Question

What is the adjustment factor for the following table?

x	10-15	17-22	24-29	31-36
Frequency	2	1	4	5

Answer 1

In the above question, we are given a table of interval and frequency. We have to find the adjustment factor of the given table. The given intervals are discontinuous, we have to adjust it into continuous form. The formula for adjustment factor is given by
$Adjustment{\text{ }}factor = \dfrac{1}{2} \times Distance{\text{ }}between{\text{ }}2{\text{ }}intervals$

Complete step by step solution:
Given frequency distribution table is,

x	10-15	17-22	24-29	31-36
Frequency	2	1	4	5

We have to find its adjustment factor. In statistics tables, discontinuous class are the ones in which upper limits does not continue in succeeding class. It is also called as an inclusive form as both upper and lower limits are included while recording the observations.
To convert data given in discontinuous form to the continuous form, we subtract the adjustment factor from each lower limit and add the adjustment factor to each upper limit to get the true limits.
In the given table, the intervals are discontinuous, but the class size is constant which is $5$ .
So we only have to find the distance between $2$ successive class intervals.
From the table, we can observe that the upper limit of each class interval is two units less than the lower limit of the succeeding class interval as $15 - 17,{\text{ }}22 - 24,{\text{ }}29 - 31$ . Here the difference is two.
Now we can obtain the adjustment factor by using the formula,
$Adjustment{\text{ }}factor = \dfrac{1}{2} \times Distance{\text{ }}between{\text{ }}2{\text{ }}intervals$
Therefore, putting the distance between two consecutive intervals as $2$ in the above formula we get,
$Adjustment{\text{ }}factor = \dfrac{1}{2} \times 2$
Therefore,
$Adjustment{\text{ }}factor = 1$
That is the required adjustment factor for the above table.
Therefore, the adjustment factor for the given table is $1$ .

Note: Since, that we know the adjustment factor is $1$ . Now we can convert the discontinuous class intervals into continuous form by subtracting $1$ from each lower limit and adding $1$ to each upper limit.
So the new intervals are,
$9 - 16,{\text{ }}16 - 23,{\text{ }}23 - 30,{\text{ }}30 - 37$
Now the new class intervals are continuous.
Putting the new intervals in the frequency distribution table, we get the new continuous table as:

x	9-16	16-23	23-30	30-37
Frequency	2	1	4	5