Here you will learn what is the formula for mean of grouped and ungrouped data and how to find mean with examples.
Let’s begin –
The formula for mean median and mode for grouped and ungrouped frequency distribution is given below.
Formula for Mean :
(i) For ungrouped distribution : If \(x_1\), \(x_2\), …… \(x_n\) are n values of variate \(x_i\) then their mean \(\bar{x}\) is defined as
\(\bar{x}\) = \(x_1 + x_2, …… + x_n \over n\) = \({\sum_{i=1}^{n}x_i}\over n\)
\(\implies\) \(\sum x_i\) = n\(\bar{x}\)
Example : Neeta and her four friends secured 65, 78, 82, 94 and 71 marks in a test of mathematics. Find the average (arithmetic mean) of their marks.
Solution : Arithmetic mean or average = \(65 + 78 + 82 + 94 + 71\over 5\) = \(390\over 5\) = 78
Hence, arithmetic mean = 78
(ii) For ungrouped and grouped frequency distribution :
(a) By Direct Method :
If \(x_1\), \(x_2\), …… \(x_n\) are values of variate with corresponding frequencies \(f_1\), \(f_2\), …… \(f_n\), theb their A.M. is given by
\(\bar{x}\) = \(f_1x_1 + f_2x_2 + …… + f_nx_n \over f_1 + f_2 + …… + f_n\) = \({\sum_{i=1}^{n}f_ix_i}\over N\), where N = \({\sum_{i=1}^{n}f_i}\)
Example : Find the mean of the following freq. dist.
| \(x_i\) | 5 | 8 | 11 | 14 | 17 |
| \(f_i\) | 4 | 5 | 6 | 10 | 20 |
Solution : Here N = \(\sum f_i\) = 4 + 5 + 6 + 10 + 20 = 45
\(\sum f_ix_i\) = 606
\(\therefore\) \(\bar{x}\) = \(\sum f_ix_i\over N\) = \(606\over 45\) = 13.47
(b) By short method or assumed mean method :
If the value of \(x_i\) are large, then calculation of A.M. by using mean formula is quite tedious and time consuming. In such case we take deviation of variate from an arbitrary point a.
Let \(d_i\) = \(x_i\) – a
\(\therefore\) \(\bar{x}\) = a + \(\sum f_id_i\over N\), where a is assumed mean
Example : Find the A.M. of the following freq. dist.
| Class Interval | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
| \(f_i\) | 17 | 35 | 43 | 40 | 21 | 24 |
Solution : Let assumed mean a = 175
| Class Interval | mid value \((x_i)\) | \(d_i\) = \(x_i – 175\) | frequency \(f_i\) | \(f_id_i\) |
| 0-50 | 25 | -150 | 17 | -2550 |
| 50-100 | 75 | -100 | 35 | -3500 |
| 100-150 | 125 | -50 | 43 | -2150 |
| 150-200 | 175 | 0 | 40 | 0 |
| 200-250 | 225 | 50 | 21 | 1050 |
| 250-300 | 275 | 100 | 24 | 2400 |
| \(\sum f_i\) = 180 | \(\sum f_id_i\) = -4750 |
Now, a = 175 and N = \(\sum f_i\) = 180
\(\therefore\) \(\bar{x}\) = a + (\(\sum f_id_i\over N\)) = 175 + \((-4750)\over 180\) = 175 – 26.39 = 148.61
(c) By step deviation method :
Sometime during the application of short method (given above) of finding the A.M. If each deviation \(d_i\) are divisible by a common number h(let)
Let \(u_i\) = \(d_i\over h\) = \(x_i – a\over h\), where a is assumed mean.
\(\therefore\) \(\bar{x}\) = a + (\(\sum f_iu_i\over N\))h
Example : Find the A.M. of the following freq. dist.
| \(x_i\) | 5 | 15 | 25 | 35 | 45 | 55 |
| \(f_i\) | 12 | 18 | 27 | 20 | 17 | 6 |
Solution : Let assumed mean a = 35, h = 10
here N = \(\sum f_i\) = 100, \(u_i\) = \(x_i – 35\over 10\)
\(\sum f_iu_i\) = (12\(\times\)-3) + (18\(\times\)-2) + (27\(\times\)-1) + (20\(\times\)0) + (17\(\times\)1) + (6\(\times\)2)
= -70
\(\therefore\) \(\bar{x}\) = a + (\(\sum f_iu_i\over N\))h = 35 + \((-70)\over 100\)\(\times\)10 = 28
