Chapter 7.5, Problem 7.17E

A population consists of $N=5$ numbers: 1, 3, 5, 6, and 7. It can be shown that the mean and standard deviation for this population are $\mu =4.4$ and $\sigma =2.15,$ respectively.

a. Construct a probability histogram for this population.

b. Use the random number table. Table 10 in Appendix I, to select a random sample of size $n=10$ with replacement from the population. Calculate the sample mean. $\bar{x}$ . Repeat this procedure. calculating the sample mean $\bar{x}$ for your second sample.

(HINT: Assign the random digits 0 and 1 to the measurement $x=1;$ assign digits 2 and 3 to the measurement $x=3,$ and so on.)

c. To simulate the sampling distribution of $\bar{x}$ , we have selected 50 more samples of size $n=10$ with replacement, and have calculated the corresponding sample means. Construct a relative frequency histogram for these 50 values of $\bar{x}$ . What is the shape of this distribution?

To determine

To graph: A probability histogram for the data.

Explanation of Solution

Given information:

The given population contain $N=5$ values such that

  $1,3,5,6,\text{ and 7}$ .

The population mean $\mu =4.4$ and the population standard deviation $\sigma =2.15$ .

Calculation:

Here there are five numbers without repeating therefore the probability of each number to be equal. Hence the probability for a number is

  $\frac{1}{5}=0.2$.

Graph:

In the below histogram height of the bars equal to the probability.

  

To determine

To find: The two sets of samples of size $n=10$ using the random number Table $10$ in Appendix $\text{I}$ and theirmean values.

Answer to Problem 7.17E

The first selected sample is $1,\text{ 5, 5, 7, 3, 1, 6, 7, 1, 3}$ and $\overline{x}=3.9$ .

The second selected sample is $1,\text{ 5, 3, 6, 1, 1, 6, 5, 3, 3}$ and $\overline{x}=3.4$ .

Explanation of Solution

Given information:

The random digits $0$ and $1$ corresponds with the number $1$ ; the random digits $2$ and $3$ corresponds with the number $3$ ; the random digits $4$ and $5$ corresponds with the number $5$ ; the random digits $6$ and $7$ corresponds with the number $6$ ; the random digits $8$ and $9$ corresponds with the number $7$ .

Formula used:

  $\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$

where,

  $\begin{array}{l}{x}_i=i^{th}\text{ observation}\\ \overline{x}=\text{Mean}\\ n=\text{Sample size}\end{array}$

Calculation:

If select the first digit of the second column from the Table $10$ to generate the first sample:

  $\boxed{\begin{array}{l}SelecteddigitCorrespondingNumber\\ 1\text{ 1}\\ \text{4 5}\\ \text{4 5}\\ \text{9 7}\\ \text{3 3}\\ \text{0 1}\\ \text{7 6}\\ \text{9 7}\\ \text{1 1}\\ \text{3 3}\end{array}}$

Therefore the first selected sample is

  $1,\text{ 5, 5, 7, 3, 1, 6, 7, 1, 3}$ .

Hence the sample mean

  $\begin{array}{l}\overline{x}=\frac{1+5+5+7+3+1+6+7+1+3}{10}\\ =3.9\end{array}$

If select the second digit of the third column from the Table $10$ to generate the second sample:

  $\boxed{\begin{array}{l}SelecteddigitCorrespondingNumber\\ 1\text{ 1}\\ \text{5 5}\\ \text{2 3}\\ \text{6 6}\\ \text{1 1}\\ \text{1 1}\\ \text{6 6}\\ \text{5 5}\\ \text{3 3}\\ \text{3 3}\end{array}}$

Therefore the second selected sample is

  $1,\text{ 5, 3, 6, 1, 1, 6, 5, 3, 3}$ .

Hence the sample mean

  $\begin{array}{l}\overline{x}=\frac{1+5+3+6+1+1+6+5+3+3}{10}\\ =3.4\end{array}$

To determine

To graph: A probability histogram for the data.

Explanation of Solution

Given information:

The below is the sample means of $n=50$ selected samples.

  $\boxed{\begin{array}{rrrrrrrrrr}\hfill 4.8& \hfill 4.2& \hfill 4.2& \hfill 4.5& \hfill 4.3& \hfill 4.3& \hfill 5& \hfill 4& \hfill 3.3& \hfill 4.7\\ \hfill 3& \hfill 5.9& \hfill 5.7& \hfill 4.2& \hfill 4.4& \hfill 4.8& \hfill 5& \hfill 5.1& \hfill 4.8& \hfill 4.2\\ \hfill 4.6& \hfill 4.1& \hfill 3.4& \hfill 4.9& \hfill 4.1& \hfill 4& \hfill 3.7& \hfill 4.3& \hfill 4.3& \hfill 4.5\\ \hfill 5& \hfill 4.6& \hfill 4.1& \hfill 5.1& \hfill 3.4& \hfill 5.9& \hfill 5& \hfill 4.3& \hfill 4.5& \hfill 3.9\\ \hfill 4.4& \hfill 4.2& \hfill 4.2& \hfill 5.2& \hfill 5.4& \hfill 4.8& \hfill 3.6& \hfill 5& \hfill 4.5& \hfill 4.9\end{array}}$

Calculation:

Let the class intervals as $>2.85\text{ and }\le 3.15,>3.15\text{ and }\le 3.45,\mathrm{...}>5.85\text{ and }\le 6.15$ .

The probability is the frequency divided by the total number of samples.

The below table gives the probability for each class.

  $\boxed{\begin{array}{l}ClassMidvalueFrequencyProbability\\ >2.85\text{ and}\le 3.15\text{ 3 1 0}\text{.02}\\ >3.15\text{ and}\le 3.45\text{ 3}\text{.3 3 0}\text{.06}\\ >3.45\text{ and}\le 3.75\text{ 3}\text{.6 2 0}\text{.04}\\ >3.75\text{ and}\le 4.05\text{ 3}\text{.9 3 0}\text{.06}\\ >4.05\text{ and}\le 4.35\text{ 4}\text{.2 14 0}\text{.28}\\ >4.35\text{ and}\le 4.65\text{ 4}\text{.5 8 0}\text{.16}\\ >4.65\text{ and}\le 4.95\text{ 4}\text{.8 7 0}\text{.14}\\ >4.95\text{ and}\le 5.25\text{ 5}\text{.1 8 0}\text{.16}\\ >5.25\text{ and}\le 5.55\text{ 5}\text{.4 1 0}\text{.02}\\ >5.55\text{ and}\le 5.85\text{ 5}\text{.7 1 0}\text{.02}\\ >5.85\text{ and}\le 6.15\text{ 5}\text{.7 2 0}\text{.04}\end{array}}$

Graph:

Below is the histogram of sample means, the vertical axis represents the percentage and the horizontal axis represents the midpoints of the sample mean classes.

  

Interpretation:

From the above histogram, it can say that the distribution of the sample means is approximately mound-shaped.