Question

Find the arithmetic mean of the following data by shortcut method.

Class Interval	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	4	8	13	20	12	8

(a) 21
(b) 22
(c) 30
(d) 33

Answer 1

Hint: To solve this question, we will first all assume a mean after finding \[{{x}_{i}}\] by using \[{{x}_{i}}=\] upper limit of the class interval + lower limit of class interval and dividing by 2. And after assuming the mean, we will draw the table and use the formula \[\text{Mean}=\overline{x}=A+\left[ \dfrac{1}{N}\sum{{{f}_{i}}{{d}_{i}}} \right]\] to calculate the mean.

Complete step by step answer:
We have to find the mean using the shortcut method. It can be given as stated and explained below. If \[{{x}_{1}},{{x}_{2}},...{{x}_{n}}\] are the observations given with respective frequencies \[{{f}_{1}},{{f}_{2}},....{{f}_{n}}.\] Let the deviation A take at any point, we have, \[{{d}_{i}}={{x}_{i}}-A\] where i = 1, 2, …n. So, we can say, \[{{f}_{i}}{{d}_{i}}={{f}_{i}}\left( {{x}_{i}}-A \right)\] where i = 1, 2, … n. Then the mean \[\overline{x}\] is given by the formula,
\[\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}\]
We have our data as,

Class Interval	Frequency
0 – 10	4
10 – 20	8
20 – 30	13
30 – 40	20
40 – 50	12
50 – 60	8

Let us assume A = 13, then we will calculate \[{{d}_{i}}={{x}_{i}}-A,\] i = 1, 2, …6 and \[{{f}_{i}}{{d}_{i}}.\] But before doing so, we need \[{{x}_{i}},i=1,2,.....6\] as here class interval is given as \[{{x}_{i}}=\text{mid point of class interval}\text{.}\] This gives our new table as

Class Interval	\[{{x}_{i}}\]	\[{{f}_{i}}\]	\[{{d}_{i}}={{x}_{i}}-A\]	\[{{f}_{i}}{{d}_{i}}\]
0 – 10	5	4	– 20	– 80
10 – 20	15	8	– 10	– 80
20 – 30	25 = A	13	0	0
30 – 40	35	20	10	200
40 – 50	45	12	20	240
50 – 60	55	8	30	240
		N = 65		520

Finally, we will use the mean formula given as
\[\text{Mean}=\overline{x}=A+\left[ \dfrac{1}{N}\sum{{{f}_{i}}{{d}_{i}}} \right]\]
\[\Rightarrow \text{Mean}=\overline{x}=25+\dfrac{520}{65}\]
\[\Rightarrow \text{Mean}=\overline{x}=25+8\]
\[\Rightarrow \text{Mean}=\overline{x}=33\]
Hence, the mean of the given data is \[\overline{x}=33.\]

So, the correct answer is “Option D”.

Note: Another method is to solve this question that can be directed by using the formula, \[\text{Mean}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] and without assuming any mean although assuming a mean and solving gives a more precise value.