Question

In a large population, 67% of the households have cable tv. A simple random sample of 81 households is to be contacted and the sample proportion computed. What is the probability that the sampling distribution of sample porportions is less than 63%?

Answer 1

Answer: ---- Date: ----28/02/2019

Given: p 0.67.q-1-p 0.33 , n 81 Taking normal approximation to binomial, the sampling distribution of sample pro- portion follows normal distribution Shape: Normal p ~ N(μ, σ2), where mean, μ,-p-067 Mp =0.67 variance, σ2 = px(1-2)-0.67x 0.33 standard deviation, σ0-va- 0.0027296296 81 0.67 x0.33 ะ 0.0522458573 81 0.05722458 pN(0.67,0.05722458 2) p0.63 Normal Distribution, μ-0.67 , σ-0.05722458 we convert this to standard normal using μ 0.63-0.67 0.699000325 =-0.69 P(Z <-069-Area to the left of-0.69 0.05 722458一, 0.2451 0.69 P(β < 0.63 )-P(Z <-0.69)-0.2451 (from z-table) P(Z <-0.69): in a z-table having area to the left of z, locate-0.6in the left most column. Move across the row to the right under column 0.09 and get value 0.2451 Using technology, answer is: 0.2451 P(p < 0.63) = 0.2451