RI S3 Normal Distribution Tutorial Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ___________________________ Tutorial S3: Normal Distribution Page 1 of 11 Tutorial S3: Normal Distribution 1 Given that ~ N(1.5,2.5)X , evaluate (a) P( 0.7)X , (b) P( 3.7)X , (c) P(1.1 2.3)X  , (d) P(| | 0.5)X , (e) P(| | 0.5)X (f) P( 1.5 2.5)X  . Solution (a) to (d) can be evaluated from GC directly. (e) P(| | 0.5)X 1 P( 0.5 0.5) 0.839 (3 s.f) X      Or: P(| | 0.5) P( 0.5) P( 0.5)X X X     (f) P( 1.5 2.5)X  P( 2.5 1.5 2.5) P(1.5 2.5 1.5 2.5) 0.683 (3 s.f) X X            Or: 1.5P( 1.5 2.5) P( 1) P( 1) 0.6832.5 XX Z        2 Given that ~ N(22,5)Y and ~ N(0,1)Z , find the value(s) of k such that (a) P( ) 0.35Y k  , (b) P( ) 0.8Y k  , (c) P( 25) 0.46k Y   , (d) P(| | ) 0.27Z k  , (e) P(22 22 ) 0.67k Y k     , (f) P( ) 0.3Y k  . 1.5 1.5 +2.5 1.5 2.5

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 2 of 11 Solution ~ N(22,5)Y and ~ N(0,1)Z (a) P( ) 0.35Y k  From G.C, 21.1 (3 s.f)k k = invNorm(0.35,22,5, LEFT) = 21.1 (b) P( ) 0.8Y k  From G.C, 20.1 (3 s.f)k k = invNorm(0.8,22,5,RIGHT) = 20.1 (c) P( 25) 0.46k Y   From the diagram, P( 25) P( ) 0.46 0.91014 P( ) 0.46 P( ) 0.45014 Y Y k Y k Y k          From G.C, 21.7 (3 s.f)k k = invNorm(0.4501,22,5,LEFT) = 21.7 (d) P(| | ) 0.27Z k  From the diagram, P( ) 0.73Z k  From G.C, 1.10 (3 s.f)k k = invNorm(0.73,0,1,CENTER) =1.10 25 22 0.46 k k 22 0.8 0.35 k 22 k 0 0.135 0.135 k

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 3 of 11 (e) P(22 22 ) 0.67k Y k     From G.C, 22 19.822 and 22 24.178 2.18 (3 s.f) k k k       using invNorm(0.67,22,5,CENTER), 22 – k =19.822 and 22 + k = 24.178 (f) P( ) 0.3Y k  From G.C, P( 20.8) 0.3Y  From the diagram, 20.8 (3 s.f)k invNorm(0.3,22,5,LEFT) = 20.8 22 0.165 0.67 22+k 22 k 0.165 0.3 20.8 22

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 4 of 11 3 The random variable X follows a normal distribution with mean  and standard deviation . Given that P( 20) 0.14X  and that P( 50) 0.65X  , calculate the values of  and . Solution 2~ N( , )X   . P( 20) 0.14X  20 P 0.14Z         From GC, P( 1.0803) 0.14Z   20 1.0803    (5 s.f.) 1.0803 20    ----- (1) P( 50) 0.65X  50 P 0.65Z         From GC, P( 0.38532) 0.65Z  50 0.38532    (5s.f.) 0.38532 50    ----- (2) invNorm(0.14,0,1,LEFT) =1.0803 invNorm(0.65,0,1,LEFT) =0.38532 Solving (1) and (2), 42.1 (3 s.f.) and 20.5 (3 s.f.). 0.38532 0.65 0 0.14

