2023 S6 Sampling and Estimation APQ Solutions (RVHS)

Statistics 6 Tutorial: Sampling Additional Practice Questions 1. [N05/P2/26] In a country, 75% of the population have height exceeding 1.50m and 10% have height exceeding 1.90m. Assuming a normal distribution of heights, show that the height exceeded by 20% of the population is 1.81m, correct to 3 significant figures. A random sample of 80 people is taken from the population. Find the probability that the sample mean exceeds 1.69m. Solution Let X be the r.v. that denotes the height of a randomly chosen person from the population. ( ) 2~,XN  Want to find a when ( ) 0.20P X a= ( ) 0.80 1.809811542 1.81 P X a a = = (note: answer should be 0.0111) 2. [UCLES] The random variable X has a normal distribution with mean 2 and variance 3. The random variable S is the sum of 200 independent observations of X, and the random variable T is the sum of a further 300 independent observations of X. Find (i) , (ii) , (iii) The random variable N is the sum of n independent observations of X. Find the approximate values of and as n becomes very large. ( )1.50 0.75 & ( 1.9) 0.1 1.5 1.9( ) 0.25 & ( ) 0.9 1.5 1.9 0.6745 & 1.282 1.6377 & 0.2045 P X P X P Z P Z       =  = −− =  = −− =− = == 20.2045~ 1.6377, 80XN   ( )1.69 0.111PX = ( )P 405S  ( )P 3 40 2ST+ ( )P 2.5 ,Nn ( )P 1.7Nn ( )P 1.7 2.5n N n

Solution (i) ( )1 200... ~ 400,600S X X N= + + ( 405) 0.419PS = (ii) 3 ~ (1200,5400) 2 ~ (1200,3600) 3 2 ~ (0,9000) S N T N S T N− (3 2 40) 0.663P S T−  = (iii) ~ (2 ,3 ) ( 2.5 ) 0.5( ) 0 as 3 N N n n P N n nP Z n  =  → → ( 1.7 ) 0.3( ) 0 as 3 P N n nP Z n  −=  → → (1.7 2.5 ) 0.3 0.5( ) 1 as 33 P n N n nnP Z n  −=   → → 3. [2011 RI/II/12(modified)] The weight, x kg, of each student in a random sample of 120 students from a secondary school is measured, and the results are summarized by , (i) Find unbiased estimates of the population mean and variance. (ii) Another random sample of n students (n ≥ 50) is taken from the school. Given that the probability of the sample mean weight exceeding 49.9 kg is at most 0.01, find the least value of n. Solution (i) Unbiased estimate of population mean = Unbiased estimate of population variance ( 50) 100x− =− 2( 50) 1158x−= ( 50) 10050 50120 149 or 49.167 (5 s.f ) 49.2 (3 s.f )6 x n − −+ = + ==  ( ) 2 2 2 ( 50)1 ( 50)1 1 ( 100) (1158 ) 9.0308 9.03 (3 s.f )119 120 x xnn  −= − −−  −= −  = 

(ii) Since n is large, by Central Limit Theorem, Method 2 : Using table of values Key in Y1 = normalcdf(49.9, E99, 49.167, )  Least value of n = 91 9.0308(49.167, ) approximately.XN n ( 49.9) 0.01 ( 49.9) 0.99 49.9 49.167( ) 0.99 9.0308 49.9 49.167From , 2.3263 9.0308 90.960 Least value of 91 Method 1 : Using Standardization PX PX PZ n GC n n n   − −   = 9.0308 /X n 90 0.01033 > 0.01 91 0.00999 < 0.01