Question

In a large population, 82% 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

Mean = 0.82

Standard deviation = 0.0256

Explanation:

From the question, we are given:

n= 225

p = 0.82

For a normally distributed random variable X, the sample means can be said to be normal distribution with mean and standard deviation..

Therefore, the mean of the sampling distribution of the sample proportions will be:

p = 0.82

Now, the standard deviation of the sampling distribution of the sample proportion, will be:

`= \sqrt(p(1-p))/(n)`

`= \sqrt(0.82(1-0.82))/(225)`

= 0.0256

Therefore, the mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.

Answer 2

The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.

Explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, the mean is
`\mu = p` and the standard deviation is
`s = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`p = 0.82, n = 225`.

So

`\mu = 0.82`

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.82*0.18)/(225)} = 0.0256`

The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.