4. Normal Distribution and Sampling Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T1 Normal Distribution and Sampling 1(a) By symmetry, 155 185 1702 +== ( )P 155 0.025 155 170P 0.025 15 1.95996 7.6532 X Z    = −  = − =− = 2 58.6= (3 s.f.) (i) ( ) 2 1 2 3 2 ~ 0,6X X X N +− ( ) ( ) ( ) ( )1 2 3E 2 E E 2E 0X X X X X X+ − = + − = ( ) ( ) ( ) 2 1 2 3Var 2 2Var 4Var 6X X X X X + − = + = ( )1 2 3P 2 5X X X+  + ( )1 2 3P 2 5 0.39484 X X X= + −  = = 0.394 (3 s.f.) (ii) 2 ~, 50XN   (Since X is normally distributed, there isn’t a need to use CLT.) ( )P 172 0.96769 X  = = 0.968 (3 s.f.) (b) Since n is large, by CLT 25~,XN n  approx. ( )P 1 0.99 1P 0.99 5 11P 0.9955 0.2 2.5758 0.2 2.5758 165.87 X Z n Z nn n n n −         −    − −   Least n = 166

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T2 2(a) Let A be the random variable that denotes the weight, in grams, of a bar of Brand A chocolate ( )100,180~ NA Let B be the random variable that denotes the weight, in grams, of a bar of Brand B chocolate ( )400,240~ NB ( )1900,60~2321 NBAAA −++ ( ) ( )022 321321 −++=++ BAAAPBAAAP ( ) )s.f. 3 to(916.002321 =−++ BAAAP (b)(i)      60 100,180~ NA ( ) )s.f. 3 to(781.0179 =AP (ii) No need to use Central Limit Theorem because the population follows a Normal Distribution (c)(i) ( )( ) ( ) ( )( )10002.0,18002.0~02.0 2 NA ( )04.0,6.3~02.0 NA ( ) )s.f. 3 to(159.01586552596.080.302.0 ==AP 3(i) Y = aX + b ~ N ( 230 ,16a b a+ ) P( 4) =P( > 16) 0.06681YY= By symmetry(since the two end tailed areas are equal), 4 1630 10 (1) 2ab ++ = = −−−− 4 30 = 1.5000 24 4 0....(2)4 ab aba −− −  + − = From (1), 10 30ba=− subt. into (2) 24a + 10 – 30a – 4 = 0 1 and 20ab= =− (Shown) Alternative solution : 4 30P ( < ) = 0.066814 abZ a −− 4 30 = 1.5000 24 4 0....(1)4 ab aba −− −  + − = P( > 16) = 0.06681 P( 16) = 0.93319YY  16 30P ( < ) = 0.933194 abZ a −− 16 30 = 1.5000 36 16 0....(2)4 ab aba −−  + − =

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T3 (ii) Y = X – 20 ~ N ( 10 ,16 ) 16~ (10, )YN n P ( 8Y  ) = 0.05 P ( 8 10 4z n − ) = 0.05 8 10 1.6454 10.8 11 n n n − =−  = 4(i) Let X = duration of calls to City A. Let Y = duration of calls to City B. ( ) 2~ 8,2.5XN and ( ) 2~ 10,3.2YN ( ) 2 22 1 2 3 3 ~ 2(8) 3(8),2(2.5 3 (2.5) )X X X N+ − − + ( )1 2 3 3 ~ 8,68.75X X X N + − − ( )1 2 3 33P X X X+ −  ( ) ( )1 2 3 1 2 3 3 3 3 3P X X X P X X X= + −  + + − − = 0.0923122 + 0.726753 = 0.819 (ii) Total cost C = ( )1222 30X X Y++ +30 ( ) ( )( )1222 30 30E C E X X Y= + + +