Question #337111

Given the population 3 5,7,9,11,13





How many samples can be made from the population with sample size of 3?





Calculate the mean of the sampling distribution





Compute the variance o the sampling distribution

1
Expert's answer
2022-05-05T09:49:06-0400

The number of possible samples which can be selected without replacement is

(Nn)=N!n!(Nn)!==6!3!3!=45623=20.\begin{pmatrix} N \\ n \end{pmatrix}=\cfrac{N! } {n! \cdot(N-n)! }=\\ =\cfrac{6! } {3! \cdot3! }=\cfrac{4\cdot5\cdot6}{2\cdot3}=20.



Population mean:

μ=3+5+7+9+11+136=8.\mu=\cfrac{3+5+7+9+11+13}{6}=8.


Population variance:

σ2=(xiμ)2P(xi),\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),

Xμ=={38,58,78,98,118,138}=X-\mu=\\ =\begin{Bmatrix} 3-8,5-8,7-8,9-8,11-8,13-8 \end{Bmatrix}=

={5,3,1,1,3,5},=\begin{Bmatrix} -5, - 3,-1,1,3,5 \end{Bmatrix},

σ2=(5)216+(3)216++(1)216+1216+3216+5216=11.667.\sigma^2=(-5)^2\cdot \cfrac{1}{6}+(-3)^2\cdot \cfrac{1}{6}+\\ +(-1)^2\cdot \cfrac{1}{6}+1^2\cdot \cfrac{1}{6}+3^2\cdot \cfrac{1}{6}+5^2\cdot \cfrac{1}{6}=11.667.



Mean of the sampling distribution of sample means:

μxˉ=μ=8.\mu_{\bar x} =\mu=8.


Variance of the sampling distribution of sample means:

σxˉ2=σ2n=11.6673=3.889.\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{11.667}{3}=3.889.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS