Normal Distribution Qns and Solns

Chapter 4: Normal Distribution 1. ACJC JC2 Prelim 8865/2019/Q8 The masses, in kg, of lemons and tangerines sold by a wholesaler have independent normal distributions with means and standard deviations as shown in the following table. Mean Standard deviation Lemons 0.18 0.02 Tangerines 0.22 0.03 (i) Find the probability that the mass of a randomly chosen lemon is less than 3% below the mean mass. [2] (ii) Find the mass m such that only 2% of the tangerines have mass more than m kg. [2] (iii) By first stating clearly the mean and variance of the distribution that you use, find the probability that 2 randomly chosen lemons have a total mass not exceeding twice the mass of a randomly chosen tangerine. [3] (iv) A fruit stall owner prepares fruit juice by extracting the pulp of the two fruits. The mass of each lemon pulp is 90% of the mass of each lemon while the mass of each tangerine pulp is 80% of the mass of each tangerine. By first stating clearly the mean and variance of the distribution that you use, find the probability that the total mass of the pulp from 3 randomly chosen lemons and 4 randomly chosen tangerines is between 1.1 kg and 1.2 kg. [3] Answer: (i) 0.394 (ii) 0.282 (iii) 0.886 (iv) 0.511 1. ACJC JC2 Prelim 8865/2019/Q8 (Solutions) (i) Let L be the random variable ‘mass of a lemon’. ( ) 2~ N 0.18,0.02L ( ) ( )P 0.97 0.18 P 0.1746 0.394LL  =  = (ii) Let T be the random variable ‘mass of a tangerine’. ( ) 2~ N 0.22,0.03T ( )P 0.02 0.282T m m =  = (iii) ( ) ( ) ( ) ( ) 2 2 2 12 12 1 2 1 2 2 ~ N 2 0.18 2 0.22, 2 0.02 2 0.03 2 ~ N 0.08,0.0044 P 2 P 2 0 0.886 L L T L L T L L T L L T + −  −   +  + − − +  = + −  = (iv) Let X = 0.9L and Y = 0.8T Let J be the total mass of the pulp from 3 lemons & 4 tangerines. ( ) ( ) ( ) 1 2 3 1 2 3 4 2 2 2 2~ N 3 0.9 0.18 4 0.8 0.22,3 0.9 0.02 4 0.8 0.03 ~ N 1.19,0.003276 P 1.1 1.2 0.511 J X X X Y Y Y Y J J J = + + + + + +   +     +     =

2. ASRJC JC2 Prelim 8865/2019/Q11 A factory manufactures paperweights consisting of glass mounted on a wooden base. The volume of glass, in cm3 and the volume of the wooden base, in cm3 for a randomly selected paperweight follows independent normal distributions with means and standard deviations as shown in the table below: Mean Standard deviation Glass 58.6 2.1 Wooden base 38.5 3.3 (i) Find the probability that the total volume of a randomly selected paperweight is more than 98 cm3 [2] (ii) Find the probability that two paperweights selected at random have volumes that differ by less than 2 cm3 [2] Glass weighs 4.4 grams per cm3 and wood weighs 0.85 grams per cm3. (iii) Find the probability that the total weight of ten randomly selected paperweights is between 2800 and 2900 grams. [4] An events company has a long-standing relationship with the factory. The delivery time (in mins) of a randomly selected trip to the events company follows a normal distribution with mean  and standard deviation  . 10% of the trips take less than 24.3 minutes, while 18% of the trips take longer than 33.3 minutes. (iv) Find the values of  and  . [3] The events company is given a discount if the delivery time is more than k mins. (v) Find the minimum value of k if the events company gets a discount on 5 % of the trips. [1] Answer: (i) 0.409 (ii) 0.282 (iii) 0.426 (iv) 36.3 2. ASRJC JC2 Prelim 8865/2019/Q11 (Solutions) (i) Let X be random variable denoting the volume of glass, and Y random variable denoting the volume of wood, of a randomly selected paperweight. 2(58.6,2.1 )XN and 2(38.5,3.3 )YN Let 22(58.6 38.5,2.1 3.3 ) i.e (97.1,15.3)W X Y N N= + + + Prob ( 98) 0.409PW=  = (ii) 12 (0,15.3 15.3) i.e (0,30.6)W W N N−+ 12( 2 2) 0.282P W W−  −  = (iii) 1 2 10 Mass of one paperweight, 4.4 0.85 Total mass of one paperweights, .... M X Y T M M M =+ = + + ( ) 2 2 2 2(10 4.4 58.6 0.85 38.5 , 10(4.4 2.1 0.85 3.3 ))TN  +   +  (2905.65, 932.45625)TN (2800 2900) 0.426PT   = (iv) Let V be the random variable denoting time taken for a randomly selected trip

