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