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 probability that the sampling distribution of sample porportions is less than 82%?

Answer 1

Answer:

0.017559

Step-by-step explanation:

Data provided:

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 = $\sqrt{\frac{p_0(1-p_0)}{n}}$

or

SE = $\sqrt{\frac{0.76(1-0.76)}{225}}$

or

SE = 0.02847

and,

$Z=\frac{p-p_0}{SE}$

or

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

or

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