Answer to Question #208323 in Statistics and Probability for Erika Mae Bertiz

Question

Dan works in an insurance company. Last January, he was able to insure 3 persons. Last February, he was able to insure 8 persons. Last March, he was able to insure 10 persons. Last April, he was able to insure 15 persons. Assume that the samples of size are randomly selected without replacement. Find the following:

a. The population mean

b. The population variance

c. The population standard deviation

d. The mean of the sampling distribution of means

e. The standard deviation of the sampling distribution of the sample means

Answer 1

We have population values $3,8,10,15$ population size $N=4$ and sample size $n=2.$

a.

Population mean

$mean=\mu=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}$

$\mu=\dfrac{3+8+10+15}{4}=9$

b.

Population variance

$\sigma^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\mu)^2}{n}$

$\sigma^2=\dfrac{1}{4}\big((3-9)^2+(8-9)^2+(10-9)^2$

$(15-9)^2\big)=\dfrac{74}{4}=18.5$

c.

Population standard deviation

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{74}{4}}=\dfrac{\sqrt{74}}{2}\approx4.301163$

$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$

d.

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

$\dbinom{N}{n}=\dbinom{4}{2}=6$

2.

$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 3,8 & 5.5 \\ \hdashline 2 & 3,10 & 6.5 \\ \hdashline 3 & 3,15 & 9.0 \\ \hdashline 4 & 8,10 & 9.0\\ \hdashline 5 & 8,15 & 11.5 \\ \hdashline 6 & 10,15 & 12.5 \\ \hline \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 5.5 & 1 & 1/6 & 11/12 & 121/24 \\ \hdashline 6.5 & 1 & 1/6 & 13/12 & 169/24 \\ \hdashline 9.0 & 2 & 1/3 & 36/12 & 648/24 \\ \hdashline 11.5 & 1 & 1/6 & 23/12 & 529/24 \\ \hdashline 12.5 & 1 & 1/6 & 25/12 & 625/24 \\ \hdashline Total & 6 & 1 & 9 & 523/6 \\ \hline \end{array}$

The mean of the sampling distribution of means

$E(\bar{X})=\sum\bar{X}f(\bar{X})=9$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$E(\bar{X})=9=\mu$

e.

$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$

$=\dfrac{523}{6}-(9)^2=\dfrac{37}{6}$

The standard deviation of the sampling distribution of the sample means

$\sqrt{Var(\bar{X})}=\sqrt{\dfrac{37}{6}}\approx2.483277$

Verification:

$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{18.5}{2}(\dfrac{4-2}{4-1})$

$=\dfrac{37}{6}, True$