Question

Find the mean of the following data by the assumed mean method:

Class interval	Frequency
\[150 - 200\]	14
\[200 - 250\]	26
\[250 - 300\]	17
\[300 - 350\]	23
\[350 - 400\]	20

Answer 1

Hint: To find the mean of the data given by assumed mean method, we will first have to find the class mark or the middle value of each interval. Using that we can find the deviation of each interval. After we have these values we can easily substitute them in the formula for finding mean by assumed mean method.

Complete step-by-step answer:
We will first have to find the class mark of each interval. To find the class mark of an interval we can simply find the average of the upper and lower limit of that interval. For the first interval \[150 - 200\], the class mark will be given by the average of their upper and lower limits, that is the average of 150 and 250. Thus, their class mark will be \[\dfrac{{150 + 250}}{2} = \dfrac{{400}}{2} = 200\]. Similarly we can find the class mark of each of the intervals.
Now as we need to find the class marks of each interval and then the deviation of each interval, it is best if we put all the data in the form of a table.
Let the class mark be represented by \[x\], let the frequency be denoted by \[f\] and the deviation be denoted by \[d\].
The assumed mean will be denoted by \[a\]. The value of assumed can be whatever we want, however it is best to choose one of the values from the class mark as assumed mean and better to choose the middle values of the class mark.
The formula for deviation is \[d = x - a\]
The formula for finding mean by assumed mean method is \[\overline x = a + \dfrac{{\sum {fd} }}{h}\], where \[\overline x \] is the actual mean for the data. Also \[h\] represents the class width which is 100 for the given data.
So, the table formed will be as follows

Class interval	Frequency(\[f\])	Class mark(\[x\])	Deviation(\[d = x - a\])	\[fd\]
\[150 - 200\]	14	175	-100	-1400
\[200 - 250\]	26	225	-50	-1300
\[250 - 300\]	17	\[275 = a\]	0	0
\[300 - 350\]	23	325	50	1150
\[350 - 400\]	20	375	100	2000
				\[\sum {fd = 450} \]

Now, we have all the required values for the formula.
We will find the mean by substituting the values in the formula.
\[
\Rightarrow \overline x = a + \dfrac{{\sum {fd} }}{h} \\
\Rightarrow \overline x = 275 + \dfrac{{450}}{{100}} \\
\Rightarrow \overline x = 275 + 4.5 \\
\Rightarrow \overline x = 279.5 \\
\]
Thus, the mean of the given data will be 279.5.
Note: The formula for finding mean can be used only if the intervals are of equal width that is the intervals must have the same class width. As it can be seen that we have to find the deviation for each interval and then multiply with their respective frequencies and then add the products, it is best to organize the data in the form of a table to ease the calculations and avoid mistakes.