Answer to Question #300651 in Statistics and Probability for Owen

Question

1. Let X be random variable with values 4,5,6,7,8 and 9

a. Find the mean and the variance of the sample without replacement of n=2.

b. Find a the sampling distribution of the mean and construct it's histogram

2. An electrical firm manufactures light bulbs that have mean life span of 750 hours with a standard deviation of 30 hours. Assume that the mean life span of manufactured light bulbs is almost normal distributed. If a sample of 25 light bulbs are random selected, find the:

a. Probability that their mean life span is less than 740 hours.

b. Probability that their mean life span is greater than 775 hours .

Draw the normal distribution curve for each item.

Answer 1

1. We have population values $4,5,6,7,8,9,$ population size $N=6$  and sample size $n=2.$

Thus, the number of possible samples which can be drawn without replacements is $\dbinom{6}{2}=15.$

a.

$$\def\arraystretch{1.5} \begin{array}{c:c} Sample\ values & Sample\ mean\ (\bar{x}) \\ \hline 4,5 & 4.5 \\ \hdashline 4,6 & 5 \\ \hdashline 4,7 & 5.5 \\ \hdashline 4,8 & 6 \\ \hdashline 4,9 & 6.5 \\ \hdashline 5,6 & 5.5 \\ \hdashline 5,7 & 6 \\ \hdashline 5,8 & 6.5 \\ \hdashline 5,9 & 7 \\ \hdashline 6,7 & 6.5 \\ \hdashline 6,8 & 7 \\ \hdashline 6,9 & 7.5 \\ \hdashline 7,8 & 7.5 \\ \hdashline 7,9 & 8 \\ \hdashline 8,9 & 8.5 \\ \hdashline \end{array}$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{x} & f & f(\bar{x}) & \bar{x}f(\bar{x})& \bar{x}^2 f(\bar{x})\\ \hline & 4.5 & 1 & 1/15 & 9/30 & 81/60 \\ \hdashline & 5 & 1 & 1/15 & 10/30 & 100/60 \\ \hdashline & 5.5 & 2 & 2/15 & 22/30 & 242/60 \\ \hdashline & 6 & 2 & 2/15 & 24/30 & 288/60 \\ \hdashline & 6.5 & 3 & 3/15 & 39/30 & 507/60 \\ \hdashline & 7 & 2 & 2/15 & 28/30 & 392/60 \\ \hdashline & 7.5 & 2 & 2/15 & 30/30 & 450/60 \\ \hdashline & 8 & 1 & 1/15 & 16/30 & 256/60 \\ \hdashline & 8.5 & 1 & 1/15 & 17/30 & 289/60 \\ \hdashline Sum= & & 15 & 1 & 6.5 & 521/12 \\ \hdashline \end{array}$$

$$\mu_{\bar{x}}=E(\bar{X})=6.5$$

$$Var(\bar{X})=\sigma_{\bar{x}}^2=E(\bar{X}^2)-(E(\bar{X}))^2$$

$$=521/12-(13/2)^2=7/6$$

b.

$$\begin{matrix} x\ \ \ \ 4.5 & 5 & 5.5 & 6 & 6.5 & 7 & 7.5 & 8 & 8.5\\ \\ p(x)\ \dfrac{1}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{2}{15} & \dfrac{1}{5} &\dfrac{2}{15} & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{1}{15} \end{matrix}$$

2. Let $X=$ the mean life span: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=750 \ h, \sigma=30\ h, n=25.$

a.

$$P(X<740)=P(Z<\dfrac{740-750}{30/\sqrt{25}})$$

$$=P(Z<-5/3)\approx0.0478$$

b.

$$P(X>775)=1-P(X\le 775)$$

$$=1-P(Z\le\dfrac{775-750}{30/\sqrt{25}})$$

$$=1-P(Z\le-25/6)\approx0.000015$$