Answer to Question #311890 in Statistics and Probability for opalyn

Question

A test taken by five people has a maximum score of 100. Their scores are 78, 65, 86, 91 and 68. 

1. Compute for the mean of the population.

2. Compute for the variance of the population.

3. List all the possible samples of size 3 that can be drawn from the population and compute for the mean of each sample. Use SRSWOR method.

4. Construct the probability distribution of the sample means.

5. Compute the mean of the sampling distribution of the sample means.

6. Compute the variance of the sampling distribution of the sample means

Answer 1

Solutions

Population size $N=5$

Sample size $n=3$

1. mean of the population

Mean $\mu=\dfrac{\sum X}{N}$

$\mu=\dfrac{65+68+78+86+91}{5}$

$\mu=77.6$

2. Variance of the population

Var $\sigma^2 =\dfrac{\sum (X-\mu)^2}{N}$

$\sigma^2=\dfrac{501.2}{5}=100.24$

3. Possible samples of size 3 without replacement.

$=C_5^3= \binom{5}{3}=10$ samples

4. Probability distribution of the sample means.

5. Mean of the sampling distribution of the sample means.

Mean

$E(\bar X)=\sum\bar Xf(\bar X)$

$E(\bar X) =77.6$

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

6. Variance of the sampling distribution of the sample means

Variance $(\bar X)$

$=\sum \bar X^2f(\bar X)-(\sum\bar Xf(\bar X))^2$

$=6038.47-(77.600)^2$

$=16.71$

Verification

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

$Var (\bar X)=\dfrac{100.24}{3}.(\dfrac{5-3}{5-1})$

$Var (\bar X)= 16.71$

True.