The proportion of left handed people in the population is about 0.10. Suppose a random sample of 300 people is observed. (a) What is the sampling distribution of the sample proportion (p-hat)

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=0.10sample size : n= 300Then , 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]