Question

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 mean and standard deviation of the sampling distribution of the sample proportions

Answer 1

0.017559

It is given that:

Probability of Households Having cable TV, p₀ = 76% = 0.76

Therefore,

The probability that the Households not having cable TV = 1 - 0.76 = 0.24

Sample size, n = 225 households

sample proportions is less than 82% i.e p = 0.82

Now,

The standard error, SE = $\sqrt{} \frac{0.76(1-0.76)}{225}$

= 0.02847

Z = $\frac{0.82-0.76}{0.02847}$

Z = 2.107

therefore, P(sample porportions < 0.82) = P(Z < 2.107)

now from the p value from the Z table we get P(sample porportions < 0.82) =  0.017559

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