Chapter 4, Problem 1TP

1

To determine

To construct: The frequency distribution and histogram for a data of set of 50 comprehension score.

Answer to Problem 1TP

The frequency distribution of given table is

Class-interval	Class Boundaries	Width	Frequency	Cumulative Frequency
2-11	1.5-11.5	9	5	5
12-21	11.5-21.5	9	9	14
22-31	21.5-31.5	9	8	22
32-41	31.5-41.5	9	5	27
42-51	41.5-51.5	9	12	39
52-61	51.5-61.5	9	11	50
Total			50	157

The sketch of histogram is,

Explanation of Solution

Given info:

The data shows the comprehension of 50 scores

12	22	54	43	56
15	26	55	31	57
11	27	51	12	59
16	36	56	14	54
21	34	53	15	56
25	33	57	16	43
21	38	49	22	44
8	42	45	29	41
6	44	45	29	42
2	47	47	54	7

Calculation:

The class boundaries for any class are given by:

$\begin{array}{l}\text{Upper}\text{class}\text{boundary}=\text{Upper}\text{class}\text{limit}+\frac{d}{2}\\ \text{Lower}\text{class}\text{boundary}=\text{Lower}\text{class}\text{limit}-\frac{d}{2}\end{array}$

Where d is the gap between the upper limit of any class and the lower limit of the succeeding class.

The grouped frequency distribution is as follows:

Class-interval	Class Boundaries	Frequency
2-11	1.5-11.5	5
12-21	11.5-21.5	9
22-31	21.5-31.5	8
32-41	31.5-41.5	5
42-51	41.5-51.5	12
52-62	51.5-61.5	11

The histogram is a graph that displays the data by using contiguous vertical bars of various heights to represent the frequencies of the classes.

Upper Bound	Frequency
11.5	5
21.5	9
31.5	8
41.5	5
51.5	12
61.5	11

Construction steps for a histogram:

• Press [Ctrl]-N for a new workbook.
• Enter the data in column A, one number per cell.
• Enter the upper boundaries into column B.
• From the toolbar, select the Data tab, then select Data Analysis.
• In Data Analysis, select Histogram and click [OK].
• In the Histogram dialog box, select relative column in the Input Range box and select upper limit column in the Bin Range box.
• Select New Worksheet Ply and Chart Output. Click [OK].

The screen will show the histogram as shown below.

2

To determine

To explain: The selection of class interval for frequency distribution table.

Answer to Problem 1TP

The class width is 9 then the class interval is 1.5-11.5, 11.5-21.5 and so on upto 7 class interval.

Explanation of Solution

Step for constructing class intervals

• The largest value is 59 and the smallest value is 2.

  $\begin{array}{c}\text{Range}=\text{largest}\text{value}-\text{smallest}\text{value}\\ =\text{59-2}\\ =\text{57}\end{array}$

• Let the class interval is 7.

  $\begin{array}{c}\text{Class}\text{width}=\frac{\text{Range}}{\text{class}\text{interval}}\\ =\frac{57}{7}\\ =8.14\left[\text{Rounding}\text{up}\right]\\ =9\end{array}$

• Let the starting point be 1.5.
• Now, the 1^st class interval has lower limit 1.5 and upper limit is 11.5 because the class width is 9 and for 2^nd class interval the lower limit is the upper limit of 1^st class interval after adding class width then get the upper limit of 2^nd class interval.
• After adding 7 times the class width in preceding upper limit of class interval to get the 7 class interval that are constructed.

3.

To determine

To explain: The skewness of the distribution.

Answer to Problem 1TP

The distribution of 50 score comprehension is almost negatively skewed.

Explanation of Solution

If the most of the data is on the right with few smaller values showing up on the left side of the histogram then data are positively skewed and if most of the data is on the left with few smaller values showing up on the right side of the histogram then data are negatively skewed.

Here the distribution of the 50 score comprehensions on the histogram has more fluctuation from both the side.

Hence, on resulting the data are negatively skewed.