2023 S6 Sampling and Estimation BMQ Solutions (RVHS)

Statistics 6 Tutorial: Sampling Basic Mastery Questions 1. A random sample of 250 adult men undergoing a routine medical inspection had their heights (x cm) measured to the nearest centimeter, and the following data was obtained . (a) Calculate an unbiased estimates of the population mean and variance. (b) Describe the process of how the random sample could be obtained, assuming that the random sample was from a population of 1000 adult men, context being from a particular company’s routine medical check-up. Solution (a) 43205 172.82 173250 xx n= = =  ( ) ( ) 2 2 22 4320511 7469107 9.7144 9.711 249 250 x sx nn  = − = − =  −    (b) The random sample could be obtained as follows: (1) Create a complete list o f all members of the population from the human resources department. (2) Assign each member a unique number (1,2,3, … , N) (3) Choose a random sample by generating 250 random numbers (via a GC, computer or handheld calculator). Note the numbers and collect data of the men assigned those selected numbers. 2. (i) The time taken for a group of people to complete a survey is normally distributed with mean 10 minutes and standard deviation 2 minutes. 5 individuals are selected at random. For this sample, calculate the probability that the mean time for completing the survey is shorter than 9 minutes. (ii) A random sample of 40 observations is selected from a population with mean 20 and standard deviation 2.9. Given that is the sample mean, find (a) , (b) , (c) the value of a if .  == 1074697and20543 2xx X ( )P 18.5X  ( )P 14 19.5X ( )P 0.05Xa=

Solution (i) Let T be the time taken for a person to complete a survey. Since T is normally distributed, then T is normally distributed as well. 22~ N 10, 5T   ( )P 9 0.131776... 0.132T  = = (3 s.f.). (ii) {Note that the population distribution is not known. But n is large so CLT applies.} By CLT, 22.9~ N 20, 40X   approximately (a) (3 s.f.) (b) (3 s.f.) (c) a = 20.8 (3 s.f.) a = 20.8 P( 18.5) 0.9994647 0.999X  = = P(14 19.5) 0.1377595 0.138X  = = P( ) 0.05 P( ) 0.95X a X a =   =