Question

4. In a local university, 10% of the students live in the dormitories. A random sample of 100 students is selected for a particular study

Answer 1

Given that, n=100, p = 0.10 Mean of the sampling distributionis, ll= p = 0.10 The Standard error of the sampling distribution of sampling proportion is, p(1- p) 0.10(1 -0.10) 100 = 0.03 n a) The probability that the sample proportion is between 0.172 and 0.178 is, 0.172 -0.10 P-H P(0.172 < p < 0.178) =P 0.03 = P(2.40 < Z < 2.60) 0.178 -0.10 0.03 69 = P(Z <2.60)- P(Z <2.40) Using the excel function, (NORMSDIST(-)) = 0.9953 -0.9918 - = 0.0035

Plot: 0.0035 0.172 0.178 0.1 х b) The probability that the sample proportion is greater than 0.025 is, P(>0.025) = 1-PCP 50.025) Ô-l 0.025 – 0.10 = 1-P o 0.03 Using the excel function, = 1- P(Z S-2.50) (NORMSDIST (-)) of

= 1-0.0062 = 0.9938 Plot: 0.9938 0.025 0.1