Question

In a large population, 76% of the households own microwaves. A simple random sample of 100 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

Answer:

Mean and standard deviation of the sampling distribution of the sample proportions are 76 and 0.0427 respectively.

Step-by-step explanation:

The mean for a sample proportion is given by μ = np

n = sample size = 100

p = fraction of the sample proportion that have what is being tested = 76% = 0.76

μ = 0.76 × 100 = 76.

Standard deviation of a sample proportion = σ = √[p(1-p)/n] = √(0.76×0.24/100) = 0.0427

Answer 2

Answer:

Mean = 76

Standard deviation = 0.043

Step-by-step explanation:

We are given that in a large population, 76% of the households own microwaves. Also, a simple random sample of 100 households is to be contacted and the sample proportion computed.

The above situation can be represented as a Binomial distribution where;

Let X = Proportion of households who own microwaves, i.e;

X ~ $Binom(n=100,p=0.76)$

Here, n = sample size

        p = probability of success

So, the mean of the sampling distribution of the sample proportions is given by, $\mu$ = n * p = 100 * 0.76 = 76 .

Standard deviation of the sampling distribution of the sample proportions, σ = $\sqrt{\frac{p(1-p)}{n} }$ = $\sqrt{\frac{0.76(1-0.76)}{100} }$ = 0.043 .

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