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 co