Variance and standard deviation for grouped data

Date: 2015-10-07; view: 438.

Answers

1.12.8;12.57; 14; 2. 27.20; 26.7; 25; 3. 74.40; 74.67; 80; 4. 23.72; 23.417; 23; 5. 33.8; 31.5; modal class =27-33.

Suppose that we have data grouped into K classes, with frequencies

. If the midpoints of these classes are , the mean and variance of the grouped data are estimated by using following formulas

1. For a population of observations, so that

The variance is

The standard deviation is .

2. For a sample of observations, so that

The variance is

The standard deviation is .

Example:

The following table gives the distribution of the number of days for which all 40 employees of a company were absent during the last year

Calculate the variance and standard deviation.

Solution:

Let us apply

First we need to find and

Class	Number of days absent	Number of employees	Class mark
	0-2 3-5 6-8 9-11 12-14

Now we need to find .

In order to find products we must first find the square quantities . We need three columns to display the computation of the quantities a column for , a column for , and a column for the . We also need a column for and a final column for the products . The necessary computations are shown below in the table 1.13.

Table 1.13.

Class	Class mark
			1-4.75=-3.75 4-4.75=-0.75 7-4.75=2.25 10-4.75=5.25 13-4.75=8.25	14.06 0.56 5.06 27.56 68.06	182.81 7.88 30.38 110.25 204.19
					535.51

In the end,

and

Example:

The following table gives the frequency distribution of the number of orders received each day during the past 50 days at the office of a mail-order company. Calculate the variance and standard deviation.

Solution:

Because the data includes only 50 days,

it represents a sample. Hence, we will use sample formulas to calculate the variance and standard deviation.

Let us apply

All the information required for the calculation of the variance and standard deviation appears in the following table

Number of orders
10-12 13-15 16-18 19-21
	N=50

By substituting the values in the formula for the sample variance, we obtain

Hence, the standard deviation is .

Thus, the standard deviation of the number of orders received at office of this mail-order company during the past 50 days is 2.75.