SAJC 2021 Check your understanding (Sampling) Solutions

1 Check your Understanding (Sampling) Distributions of Sample Mean and Sample Sum from a Normal Distribution 1. A random sample of size 15 is taken from a normal distribution with mean 60 and standard deviation 4. Find the probability that the sample mean is less than 58. [0.0264] Solution: 2 2 ~ (60, 4 ) 4~ (60, ) 15 ( 58) 0.0264 XN XN PX  2 PJC H1 Promo 2017/Q5 The mass of a student in Aishan Secondary School is normally distributed and denoted by X kilograms. The masses of the population mean is 56kg and variance is 37.31544. (iii) In the school, every class is made up of 30 students. Find the probability that a randomly-chosen class has mean weight more than 54kg. [2] (iv) A sample of n students is chosen and their weights recorded. Find the least sample size n such that the probability that the mean weight of the students weighing more than 57kg is less than 0.3. Explain briefly if n needs to be large. [4] Solutions: (iii) For a sample of 30, 37.31544~ N 56, 30X    P 54 0.964X  (iv) For a sample of n, 37.31544~ N 56,X n    P 57 0.3 57 56P 0.3 37.31544 X Z n       

2 P 0.3 37.31544 1 P 0.3 37.31544 P 0.7 37.31544 nZ nZ nZ        Using inverse norm,  P 0.5244 0.7Z  Thus, 0.5244 37.31544 10.3 n n   Least n is 11. n need not be large as Central Limit Theorem is not used since X is normally distributed. 3. 2017/IJC/Prelim/Paper I/11(i) A supermarket sells Thailand guava. The masses, in kilograms, of the guava have independent normal distributions. The means and standard deviations of these distributions, and the selling prices, in $ per kilogram, are shown in the following table. Mean (kg) Standard deviation (kg) Selling price ($ per kg) Thailand 0.19 0.01 4.20 Stating clearly the mean and variance of all distributions that you use, find the probability that the average mass of 8 randomly chosen Thai guavas is less than 0.18 kg, [2] Ans: 0.00234

3 Distributions of Sample Mean and Sample Sum from a Non-Normal Distribution 4. A random sample of size 50 is taken from a binomial distribution with number of trials as 9 and probability of success as 0.5. Find the probability that the sample mean exceeds 5. [0.00921] Solution: Since 50n is large, by Central Limit Theorem, 9(0.5)(0.5)~ (9(0.5), ) 50XN 2.25i.e. ~ (4.5, ) 50XN approximately. ( 5) 0.00921PX  5. DHS/2018/Q6 The length of a string is a random variable with mean 15 cm and standard deviation 52 cm. A random sample of 30 strings is taken. Find the probability that the sample mean length lies between 180 mm and 270 mm. [0.0113] Solution: Let X be the length of a string (in cm). E( ) 15,X  Var( ) 52X  Since n is large, by Central Limit Theorem, 52~ N 15, 30X   approximately. P(18 27) 0.0113X   6. The rate of consumption of petrol of a certain model of a car is known to have a mean of 13km per litre and standard deviation of 2km per litre. (i) A random sample of 50 of the same model of cars is taken. Is this sampling distribution of the mean normally distributed? Explain briefly. State the mean and variance of this sampling distribution. (ii) Find the probability of the mean consumption rate to be more than 13.8. 3 Let T be random variable denoting the mass (in kg) of a randomly chosen Thai guava respectively. 2~ N(0.19,0.01 )T 20.01~ N(0.19, ) 8T ~ N(0.19,0.0000125)T P( 0.18) 0.00234T 

4 [(i)13, 2/25 (ii) 0.00234] Solution: (i) Let X be the random variable “rate of consumption of petrol of the car.” Since X is not normally distributed and n = 50 is large, by Central Limit Theorem, X is normally distributed. 22~ (13, ) 50XN approximately (ii) ( 13.8) 0.00234PX  7. David spends time relaxing by watching television everyday. The time X, in hours, which he spends watching television has mean 1.6 hours and standard deviation 1.1 hours. (i) If X denotes the average time over a randomly chosen period of 100 days state the approximate distribution of X and find P( 1.85X  ). (ii) Find the probability that the total time David spend watching television in a year is more than 590 hours. [(i) 0.0115 (ii) 0.00215] Solution: (i) By Central Limit Theorem, 21.1~ (1.6, ) 100XN approximately i.e. ~ (1.6,0.0121)XN approximately ( 1.85) 0.011521 0.0115 PX   (ii) Let Y denote the random variable denoting the total time David spend watching television in a year. ~ (1.6, 365,0.0121 365)YN  ~ (584,4.4165)YN P ( 590) 0.00215Y  8. Rachel is a softball player for her school team, and everyday she attends a training session to train on batting softballs. The probability that she successfully bats a softball is p, and each hit is independent of one another. In each training session, Rachel will do 60 batting attempts. (i) Let X denotes the number of successful hits out of 60 batting attempts in each training session. State, with a reason, the distribution of the mean number of successful hits, X , in 85 training sessions. (ii) It is given that the probability Rachel will have at most 17 successful hits on average in 85 training sessions is 0.5. Find the value of p, and the value of the standard deviation of X . `[(ii) 0.283, 0.379]

5 Solutions: (i) Let Y be the random variable “the number of successfully bats a softball out of 60 attempts.” Then ~ B(60, )Yp E(Y) = 60p Var(Y) = 60p(1 – p) Since the sample size 85 is large, therefore by Central Limit Theorem, 60 (1 )~ N 60 , 85 ppXp     approximately. (ii) Since ( 17) 0.5PX  , therefore E( ) 17X  . Therefore 60 17p 0.283p . Since 0.283p , therefore 60(0.283)(1 0.283)Var( ) 0.143 85X  Therefore standard deviation of X = 0.143 = 0.379 9 [For this question, define all random variables clearly and show all working clearly.] An experiment is conducted where a fair 6-sided die is thrown 8 times and the number of sixes obtained is recorded. Find the expected number of sixes obtained and the variance of the number of sixes obtained. [2] 60 such experiments are conducted. Find the probability that the mean number of sixes obtained is more than 1.5. [3] 9 Let X be the r.v no of sixes obtained for a fair die in 8 throws. 1~ B 8, 6X       14E8 63 1 5 10Var 8 6 6 9 X X         10 4 4 19~ N , ~ N ,3 60 3 54X      approx by Central Limit Theorem since n = 60 large

6  P 1.5 0.110X  (to 3 sf) 10. 2016/FE/IJC/Qn1 At a nature reserve, the weight of sloths is found to have mean 4 kg and standard deviation 0.9 kg. (i) Find the probability that the total weight of 60 randomly selected sloths is at most 250 kg. [3] (ii) If there is a probability o f more than 0.83 that the mean weight of a large sample of n randomly selected sloths is less than 4.1 kg, find the least value of n. [3] Ans: (i) (ii) 74 10(i) Let X be the random variable denoting the weight of a randomly chosen sloth. Consider 1 2 60 ...S X X X    E( ) 60 4 240S    2Var( ) 60 0.9 48.6S    Since the sample size is large (i.e. 60n ), by Central Limit Theorem,  ~ N 240, 48.6S approximately.  P( 250) 0.92428 0.924S    (3 sig fig) (ii) Consider 12 ... nX X XX n    Since the sample size is assumed to be large, by Central Limit Theorem, 20.9~ N 4,X n   Method 1: Setting up a table Using GC, When 73n , P( 4.1) 0.82877 ( 0.83)X    When 74n , P( 4.1) 0.8