Here you will learn mean square deviation formula and relation between mean square deviation and variance with example.
Let’s begin –
Mean Square Deviation Formula
The mean square deviation of a distribution is the mean of the square of deviations of variate from assumed mean. It is denoted by \(S^2\).
Hence \(S^2\) = \(\sum{x_i – a}^2\over n\) = \(\sum{d_i}^2\over n\) (for ungrouped dist.)
\(S^2\) = \(\sum{x_i – a}^2\over N\) = \(\sum{f_id_i}^2\over N\) (for frequency dist.), where \(d_i\) = \(x_i – a\)
Relation between variance and mean square deviation
\(\because\) \({\sigma}^2\) = \(\sum{f_id_i}^2\over N\) – \(({\sum f_i{d_i}\over N})^2\)
\(\implies\) \({\sigma}^2\) = \(s^2\) – \(d^2\), where d = \(\bar{x} – a\) = \({\sum f_i{d_i}\over N}\)
\(\implies\) \(s^2\) = \({\sigma}^2\) + \(d^2\), \(\implies\) \(s^2\) \(\geq\) \({\sigma}^2\)
Hence the variance is the minimum value of mean square deviation of a distribution.
Example : Find the variance of the following freq. dist.
| class | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 |
| \(f_i\) | 2 | 7 | 12 | 19 | 9 | 1 |
Solution : Let a = 7 and h = 2
| class | \(x_i\) | \(f_i\) | \(u_i\) = \(x_i – a\over h\) | \(f_iu_i\) | \(f_iu_i^2\) |
| 0 – 2 | 1 | 2 | -3 | -6 | 18 |
| 2 – 4 | 3 | 7 | -2 | -14 | 28 |
| 4 – 6 | 5 | 12 | -1 | -12 | 12 |
| 6 – 8 | 7 | 19 | 0 | 0 | 0 |
| 8 – 10 | 9 | 9 | 1 | 9 | 9 |
| 10 – 12 | 11 | 1 | 2 | 2 | 4 |
| N = 50 | \(\sum{f_iu_i}\) = -21 | \(\sum{f_iu_i^2}\) = 71 |
\(\because\) \({\sigma^2}\) = \(h^2\)[\(\sum
f_i{u_i}^2\over n\) – \(({\sum f_i{u_i}\over n})^2\)]
= 4[\(71\over 50\) – (\({-21\over 50}^2\))]
= 4[1.42 – 0.1764] = 4.97
Mathematical Properties of Variance
(i) Var.\((x_i + p\)) = Var.(\(x_i\))
(ii) Var.\((px_i\)) = \(p^2\)Var.(\(x_i\))
(iii) Var\((ax_i + b\)) = \(a^2\).Var(\(x_i\))
where p, a, b are constants.
