Chapter 6.1, Problem 1AC

Many times in statistics it is necessary to see if a set of data values is approximately normally distributed. There are special techniques that can be used. One technique is to draw a histogram for the data and see if it is approximately bell-shaped. (Note: It does not have to be exactly symmetric to be bellshaped.)

The numbers of branches of the 50 top banks are shown.

1. Construct a frequency distribution for the data.

2. Construct a histogram for the data.

3. Describe the shape of the histogram.

4. Based on your answer to question 3, do you feel that the distribution is approximately normal?

In addition to the histogram, distributions that are approximately normal have about 68% of the values fall within 1 standard deviation of the mean, about 95% of the data values fall within 2 standard deviations of the mean, and almost 100% of the data values fall within 3 standard deviations of the mean. (See Figure 6–5.)

5. Find the mean and standard deviation for the data.

6. What percent of the data values fall within 1 standard deviation of the mean?

7. What percent of the data values fall within 2 standard deviations of the mean?

8. What percent of the data values fall within 3 standard deviations of the mean?

9. How do your answers to questions 6, 7, and 8 compare to 68, 95, and 100%, respectively?

10. Does your answer help support the conclusion you reached in question 4? Explain.

To determine

To construct: The frequency distribution for the data.

Answer to Problem 1AC

The frequency distribution for the data is,

Limits	Frequency
0-9	1
10-19	14
20-29	17
30-39	7
40-49	3
50-59	2
60-69	2
70-79	1
80-89	2
90-99	1

Explanation of Solution

Given info:

The data shows that the number of branches of the 50 top banks.

Calculation:

Answers may vary; one of the possible answers is given below.

First consider the limits with the difference of 9.

The limits are 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89 and 90-99.

From the data, it can be observed that the number banks having branches between 0-9 is 1. Also, the number banks having branches between 10-19 is 14.

Similarly, the frequencies of other classes are obtained as follows.

Limits	Frequency
0-9	1
10-19	14
20-29	17
30-39	7
40-49	3
50-59	2
60-69	2
70-79	1
80-89	2
90-99	1

To determine

To construct: The histogram for the data.

Answer to Problem 1AC

The histogram for the data is,

Explanation of Solution

Calculation:

Step-by-step procedure to obtain histogram using the MINITAB software:

• Choose Graph > Histogram.
• Choose Simple, and then click OK.
• In Graph variables, enter the column of Number of branches of the bank.
• Click OK.

To determine

To describe: The shape of the histogram.

Answer to Problem 1AC

The shape of the histogram is unimodel and skewed to the right.

Explanation of Solution

Justification:

From the histogram of part (2), it can be observed that the data has single group and the bar in histogram starts with a high value which is gradually decreasing to the minimum value. Thus, it can be conclude that shape of the histogram is unimodel and skewed to the right.

To determine

To check: Whether the distribution is approximately normal or not based on the answer of question 3.

Answer to Problem 1AC

No, the distribution is not approximately normal.

Explanation of Solution

Justification:

From the part (3), it can be observed shape of the histogram is unimodel and skewed to the right. Thus, it can be conclude that the distribution does not appear to be normal.

To determine

To find: The mean and standard deviation for the data.

Answer to Problem 1AC

The mean and standard deviation for the data are 31.4 and 20.6.

Explanation of Solution

Calculation:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns Variable.
• Choose option statistics, and select Mean and Standard deviation.
• Click OK.

Output using Minitab software is,

Form the Minitab output, it can be observed that the mean and standard deviation for the data are approximately 31.4 and 20.6.

To determine

To find: The percent of the data values fall within 1 standard deviation of the mean.

Answer to Problem 1AC

The percent of the data values fall within 1 standard deviation of the mean is 80%.

Explanation of Solution

Calculation:

Here, 1 standard deviation of the mean represents that $\mu \pm \sigma$.

Substitute 31.4 for $\mu$ and 20.6 for $\sigma$

For $\mu -\sigma$:

$\begin{array}{c}\mu -\sigma =31.4-20.6\\ =10.8\end{array}$

For $\mu +\sigma$:

$\begin{array}{c}\mu +\sigma =31.4+20.6\\ =52\end{array}$

The percentage of the data values fall within 1 standard deviation of the mean represents the percentage of the data values lies between 10.8 and 52.