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 5 of 11 4 The independent random variables X and Y are each normally distributed with means 6 and 8 respectively, and variances 9 and 16 respectively. Find (a) (i) P( )X Y (ii) P( 1.5)X Y  . (b) 1 2, X X and 3X are three independent observations of X and 1 2 and Y Y are two independent observations of Y. Find (i) 1 2 3P( 15)X X X   , (ii) 1 2 1 2P( )X X Y Y   , (iii) 1 2P(2 )X Y Y  . Solution Given 2~ N(6,3 )X and 2~ N(8,4 )Y (a)(i) E( ) E( ) E( ) 6 8 2X Y X Y      Var( ) Var( )X Y X  +Var( ) 9 16 25Y    ~ N( 2, 25)X Y   P( )X Y P( 0) 0.655 (3 s.f)X Y    (ii) P(| | 1.5)X Y  P( 1.5 1.5) 0.218 (3 s.f) X Y      (b)(i) 1 2 3E( ) 3E( ) 18X X X X    1 2 3Var( ) 3Var( ) 27X X X X    1 2 3 ~ N( 18, 27 )X X X   1 2 3P( 15)X X X   0.282 (3 s.f) (ii) 1 2 1 2E ( ) 2E( ) 2E( ) 12 16 4X X Y Y X Y        1 2 1 2Var ( ) 2Var( )X X Y Y X    +2Var( ) 2(9) 2(16) 50Y    1 2 1 2 ( ) ~ N( 4,50)X X Y Y     1 2 1 2P( )X X Y Y   1 2 1 2P( ( ) > 0)X X Y Y    0.286 (3 s.f) (iii) 1 2E 2 ( ) 2E( ) 2E( ) 12 16 4X Y Y X Y       2 1 2Var 2 ( ) 2 Var( )X Y Y X   +2Var( ) 4(9) 2(16) 68Y    1 2 2 ( ) ~ N( 4,68)X Y Y    1 2P(2 )X Y Y  1 2=P(2 ( ) 0)X Y Y   0.314 (3 s.f)

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 6 of 11 5 HCI Prelim 9740/2007/02/Q7 (ii) (modified) The random variables X and Y are normally distributed with known means and variances. A student was asked to find the probability that sum of 3 independent observations of X differ from 3 times an observation of Y by at least 1. State an assumption that the student has to make to solve the question. Part of the student’s solution to this question is given as follows: 3 3 ~ N 3E 3E , 9Var 9VarX Y X Y X Y   P 3 3 1 ....X Y   Point out 3 mistakes that the student has made. [3] Solution The student has to assume that X and Y are independent. (1) It should be 1 2 3X X X  instead of 3X (2) It should be 3Var(X) + 9 Var(Y) instead of 9Var 9VarX Y (3) It should be 1 2 3P 3 1X X X Y    instead of P 3 3 1X Y  6 9233/2005/02/Q23 The pineapples in a large warehouse have masses with a normal distribution. The mean of this distribution is 1.52 kg and the standard deviation is 0.070 kg. Six randomly chosen pineapples are packed in a box of mass 2.15 kg. Find the probability that the total mass of the filled box is less than 11.2 kg. Find the probability that, out of 3 randomly chosen filled boxes, at least 1 has total mass less than 11.2 kg. [4] [0.342 , 0.715] Solution Let X be the mass of a pineapple in kg. Then 2N(1.52,0.07 )X  . Let Ybe the mass of a filled box of pineapples in kg. Then 1 2 6 2.15Y X X X         2 E 6E 2.15 6 1.52 2.15 11.27 Var 6Var 6 0.07 0.0294 N(11.27,0.0294) Y X Y X Y          P( 11.2) 0.34155 (5s.f.) 0.342 (3s.f.)Y   . Let Wbe the number of boxes, out of 3, with total mass less than 11.2kg. B(3,0.34155)W P( 1) 1 P( 0) 0.715 (3 s.f.). W W      OR: Let 1 2 6T X X X    Then N(9.12,0.0294)T  P( 11.2 2.15) P( 9.05) 0.342 (3s.f.) T T     

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________________ Tutorial S3: Normal Distribution Page 7 of 11 7 VJC Prelim 9758/2020/02/Q8 A company sells bags of cement in two sizes. The mass of a large bag of cement can be modelled by a normal distribution with mean 50 kg and standard d