2023 JC2 H2 Normal Distribution APQ Solutions (RVHS)

Statistics 5 Tutorial: Normal Distribution Additional Practice Questions 1. 2011 Prelim/DHS/P2/Q9 In a certain junior college, the marks (out of 100) scored by a JC 1 student in a Class Test, Common Test and Promotional Examination are denoted by C, T and S respectively. C, T and S may be modelled by normal distributions with means and standard deviations as shown in the table below. Type of assessment Mean Standard deviation Class Test, C 68  Common Test, T 65 8 Promotional Examination, S 70 10 (i) Given that P( 85) 0.05,C= determine the value of . In a particular year, a student sits for five Class Tests, a Common Test and a Promotional Examination. The average mark of the five Class Tests constitutes 20% of the overall assessment mark for the year. The Common Test and Promotional Examination consti tute 20% and 60% of the overall assessment mark for the year respectively. For the following parts, assume 10 = . (ii) Find the probability that the average mark scored by a student in the five Class Tests is more than 75. (iii) Find the probability that a student scores an overall mark of more than 80 for the year. (iv) State an assumption used in your calculations for (iii). Solution: (i) 2 2 2 ~ N(68, ) ~ N(65,8 ) ~ N(70,10 ) C T S  P( 85) 0.05 85 68P 0.05 85 68 1.6449 10.335 10.3 (3 s.f.) C Z    = − = − = == (ii) 2 2 1 2 5 ~ N(68,10 ) 10~ N 68,55 C C C CC + + +=  P( 75) 0.058762 0.0588 (3 s.f.)C = =

(iii) Let 0.2 0.2 0.6 E( ) 0.2(68) 0.2(65) 0.6(70) 68.6 X C T S X = + + = + + = 2 2 2 2 2 210Var( ) 0.2 0.2 (8 ) 0.6 (10 ) 39.365 ~ N(68.6,39.36) X X = + + =  P( 80) 0.034601 0.0346 (3 s.f.)X  = = (iv) or , and are independent.C C T S 2. 2010 Prelim/CJC/P2/Q8 The masses of snapper fish and pomfret fish sold by a fishmonger are normally distributed and independent of each other. The mean mass, standard deviation and selling price of snapper fish and pomfret fish are given in the following table: Snapper fish Pomfret fish Mean mass in kg 1 0.6 Standard deviation in kg 0.1 0.05 Selling price per kg in $ 12 7 Find the probability that the (i) total mass of 3 snapper fish and 2 pomfret fish is more than 4.5 kg; (ii) mass of 3 snapper fish exceeds twice the mass of a pomfret fish by more than 1.85 kg; (iii) total selling price of a snapper fish and 2 pomfret fish is more than $21. A customer buys 15 fish, out of which n are snapper fish and the rest are pomfret fish. The probability that the customer pays more than $150 is less than 0.7. Find the largest value of n. Solution: (i) Let X and Y be the r.v. mass of a snapper fish and a pomfret fish respectively. X ~ N(1, 0.12); Y ~ N(0.6, 0.052) X1 + X2 + X3 + Y1 +Y2 ~ N(4.2, 0.035) P[X1 + X2 + X3 + Y1 +Y2 > 4.5] = 0.0544 (ii) X1 + X2 + X3 – 2Y ~ N(1.8, 0.04) P[X1 + X2 + X3 – 2Y > 1.85] = 0.401 (iii) 12X + 7( Y1 +Y2) ~ N(20.4, 1.685) P[12X + 7( Y1 +Y2) > 21] = 0.322 12(X1 + X2 + …+ Xn) + 7(Y1 +Y2 + ….+ Y15 – n) ~ N(63 + 7.8n, 1.8375 + 1.3175n) P[12(X1 + X2 + …+ Xn) + 7(Y1 +Y2 + ….+ Y15 – n)> 150 ] < 0.7. Largest n = 11

