Find the mean, variance and standard deviation using short-cut method.
Height in cms | 70-75 | 75-80 | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
No. of children | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Class Interval | Frequency \(f_i\) | \(mid-point\,x_i\) | \(y_i=\frac{x_i-92.5}{5}\) | \(f_iy_i\) | \(f_i^2\) | \(f_iy_1^2\) |
70-75 | 3 | 72.5 | -4 | -12 | 16 | 48 |
75-80 | 4 | 77.5 | -3 | -12 | 9 | 36 |
80-85 | 7 | 82.5 | -2 | -14 | 4 | 28 |
85-90 | 7 | 87.5 | 1 | -7 | 1 | 7 |
90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
95-100 | 9 | 97.5 | 1 | 9 | 9 | 9 |
100-105 | 6 | 102.5 | 2 | 12 | 12 | 24 |
105-110 | 6 | 107.5 | 3 | 18 | 18 | 54 |
110-115 | 3 | 112.5 | 4 | 12 | 12 | 48 |
60 | 6 | 6 | 254 |
Mean, \(\bar{x}=A\frac{\sum_{i=1}^9f_ix_i}{n}×h=92.5+\frac{6}{60}×5=92.5+0.5=93\)
Variance (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^9f_iy_i^2-(\sum_{i=1}^9f_iy_i)^2]\)
\(=\frac{(5)^2}{(60)^2}[60×254-(6)^2]\)
\(=\frac{25}{3600}(15204)=105.58\)
\(Standard\,\,deviation\.(σ) =√105.58=10.27\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
According to layman’s words, the variance is a measure of how far a set of data are dispersed out from their mean or average value. It is denoted as ‘σ2’.
Read More: Difference Between Variance and Standard Deviation
The spread of statistical data is measured by the standard deviation. Distribution measures the deviation of data from its mean or average position. The degree of dispersion is computed by the method of estimating the deviation of data points. It is denoted by the symbol, ‘σ’.
1. Population Standard Deviation
2. Sample Standard Deviation