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= + + + ( )( ) ( )( )22 2 8 30 10 30E= + + = 682 Expected amount of money = 682 cents or $6.82 (iii) ( ) ( )( )1222 30 30Var C Var X X Y= + + + ( )( ) ( )( ) 222222 2 2.5 30 3.2Var=+ = 15266 ( )~ 682,15266CN ( )850 0.0869606 0.0870PC  = = 5(i)(a) Let X be the random variable for the length of a pickle X ~ N(9 , 0.82) P(X < 8 ) = 0.105650 = 0.106 (to 3sf) (b) P(X >10.5) = 0.0303963 = 0.0304 (to 3sf) (ii) Since 10.6% of the pickles are already discarded because they are too short, it is impossible to reduce the total rejects to 5%

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T4 (iii) P(both pickles to meet his guaranteed standard | X1 + X2 < 16 cm) = 12 12 P(both pickles fail his g'teed standard & 16) P( 16) XX XX + + = 2 12 (P( 8)) P( 16) X XX  + …….(***) = 03854989.0 )1056498.0( 2 = 0.290 6(i) Let X denotes the height of a girl ~ (155,100) P(155 155 ) 0.4 P( 155 ) 0.3 P(Z<- ) 0.310 0.5244 5.2410 XN a X a Xa a a a −   + =  − = = − =−  = range of girls’ height is (149.76,160.24) (ii) 22 170 155P( 170) P(Z> ) 0.06680710 4!required probabilty (0.066807) (1 0.066807) 2!2! 0.0233 X − = = =− = (iii) ~ (155 ,100 ) 150 155P( 150 ) P(Z ) 100 1P(Z ) 2 1 (since is large) T T X N n n nnXn n n n − =  = − =

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T5 7 8(i) Let C and W be the random variables denoting the mass of a roll of Cleanex and WoolSoft toilet paper in grams, respectively. ( ) 2N 200,10C ( ) 2N 220,15W ( )P 202 208 0.20888 0.209  = =C (ii) Let 1 2 10 ...A C C C= + + + and 1 2 10 ...B W W W= + + + ( ) ( )( ) ( ) ( )( )( ) 2 2 21.05 N 10 220 1.05 10 200 ,10 15 1.05 10 10BA− − + ( )N 100,3352.5 ( ) ( )P 1.05 P 1.05 0 0.95792 0.958 B A B A = −  = = (iii) ( ) ( )( ) ( ) ( )( )( ) ( ) 22 1 2 3 ~ 10 200 3 10 220 , 10 10 3 10 15 ~ 8600, 7750 A B B B N N + + + + + ( )1 2 3P 0.4A B B B k+ + +  = 8577.7 8580 (to 3 s.f.) k = = Let U be the random variable denoting the mass of a roll of 4-ply toilet paper in grams, respectively.

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T6 ( ) 2N,U  ( )P 220 0.04U = 220P 0.04Z   − = 220 1.7507  − =− 1.7507 220−= ------(1) ( )P 230 0.80U = 230P 0.80Z   − = 230 0.84162  − =− 0.84162 230−= ------(2) 239.26 239== 11.000 11.0 == 9(i) (ii) 10(i) Let X be the waiting time on a randomly chosen day. X~ N(8,5) Let Y be the journey time on a randomly chosen day. Y~N(11,4) Let T be the total time taken on a randomly chosen day. T=X+Y~N(19,9) P(T>20) = 0.369 (to 3 sf) (ii) Expected no. of days late in a month = 30 x 0.369 = 11.07  11 days

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T7 (iii) Let c be the time taken. P(T > c) < 0.05 P(T < c) > 0.95 c > 23.9346 Least value of c = 24 min Thus, the latest time he can leave his house = 7:40am – 24 min = 7:16am (iv) Let T be the average time taken in a month. T ~N(19, 9 / 30 ) P(15< T <20) = 0.966 (to 3sf) 11(a) No. The distribution would be asymmetric or skewed. If it can be modeled by Normal distribution, then there will be approximately the same number of employees earning above and below the mean salary (b)(i) TA ~ N(55, 25), TB ~ N(53, 16) Let T ~ N(54, 10.25) where 2 ABTTT += P(50 < T <60) = 0.864 (3 sig figures). (ii) TB ~ N(53, 16) P( 53 – a  TB  53 + a)  0.6  P(TB  53 + a)  0.8  53 + a  56.366 (from GC)  a  3.366 Hence greatest value of a is 3.36 12(i) Let r.v. A be the mass of a snapper fish and r.v. B be the mass of a pomfret fish. A ~ N(1, 0.12); B ~ N(0.6, 0.052) (a)(i) A1 + A2 + A3 + B1 +B2 ~ N(4.2, 0.035) P[A1 + A2 + A3 + B1 +B2 > 4.5] = 0.0544 (ii) A1 + A2 + A3 – 2B ~ N(1.8, 0.04) P[A1 + A2 + A3 – 2B > 1.85] = 0.401 (iii) 12A + 7( B1 +B2) ~ N(20.4, 1.685) P[12A + 7( B1 +B2) > 21] = 0.322 12(A1 + A2 + …+ An) + 7(B1 +B2 + ….+ B15 – n) ~ N(63 + 7.8n, 1.8375 + 1.3175n) P[12(A1 + A2 + …+ An) + 7(B1 +B2 + ….+ B15 – n) > 150 ] < 0.7. Largest n = 11

River Valley High School, Mathematics Department 2023 Normal Distribution and Sampling T8 13(i) Let A be the mass of an avocado. ( ) 2115 9AN , ( )P 110 115 0 21074X  = . Required prob= ( )( ) 2 3 0 21074 0 5 .. = 0.158 (ii) Let K be the mass of a kiwi. ( ) 282KN  , ( )P 90 0 1055 90 82P 0 1055 8 12508 6 40 X Z = − =  = = . . . (iii) ( ) ( )1 2 1 2 3 2 11509 324 925 5 5 K K A A A N+ + + +   ,. (