In a large population, 76% of the households have cable tv. A simple random sample of 225 households is to be contacted and the sample proportion computed. What is the probability that the sampling distribution of sample porportions is less than 82%?

Answer:0.017559Step-by-step explanation:Data provided:Probability of Households Having cable TV, p₀ = 76% = 0.76Therefore, The probability that the Households not having cable TV = 1 - 0.76 = 0.24Sample size, n = 225 householdssample proportions is less than 82% i.e p = 0.82Now, The standard error, SE = [tex]\sqrt{\frac{p_0(1-p_0)}{n}}[/tex]or SE = [tex]\sqrt{\frac{0.76(1-0.76)}{225}}[/tex]orSE = 0.02847 and, [tex]Z=\frac{p-p_0}{SE}[/tex]or[tex]Z=\frac{0.82-0.76}{0.02847}[/tex]orZ = 2.107 therefore, P(sample porportions < 0.82) = P(Z < 2.107)now from the p value from the Z tablewe getP(sample porportions < 0.82) =  0.017559