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# Dispersion (Variance and Standard Deviation) - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Dispersion (Variance and Standard Deviation) is considered one the most difficult concept.

• 75 Questions around this concept.

## Solve by difficulty

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks are given?

Let  $\dpi{100} x_{1},x_{2},......................,x_{n}$ be n observations, and let $\dpi{100} \bar{x}$  be their arithmetic mean and $\dpi{100} \sigma ^{2}$ be their variance.

Statement 1: Variance of $\dpi{100} 2x_{1},2x_{2},..............,2x_{n} \: is \: 4\sigma ^{2}$.

Statement 2: Arithmetic mean of $\dpi{100} 2x_{1},2x_{2},..............,2x_{n} \: is \: 4 \bar{x}$.

The mean of the numbers $\dpi{100} a,b,8,5,10\; is\; 6$ and the variance is $\dpi{100} 6.80.$ Then which one of the following gives possible values of $a\; \; and\; \; b$ ?

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively.  Later on, three observations, 3, 4 and 5, were found to be incorrect.  If the incorrect observations are omitted, then the variance of the remaining observations is :

Let $\mathrm{a}_1, \mathrm{a}_2, \ldots \mathrm{a}_{10}$ be 10 observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{10}$ is equal to:

## Concepts Covered - 1

Dispersion (Variance and Standard Deviation)

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity which leads to a proper measure of dispersion.

The variance of n observations x1 , x2 ,..., xn is given by

$\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$

Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.

The positive square-root of the variance is called standard deviation. The standard deviation is usually denoted by σ  and it is given by

$\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}$

Variance and Standard Deviation of a Ungrouped Frequency Distribution

The given data is

$\begin{matrix} x &: &x_1, & x_2, & x_3, &\ldots &x _n & \\ & & & & & & & \\ f& : & f_1, &f_2, &f_3, &\ldots & f_n & \end{matrix}$

$\\\text{In this case, Variance}\;\left ( \sigma^2 \right ) \;\;\;=\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}\\\\\text{and, Standard Deviation}(\sigma)\;=\sqrt{\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}}\\\text{where, N}=\sum_{i=1}^{n} f_{i}$

Variance and Standard deviation of a grouped frequency distribution

The formula for variance and standard deviation are the same as in the case of ungrouped frequency distribution. Here, $x_i$ is the mid point of each class.

Another formula for Standard Deviation

$\\\text{Variance }\left ( \sigma^2 \right )=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(x_{i}-\bar{x}\right)^{2}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i}\left(x_{i}^{2}+\bar{x}^{2}-2 \bar{x} x_{i}\right)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\sum_{i=1}^{n} \bar{x}^{2} f_{i}-\sum_{i=1}^{n} 2 \bar{x} f_{i} x_{i}\right]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} \sum_{i=1}^{n} f_{i}-2 \bar{x} \sum_{i=1}^{n} x_{i} f_{i}\right]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} N-2 \bar{x} \cdot N\bar x\right]$

$\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{N}\left[\sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2} N-2 \bar{x} \cdot N\bar x\right]\\\\\left [ \because \frac{1}{N} \sum_{i=1}^{n} x_{i} f_{i}=\bar{x} \text { or } \sum_{i=1}^{n} x_{i} f_{i}=\mathrm{N} \bar{x} \right ]\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i} x_{i}^{2}+\bar{x}^{2}-2 \bar{x}^{2}=\frac{1}{\mathrm{N}} \sum_{i=1}^{n} f_{i} x_{i}^{2}-\bar{x}^{2}$

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Dispersion (Variance and Standard Deviation)

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