Question

Suppose 56V% of politicians are lawyers. If a random sample of size 787787 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 4%4%

Answer 1

Answer: 0.0238

Step-by-step explanation:

Given : Proportion of politicians are lawyers: p= 0.56

Sample size : n= 787

Let $\hat{p}$ be th sample proportion.

The the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 4% will be :-

$P(|\hat{p}-0.56|>0.04)=P(-0.04>\hat{p}-0.56>0.04)=\\\\ P(-0.04+0.56>\hat{p}>0.04+0.56)\\\\=P(0.52>\hat{p}>0.60)\\\\=P(\dfrac{0.52-0.56}{\sqrt{\dfrac{0.56(1-0.56)}{787}}}>\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.60-0.56}{\sqrt{\dfrac{0.56(1-0.56)}{787}}})\\\\=P(-2.26>z>2.26)\ \ \ [Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]\\\\=1-P(-2.26$

Hence, the required probability = 0.0238