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navik [9.2K]
4 years ago
15

In Metro Vancouver, the proportion of households that were renters in 2011 was 0.34. Imagine you constructed a sampling distribu

tion by repeatedly taking samples of size 500. What are the mean and standard deviation of this sampling distribution?
Mathematics
1 answer:
GenaCL600 [577]4 years ago
6 0

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

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<h3>Detailed methods used to prove response</h3>

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1. ∠PTA = ∠B {}                           1.  Given

2. ∠PAT = ∠ATB + ∠B  {}           2.  Exterior angle of a triangle theorem

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\displaystyle 1. \hspace{0.3 cm} \frac{GJ}{HK}  = \frac{GK}{GM}                            {}          1.  Given

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