3. 2011 Prelim/SAJC/P2/Q10 The operator of Queen Motorways records its weekly earnings from road toll charges according to the categories of vehicles using the road. The weekly earnings (in thousands of dollars) for each category are assumed to be normally distributed. These distributions are independent of one another and are summarised in the table below. Vehicle Category Mean (thousands) Standard deviation (thousands) Cars 120.3 10.4 Buses 69.2 12.5 Lorries 64.5 9.5 Find the probability that the difference in weekly earnings for buses and cars is not more than $60,000. Find the probability that over a 5-week period, the total earnings for lorries exceed $345,000. What assumption must be made in your calculation? Each week, the operator allocates part of the earnings for repairs. This is determined for each category of vehicle according to estimates of long -term damaged caused. It is calculated as follows: x% of earnings from cars, 8% from buses and 15% from lorries. Given that the probability that the total amount for repairs is at least $25,000 in a given week is 0.097, find x. Queen Motorways also records its weekly takings from collection of administration fees from drivers who pay the toll charges by cash. The mean weekly takings is $2000 and the standard deviation of the weekly takings is $800. State, with a reason, whether the weekly takings follows a normal distribution. Solution: Let C and B be random variables “weekly earnings (in thousands of dollars) for cars and buses respectively”. 22~ (120.3,10.4 ) and ~ (69.2,12.5 )C N B N 22~ (51.1,10.4 12.5 ) ~ (51.1, 264.41) ( 60) ( 60 60) 0.708 C B N C B N P C B P C B −+ − −  = −  −  = Let L be random variable “weekly earnings (in thousands of dollars) for lorries”. 2~ (64.5,9.5 )LN 21 2 5 ... ~ (5(64.5),5(9.5 )) (322.5,451.25)L L L N N+ + + = 1 2 5( ... 345) 0.145P L L L+ + +  = The weekly earnings for lorries are independent of one another in the 5-week period.

Let T be random variable “total amount for repairs (in thousands of dollars)”. 2 0.08 0.15100 ~ (120.3 15.211,108.16 3.030625)100 100 xT C B L xxTN = + +    ++       ( ) 2 P 25 0.097 ( 25) 0.903 25 ( )( ) 0.903 () 9.789 120.3 100 1.2988 108.16 3.030625100 From GC, 6.14 T PT ETPZ Var T x x x = = −= − =  + = If a normal distribution is used to model the distribution of weekly takings, we would expect ( )P 2 3(0.8) 2 3(0.8) 0.998X−   + = i.e. ( )P 0.4 4.4 0.998X−   = . However this is not reasonable as there is a significant range of weekly takings that are negative amounts. Hence the weekly takings should not follow a normal distribution. Alternatively, If ( ) 2~ 2000,800XN , ( )0 0.00621PX = (which is too large)

4. 2010 Prelim/TJC/P2/Q9 The times taken, in seconds, for two swimmers, A and B, to complete a 100-metre freestyle race are independent and normally distributed with means 48.0 and 47.2 and standard deviations 0.5 and 0.8 respectively. The two swimmers compete in a 100 - metre race for which the world record is 46.9 seconds. (i) Show that the probability that at least one of the two swimmers breaks the world record during the race is 0.363 correct to 3 significant figures. (ii) Find the probability of Swimmer B beating Swimmer A. (iii) If A and B are to meet 20 times for the 100m freestyle race, how many times do you expect A to beat B? Give your answer correct to the nearest integer. (iv) Find the probability that the total s um of four randomly chosen timings of A is more than 4 times a randomly chosen timing of B. Solution: (i) Let A be the timing of swimmer A for the 100m freestyle. 2~ (48.0, 0.5 )AN Let B be the timing of swimmer B for the 100m freestyle. 2~ (47.2, 0.8 )BN Required Probability ( ) ( ) ( )( ) =1 P(non among the two broke the world r ecord) 1 46.9 46.9 1 0.986097 0.646170 0.36281 0.363 P A P B − = −   = − =  (ii) ~ (0.8,0.89)A B N− ( ) ( )0 0.801781 0.802P A B P A B = −  =  (iii) ( )20 1 0.801781 4 −  Expected times out of 20 is 4 times. (iv) Let 1 2 3 4 4 ~ (3.2,11.24)W A A A A B N= + + + − ( ) ( )1 2 3 4 4 0 0.83008 0.830P A A A A B P W+ + +  =  =  5. 2009/Prelim/RJC/P2/Q8 The random variable X is normally distributed with mean 20 and variance 2 . (i) Given that P(12 ) P( ) 0.42065X k X k  =  = , show that  = 8, correct to the nearest whole number, and find the value of the constant k. X is related to a normal random variable Y by the formula 12 4X Y Y= + − , where 1Y and 2Y are two independent observations of Y. Another random variable W, where W and Y are independent, is normally distributed with mean 25 and vari