Course Notes - Department of Mathematics and Statistics

Here the mean is n × π = 100 × 0.11 = 11 and the standarddeviation is √ n × π × (1 − π) = √ 100 × 0.11 × 0.89 = 3.13. nπ ±3 √ nπ(1 − π) = 11 ± 3 × 3.13 = 1.61 and 20.39. These values arebetween 0 and n = 100, therefore appropriate to use normal approximationto the binomial here.EXAMPLESuppose that a random sample of 500 students in another 100 levelclass has 70 left handed people. Assuming that the proportion of lefthandedpeople in both papers is 11%, find the probability that 70 ormore from a sample of 500 students are left-handed. What conclusionwould you draw about the proportion of left-handed students in thedifferent papers? Justify your answer.Find the probability that 70 or more students are left handed. π =0.11 n = 500 Find P r(X ≥ 70)Mean = nπ = 500×0.11 = 55 Standard deviation = √ 500 × 0.11 × 0.89 =6.996(3dp)P r(X ≥ 69.5) = 0.0191184