Answer: [tex](\hat{p})\sim N(0.10,\ 0.017)[/tex]
Step-by-step explanation:
Sampling distribution of the sample proportion [tex](\hat{p})[/tex] :
[tex](\hat{p})\sim N(p,\ \sqrt{\dfrac{p(1-p)}{n}})[/tex]
The sampling distribution of the sample proportion [tex](\hat{p})[/tex] has mean = [tex]\mu_{\hat{p}}=p[/tex] and standard deviation = [tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex].
Given : The proportion of left handed people in the population is about 0.10.
i.e. p=0.10
sample size : n= 300
Then , the sampling distribution of the sample proportion[tex](\hat{p})[/tex] will be :-
[tex](\hat{p})\sim N(0.10,\ \sqrt{\dfrac{0.10(1-0.10)}{300}})[/tex]
[tex](\hat{p})\sim N(0.10,\ \sqrt{0.0003})[/tex]
[tex](\hat{p})\sim N(0.10,\ 0.017)[/tex] (approx)
Hence, the sampling distribution of the sample proportion[tex](\hat{p})[/tex] is [tex](\hat{p})\sim N(0.10,\ 0.017)[/tex]
