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In Metro Vancouver, the proportion of households that were renters in 2011 was 0.34. Imagine you constructed a sampling distribution by repeatedly taking samples of size 500. What are the mean and standard deviation of this sampling distribution?

User ARGO
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Answer:
\mu=170\ ;\ \sigma=10.59

Explanation:

Given : The proportion of households that were renters in 2011 : p= 0.34

Sample size : n= 500

The mean and standard deviation for binomial distribution is given by :-


\mu=np\\\\\sigma=√(np(1-p))

Then , the mean and standard deviation of the given sampling distribution :-


\mu=500(0.34)=170\\\\\sigma=√(500(0.34)(1-0.34))\approx10.59

User Brunson
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