Question:

Find the mean and variance for the following frequency distribution.

Classes0-3030-6060-9090-120120-150150-180180-210
Frequencies23510352

Updated On: Oct 20, 2023
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Solution and Explanation

ClassFrequency \(f_i\)\(mid-point\,x_i\)\(y_i=\frac{x_i-105}{30}\)\(f_iy_i\)\(y_i^2\)\(f_iy_1^2\)
0-30215-3-6918
30-60345-2-6412
60-90575-1-515
90-120101050000
120-15031351313
150-1805165210420
180-210219536918
 30  2 76

Mean,  \(\bar{x}=A\frac{\sum_{i=1}^7f_ix_i}{n}×h=105+\frac{2}{30}×30=105+2=107\)

Variance(σ2) = \(\frac{h^2}{N^2}[N\sum_{i=1}^7f_iy_i^2-(\sum_{i=1}^7f_iy_i)^2]\)

\(=\frac{(30)^2}{(30)^2}[30×76-(2)^2]\)

\(=2280-4\)

\(=2276\)

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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