RI S3 Normal Distribution_Add Prac_Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ______________________________________________________ Additional Practice Questions for Chapter S3: Normal Distribution Page 1 of 15 Additional Practice Questions for Chapter S3: Normal Distribution (Solutions) 1 8864/2007/01/Q6 A manufacturer produces packets of margarine. The mass of margarine in a packet has a normal distribution with mean 502g and standard deviation 0.8g. Find the proportion of packets which contain less than 500g of margarine. [2] The manufacturer increases the mean amount of margarine in a packet to  g. The standard deviation remains unchanged. Only 1 packet in 1000, on average, now contains less than 500g. Find  , correct to 1 decimal place. [3] Solution Let X be the mass of a packet of margarine in grams. Then  2N 502, 0.8X Required proportion= P X 500 0.00621 (3 s.f)  Let Y be the mass of a packet of margarine with new mean mass  in grams. Then 2N( ,0.8 )Y  Given that P( 500) 0.001Y    500 P 0.0010.8 From GC, P 3.09023 0.001 500 3.090230.8 500 2.4722 502.5 (1 d.p) Z Z                

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ______________________________________________________ Additional Practice Questions for Chapter S3: Normal Distribution Page 2 of 15 2 9205/1995/02/Q9 (modified) The length of time which an ordinary light-bulb will last may be taken to have a normal distribution with mean 600 hours and standard deviation 100 hours. The length of time for which a new ‘long life’ bulb will last may be taken to have a normal distribution with mean 2000 hours and standard deviation 200 hours. (i) One ordinary bulb is chosen at random. Find the probability that it will last for more than 450 hours. [2] (ii) Two ordinary bulbs are chosen at random. Find the probability that the sum of the times for which they last will be less than 1100 hours. [3] (iii) One ordinary bulb and one long-life bulb are chosen at random. Find the probability that the long-life bulb lasts for more than three times as long as the ordinary bulb. [3] The total time in which 10 ‘long life’ bulbs will last is denoted by W hours. Find the value of t (to the nearest hour) such that P( ) 0.75W t  . [3] Solution (i) Let X be the length of time for which an ordinary light-bulb will last, in hours. Then 2N 600,100X  Let Y be the length of time for which a new ‘long-life’ light-bulb will last, in hours. Then 2N 2000, 200Y  P( 450) 0.933X  (3 s.f) (ii) 2 1 2 N 2(600), 2(100) X X  1 2i.e. N 1200, 20000X X  P( 1 2 1100X X  ) = 0.240 (3 s.f) (iii)   2 2 23 N 2000 3(600), 200 3 (100 ) i.e. 3 N 200, 130000 Y X Y X       P( 3 0) 0.710Y X   (3 s.f)   1 2 10 2 Let ... . Then N 10(2000),10(200) W Y Y Y W      i.e. N 2