Find the mean and variance for the following frequency distribution.
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Class | Frequency \(f_i\) | \(mid-point\,x_i\) | \(y_i=\frac{x_i-25}{10}\) | \(y_i^2\) | \(f_iy_i\) | \(f_iy_1^2\) |
0-10 | 5 | 5 | -2 | 4 | -10 | 20 |
10-20 | 8 | 15 | -1 | 1 | -8 | 8 |
20-30 | 15 | 25 | 0 | 0 | 0 | 0 |
30-40 | 16 | 35 | 1 | 1 | 16 | 16 |
40-50 | 6 | 45 | 2 | 12 | 12 | 24 |
50 | 10 | 68 |
Mean, \(\bar{x}=A\frac{\sum_{i=1}^5f_ix_i}{n}×h=25+\frac{10}{50}×10=25+2=27\)
Variance (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^5f_iy_i^2-(\sum_{i=1}^5f_iy_i)^2]\)
\(=\frac{(10)^2}{(50)^2}[50×68-(10)^2]\)
\(\frac{1}{25}[3400-100]=\frac{3300}{25}\)
\(=132\)
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |
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