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

0.017559

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{(p_0(1-p_0))/(n)}`

or

SE =
`\sqrt{(0.76(1-0.76))/(225)}`

or

SE = 0.02847

and,

`Z=(p-p_0)/(SE)`

or

`Z=(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