1.25) -P(220) - 0.8944-0.500०. chrom 2-tables - 10.394दा

Therefore

a) P(X<40) = 0.9938

b) P(X>21) = 0.9878

c) P(30<X<35) = 0.3944

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