Problem

# At the downtown office of First National Bank there are five tellers. Last week the teller...

At the downtown office of First National Bank there are five tellers. Last week the tellers made the following number of errors each: 2, 3, 5, 3, and 5.

a. How many different samples of 2 tellers are possible?

b. List all possible samples of size 2 and compute the mean of each.

c. Compute the mean of the sample means and compare it to the population mean.

#### Step-by-Step Solution

Solution 1
(a) To find the possible different samples of 2 tellers, use the combination formula: 2,3,5,3 and 5

Combination formula:

$${ }^{n} \mathrm{C}_{r}=\frac{n !}{(n-r) ! r !}$$

The number of errors is $$2,3,5,3$$ and 5 .

Number of members of the population $$(N)=5$$

Number of size of the sample $$(n)=2$$

Find the possible different samples.

\begin{aligned} { }^{5} C_{2} &=\frac{5 !}{2 !(5-2) !} \\ &=\frac{5 !}{2 ! 3 !} \\ &=\frac{5 \times 4 \times 3 !}{2 ! 3 !} \\ &=\frac{5 \times 4}{2 \times 1} \\ &=10 \end{aligned}

Hence, the possible different samples of 2 tellers is 10 .

The required solution is:10

(b) To find the list of the possible samples of size 2 and calculate the mean of each sample, use the mean formula: 2,3,5,3 and 5

The formula of the mean is:

$$\text { Mean }=\frac{\text { Sum of the values }}{\text { Total number of the value }}$$

The number of errors is $$2,3,5,3$$ and 5 .

Using the $$1^{\text {st }}$$ value 2:

Errors $$2,3:$$

\begin{aligned} \text { Mean } &=\frac{2+3}{2} \\ &=\frac{5}{2} \\ &=2.5 \end{aligned}

Errors $$2,5:$$

\begin{aligned} \text { Mean } &=\frac{2+5}{2} \\ &=\frac{7}{2} \\ &=3.5 \end{aligned}

Errors $$2,3:$$

\text { Mean } \begin{aligned} \text { Mean } &=\frac{2+3}{2} \\ &=\frac{5}{2} \\ &=2.5 \end{aligned}

Errors $$2,5:$$

\begin{aligned} \text { Mean } &=\frac{2+5}{2} \\ &=\frac{7}{2} \\ &=3.5 \end{aligned}

Using the $$2^{\text {nd }}$$ value 3 :

Errors 3,2 :

\text { Mean } \begin{aligned} &=\frac{3+2}{2} \\ &=\frac{5}{2} \\ &=2.5 \end{aligned}

Errors 3,5 :

\begin{aligned} \text { Mean } &=\frac{3+5}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Errors 3,3 :

\begin{aligned} \text { Mean } &=\frac{3+3}{2} \\ &=\frac{6}{2} \\ &=3 \end{aligned}

Errors 3,5 :

\begin{aligned} \text { Mean } &=\frac{3+5}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Using the 5 at $$1^{\text {st }}$$

Errors 5,2 :

\begin{aligned} \text { Mean } &=\frac{5+2}{2} \\ &=\frac{7}{2} \\ &=3.5 \end{aligned}

Errors $$5,3:$$

\begin{aligned} \text { Mean } &=\frac{5+3}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Errors $$5,3:$$

\begin{aligned} \text { Mean } &=\frac{5+3}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Errors $$5,5:$$ mean $$=5$$

Using the 3 rd value 3 :

Errors 3,2 :

\begin{aligned} \text { Mean } &=\frac{3+2}{2} \\ &=\frac{5}{2} \\ &=2.5 \end{aligned}

Errors $$3,3:$$

\begin{aligned} \text { Mean } &=\frac{3+3}{2} \\ &=\frac{6}{2} \\ &=3 \end{aligned}

Errors 3,5 :

\begin{aligned} \text { Mean } &=\frac{3+5}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Errors $$3,5:$$

\begin{aligned} \text { Mean } &=\frac{3+5}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Using the $$4^{\text {th }}$$ value 5:

Errors 5,2 :

\text { Mean } \begin{aligned} \text { Mean } &=\frac{5+2}{2} \\ &=\frac{7}{2} \\ &=3.5 \end{aligned}

Errors 5,3 :

\begin{aligned} \text { Mean } &=\frac{5+3}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Errors 5,5 :

\begin{aligned} \text { Mean } &=\frac{5+5}{2} \\ &=\frac{10}{2} \\ &=5 \end{aligned}

Errors $$5,3:$$

\begin{aligned} \text { Mean } &=\frac{5+3}{2} \\ &=\frac{8}{2} \\ &=4 \end{aligned}

Hence, the required result is found.

(c) To find the mean of sample means and compare with the population mean, use the mean formula: 2,3,5,3 and 5

The formula of the mean is:

$$\text { Mean }=\frac{\text { Sum of the values }}{\text { Total number of the value }}$$

The number of errors is $$2,3,5,3$$ and 5 .

Find the sample mean.

\begin{aligned} \text { Sample Mean } &=\left(\begin{array}{l} 2.5+3.5+2.5+3.5+2.5+4+3+4+3.5+4+4+5 \\ +2.5+3+4+4+3.5+4+5+4 \end{array}\right) \\ &=\frac{72}{20} \\ &=3.6 \end{aligned}

Find the population mean.

\begin{aligned} \text { Population mean } &=\frac{(2+3+5+3+5)}{5} \\ &=\frac{18}{5} \\ &=3.6 \end{aligned}

Hence, the sample mean is same as the population mean.

The required solution is:

Both mean is same

Thus, the required result is found.