Question

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?

Answer 1

Answer: $\mu=170\ ;\ \sigma=10.59$

Step-by-step 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 :-

$\mu=np\\\\\sigma=\sqrt{np(1-p)}$

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

$\mu=500(0.34)=170\\\\\sigma=\sqrt{500(0.34)(1-0.34)}\approx10.59$