Question

How do you find the deviation from the assumed mean for the below data?

Class Interval	Frequency
10 – 25	2
25 – 40	3
40 – 55	7
55 – 70	6
70 – 85	6
85 – 100	6

Answer 1

Hint: To solve this question, firstly we will find the Mean Class Interval for every six intervals by using the mean class formula. After that, we will find the Deviation of each interval by subtracting the assumed mean by itself from their respected mean class deviation. Then, we will make the table and by using the Mean Deviation formula we will find the Mean of the given data.

Complete step-by-step solution:
Now, firstly let us find the mean class interval, using formula of Mean class Interval which is equals to,
$\text{Mean Class Interval = }\dfrac{\text{Upper Limit +}\text{Lower Limit}}{2}$
So, for interval 10 – 25, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{\text{10+}25}{2}$
On simplifying, we get
$\text{Mean Class Interval = 17}\text{.5}$
For interval 25 – 40, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{25+40}{2}$
On simplifying, we get
$\text{Mean Class Interval = 32}\text{.5}$
For interval 40 – 55, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{\text{40+}55}{2}$
On simplifying, we get
$\text{Mean Class Interval = 47}\text{.5}$
For interval 55 – 70, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{\text{55+}70}{2}$
On simplifying, we get
$\text{Mean Class Interval = 62}\text{.5}$
For interval 70 – 85, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{\text{70+}85}{2}$
On simplifying, we get
$\text{Mean Class Interval = 77}\text{.5}$
For interval 85 – 100, we have mean class interval equals to,
$\text{Mean Class Interval = }\dfrac{\text{100+}85}{2}$
On simplifying, we get
$\text{Mean Class Interval = 92}\text{.5}$
Now, let Assumed Deviation be A and is equal to mean deviation of interval 40 – 55,
So, A = 47.5
Now, we will make another column which will represent Deviation of mean deviation from the Assumed Deviation,
So, we can see that
Deviation = Mean Deviation – Assumed Deviation
So, for interval 10 – 25,
Deviation = 17.5 – 47.5
On simplifying, we get
Deviation = - 30
For interval 25 – 40,
Deviation = 32.5 – 47.5
On simplifying, we get
Deviation = - 15
For interval 40 – 55,
Deviation = 47.5 – 47.5
On simplifying, we get
Deviation = 0
For interval 55 – 70,
Deviation = 62.5 – 47.5
On simplifying, we get
Deviation = 15
For interval 70 – 85,
Deviation = 77.5 – 47.5
On simplifying, we get
Deviation = 30
For interval 85 – 100,
Deviation = 92.5 – 47.5
On simplifying, we get
Deviation = 45
So, we get a new table as,

Class Interval	Frequency	Mean Class Deviation	Deviation
10 – 25	2	17.5	- 30
25 – 40	3	32.5	- 15
40 – 55	7	47.5 = A	0
55 – 70	6	62.5	15
70 – 85	6	77.5	30
85 – 100	6	92.5	45

Then, we know that mean deviation from the assumed mean for the above data, will be given by
$\text{Mean Deviation = }\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{d}_{i}}}}{n}$ , where \[{{f}_{i}}\] represents frequency of each interval and \[{{d}_{i}}\] represents the deviation of interval from assumed mean.
Here, n = 6 and $\sum\limits_{i=1}^{6}{{{f}_{i}}{{d}_{i}}}={{f}_{1}}{{d}_{1}}+{{f}_{2}}{{d}_{2}}+{{f}_{3}}{{d}_{3}}+{{f}_{4}}{{d}_{4}}+{{f}_{5}}{{d}_{5}}+{{f}_{6}}{{d}_{6}}$
On putting values we get
$\sum\limits_{i=1}^{6}{{{f}_{i}}{{d}_{i}}}=2(-30)+3(-15)+6(15)+6(30)+6(45)$
On simplifying, we get
$\sum\limits_{i=1}^{6}{{{f}_{i}}{{d}_{i}}}=(-60)-(45)+(90)+(180)+(270)$
On solving, we get
$\sum\limits_{i=1}^{6}{{{f}_{i}}{{d}_{i}}}=435$
So, $\text{Mean Deviation = }\dfrac{435}{6}$
On solving, we get
$\text{Mean Deviation = 72}\text{.5}$

Note: Always remember that, mean class interval for any class interval is given by, $\text{Mean Class Interval = }\dfrac{\text{Upper Limit +}\text{Lower Limit}}{2}$. Also, remember that, you can assume any of the mean class intervals as assumed mean, we will get the same mean from any assumed mean. Also, remember that $\text{Mean Deviation = }\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{d}_{i}}}}{n}$ , where \[{{f}_{i}}\] represents frequency of each interval and \[{{d}_{i}}\] represents the deviation of interval from assumed mean. Try not to make any calculation errors.