2( , )VN  (0,1)VZN   −= ( 24.3) 0.10 24.3( ) 0.10 24.3 1.2815516 (1) 1.2815516 24.3 PV PZ      = −  = − =− −−−−−−  − = ( 33.3) 0.18 33.3( ) 0.18 33.3 0.915365 (2) 0.915365 33.3 PV PZ      = −  = − = −−−−−−  + =  =29.550  = 4.09665 (v) ( ) 0.05 36.288P V k k =  = Minimum k is 36.3 3. DHS JC2 Prelim 8865/2019/Q11 The speeds of an e -scooter ( km/h)X and a pedestrian ( km/h)Y measured on a particular stretch of footpath are normally distributed with mean and variance as follows: mean variance X 12.3 9.9 Y  2 It is known that P( 5.2) P( 7.0) 0.379.YY =  = (i) State the value of  and find the value of . [3] (ii) Given that the speeds of half of the e -scooters measured are found to be within a km/h of the mean, find a. [2] (iii) A LTA officer stationed himself at the footpath and measured the speeds of 50 e -scooters at random. Find the probability that the 50th e-scooter is the 35th to exceed LTA’s legal speed limit of 10 km/h. [3] Answer: (i) 6.1, 2.92 (ii) 2.12 (iii) 0.0468 3. DHS JC2 Prelim 8865/2019/Q11 (Solutions) (i) By symmetry, 15.2 7. 6.0 2 +==

2 P( 5.2) P( 7.0) 0.379 5. 2 2.9 105 2 2 6.1P 0.379 0.9 0.30 9 08 . 81 YY Z     =  = − = = − =− = (ii) ~ (12.3,9.9)XN P( | 12.3 | ) 0.5 P(12.3 12.3 ) 0.5 Xa a X a −  = −   + = From GC, 12.3 10.1777 2.1223 2.12 a a −= == Alternative P( | 12.3 | ) 0.5 9 P( 5 .9 | | ) 0. Xa aZ −  = = P(Z< ) 0.25 9.9 a−= 0.674489 9.9 2.1222 2.12 a a − =− == (iii) Let W be number of e-scooters that exceed speed limit, out of 49 ~ B(49,P( 10))WX  i.e. ~ B(49,0.76761)W Probability required 0.76761 0.0468 P( 34)W==  = 4. CJC JC2 Prelim 8865/2019/Q12 A vegetable stall in the wet market sells carrots at 25 cents per 100 grams and potatoes at 30 cents per 100 grams. The seller gives a 10% discount for customers who spend above $5 before discount. The masses of carrots and potatoes have independent normal distributions with means and standard deviations as shown in the following table. Mean (grams) Standard Deviation (grams) Carrot 65 10 Potato 170 30

(i) Find the probability that a randomly chosen carrot costs more than 20 cents and a randomly chosen potato costs more than 60 cents. [3] (ii) Find the probability that the total cost of a randomly chosen carrot and a randomly chosen potato exceeds $0.80. [4] (iii) Explain why your answers in parts (i) and (ii) differ. [1] (iv) A customer decides to purchase carrots in bulk in order to benefit from the discount. Find the minimum number of carrots he needs to buy such that the probability of qualifying for the discount exceeds 0.5. [4] (v) On a particular day, fourteen customers buy some potatoes at the stall. Find the probability that at least five customers purchase at most 140 grams of potatoes. [3] Answer: (i) 0.0106 (ii) 0.0861 (iv) 31 (v) 0.0576 4. CJC JC2 Prelim 8865/2019/Q12 (Solutions) (i) Let C and D be the random variables denoting the mass, in grams, of a randomly chosen carrot and potato respectively. ( ) ( ) 2 2 ~ N 65,10 ~ N 170,30 C D Method 1: ( ) ( ) ( )( ) 25 30P 20 P 60100 100 P 80 P 200 0.0668072287 0.1586552596 0.0105993182 0.0106 (3 s.f.) CD CD           =   = = = Method 2: ( ) ( ) 2 225 25 25~ N 65 , 10100 100 100 ~ N 16.25,6.25 C     ( ) ( ) 2 230 30 30~ N 170 , 30100 100 100 ~ N 51,81 D     ( )( ) 25 30P 20 P 60100 100 0.0668072287 0.1586552596 0.010599318