Answer to Question #314931 in Statistics and Probability for Nick

Question

Consider a population consisting of 4 members whose weights (in pounds) are 148, 156, 176 and 184.

i) Calculate the population mean and the population standard deviation (2marks)

ii) Draw all possible samples of size two, with replacement, from the population. Calculate the mean of each sample and obtain the distribution of the sample means .

Answer 1

X=(148+156+176+184)/4 = 166

$\sigma=\sqrt{\frac{324+100+100+324}{3}}=16.8$

$\mu(148,156)=152$

$\mu(148,176)=162$

$\mu(148,184)=166$

$\mu(156,176)=166$

$\mu(156,184)=170$

$\mu(176,184)=180$

$\mu(148,148)=148$

$\mu(156,156)=156$

$\mu(176,176)=176$

$\mu(184,184)=184$

$\mu=(152+162+166+166+170+180+148+156+176+184)/10=166$

$f(152)=1/10$

$f(162)=1/10$

$f(166)=2/10$

$f(170)=1/10$

$f(180)=1/10$

$f(148)=1/10$

$f(156)=1/10$

$f(176)=1/10$

$f(184)=1/10$

$\sum{\mu f(\mu)}=\sum{15.2+16.2+33.2+17+18+14.8+15.6+17.6+18.4}=166$

$\sum{\mu^2}f(\mu)=\sum{2310.4+2624.4+5511,2+2890+3240+2190+2433.6+3097.6+3385.6}=27682.8$

$Var=\sum{\mu^2f}-(\sum{\mu f})^2=27682.8-166^2=126.8$