Question:

Find the mean and variance for the following frequency distribution.

Classes0-1010-2020-3030-4040-50
Frequencies5815166

Updated On: Oct 20, 2023
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Solution and Explanation

ClassFrequency \(f_i\)\(mid-point\,x_i\)\(y_i=\frac{x_i-25}{10}\)\(y_i^2\)\(f_iy_i\)\(f_iy_1^2\)
0-1055-24-1020
10-20815-11-88
20-3015250000
30-401635111616
40-506452121224
 50   1068

Mean,  \(\bar{x}=A\frac{\sum_{i=1}^5f_ix_i}{n}×h=25+\frac{10}{50}×10=25+2=27\) 

Variance (σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^5f_iy_i^2-(\sum_{i=1}^5f_iy_i)^2]\)

\(=\frac{(10)^2}{(50)^2}[50×68-(10)^2]\)

\(\frac{1}{25}[3400-100]=\frac{3300}{25}\)

\(=132\)

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Concepts Used:

Variance and Standard Deviation

Variance:

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’.

Variance Formula:

Read More: Difference Between Variance and Standard Deviation

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’.

Types of Standard Deviation:

  • Standard Deviation for Discrete Frequency distribution
  • Standard Deviation for Continuous Frequency distribution

Standard Deviation Formulas:

1. Population Standard Deviation

2. Sample Standard Deviation