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

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aroranishant799 aroranishant799 · Answer 1 · 2022-07-17T08:11:15+02:00

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 = [tex]\sqrt{} \frac{0.76(1-0.76)}{225}[/tex]

= 0.02847

Z = [tex]\frac{0.82-0.76}{0.02847}[/tex]

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

To know more about standard deviations visit: https://brainly.com/question/475676

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