The formula for the percentage of the data values fall within 1 standard deviation of the mean is,

$\text{Percentage=}\frac{\text{Number of data values lies within 10}\text{.8 and 52}}{\text{Total number of data values}}\times 100$

Substitute 40 for “Number of data values lies within 10.8 and 52” and 50 for “Total number of data values”

$\begin{array}{c}\text{Percentage=}\frac{40}{50}\times 100\\ =80\end{array}$

Thus, the percent of the data values fall within 1 standard deviation of the mean is 80%.

To determine

To find: The percent of the data values fall within 2 standard deviation of the mean.

Answer to Problem 1AC

The percent of the data values fall within 2 standard deviation of the mean is 92%.

Explanation of Solution

Calculation:

Here, 2 standard deviation of the mean represents that $\mu \pm 2\sigma$.

Substitute 31.4 for $\mu$ and 20.6 for $\sigma$

For $\mu -2\sigma$:

$\begin{array}{c}\mu -2\sigma =31.4-2\times 20.6\\ =-9.8\end{array}$

For $\mu +2\sigma$:

$\begin{array}{c}\mu +2\sigma =31.4+2\times 20.6\\ =72.6\end{array}$

The percentage of the data values fall within 2 standard deviation of the mean represents the percentage of the data values lies between –9.8 and 72.6.

The formula for the percentage of the data values fall within 2 standard deviation of the mean is,

$\text{Percentage=}\frac{\text{Number of data values lies within }-\text{9}\text{.8 and 72}\text{.6}}{\text{Total number of data values}}\times 100$

Substitute 46 for “Number of data values lies within –9.8 and 72.6” and 50 for “Total number of data values”

$\begin{array}{c}\text{Percentage=}\frac{46}{50}\times 100\\ =92\end{array}$

Thus, the percent of the data values fall within 2 standard deviation of the mean is 92%.

To determine

To find: The percent of the data values fall within 3 standard deviation of the mean.

Answer to Problem 1AC

The percent of the data values fall within 3 standard deviation of the mean is 98%.

Explanation of Solution

Calculation:

Here, 3 standard deviation of the mean represents that $\mu \pm 3\sigma$.

Substitute 31.4 for $\mu$ and 20.6 for $\sigma$

For $\mu -3\sigma$:

$\begin{array}{c}\mu -3\sigma =31.4-3\times 20.6\\ =-30.4\end{array}$

For $\mu -3\sigma$:

$\begin{array}{c}\mu +3\sigma =31.4+3\times 20.6\\ =93.2\end{array}$

The percentage of the data values fall within 3 standard deviation of the mean represents the percentage of the data values lies between –30.4 and 93.2.

The formula for the percentage of the data values fall within 3 standard deviation of the mean is,

$\text{Percentage=}\frac{\text{Number of data values lies within }-\text{30}\text{.4 and 93}\text{.2}}{\text{Total number of data values}}\times 100$

Substitute 49 for “Number of data values lies within –30.4 and 93.2” and 50 for “Total number of data values”

$\begin{array}{c}\text{Percentage=}\frac{49}{50}\times 100\\ =98\end{array}$

Thus, the percent of the data values fall within 3 standard deviation of the mean is 98%.

To determine

To compare: The 6,7 and 8 answers with the 68, 95 and 100%.

Answer to Problem 1AC

The 6,7 and 8 answers are different with the 68, 95 and 100%.

Explanation of Solution

Calculation:

From the answers of 6,7 and 8, it is observed that the percent of the data values fall within 1,2 and 3 standard deviation of the mean are 80%, 92% and 98%. Thus, it can be conclude that the 6,7 and 8 answers are different with the 68, 95 and 100%.

To determine

To explain: Whether the answer help to support the conclusion in question 4.

Answer to Problem 1AC

Yes, the answer help to support the conclusion in question 4 because the distribution of the variable does not follow normal distribution.

Explanation of Solution

Calculation:

From the answer of (9), it can be observed that the distribution of the variable does not follow normal distribution. Thus, it can be concluding that the answer help to support the conclusion in question 4.