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Let the observations x_{i}(1\leq i\leq 10) satisfy the equations, \sum_{i=1}^{10}(x_{i}-5)=10 and \sum_{i=1}^{10}(x_{i}-5)^{2}=40. If \mu \: and\: \lambda are the mean and the variance of the observations, x_{1}-3,x_{2}-3,\cdots ,x_{10}-3, then the ordered pair \left ( \mu ,\lambda \right ) is equal to : 

Option: 1

(6,3)


Option: 2

(3,6)


Option: 3

(3,3)


Option: 4

(6,6)


Answers (1)

best_answer

 

 

Mean -

MEAN (Arithmetic Mean)

The mean is the sum of the value of each observation in a dataset divided by the number of observations. For example, to calculate the mean weight of 50 people, add the 50 weights together and divide by 50. Technically this is the arithmetic mean. 

 

Mean of the Ungrouped Data 

If n observations in data are x1, x2, x3, ……, xn, then arithmetic mean \mathit{\bar x} is given by    

\mathit{\bar x}=\frac{x_1+x_2+x_3+\ldots\dots +x_n }{n}=\frac{1}{n}\sum^{n}_{i=1}x_i

Shortcut Method

\\\text{Arithmetic\;mean}\;\;\bar x=A+\frac{\sum_{i=1}^{n}d_i}{n}\\\text{where, }A=\text{assumed mean, }\;\\d_i=\text{deviation from assumed mean}=x_i-A

 

Mean of Discrete Frequency Distribution and Continuous Frequency Distribution

If observations in data are x1, x2, x3, ……, xn with respective frequencies f1, f2, f3, ……, fn ; then 

Sum of the value of the observation = f1x1 + f2??????x2 + f3?????x3 + …….. + fnxn

and            Number of observations = f1 + f2 + f3 + ……+ fn

The mean in this case is given by 

\bar x=\frac{f_1x_1+f_2x_2+f_3x_3+\ldots\ldots +f_nx_n}{f_1+f_2+f_3+\ldots\ldots +f_n}=\frac{\sum_{i=1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}

Shortcut Method

\\\text{Arithmetic\;mean}\;\;\bar x=A+\frac{\sum_{i=1}^{n}f_i\left ( x_i-A \right )}{n}\\\text{where, }A=\text{assumed mean. }

-

 

 

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}

 


Variance and Standard Deviation of a Discrete Frequency Distribution

The given discrete frequency distribution be

         \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 continuous frequency distribution 

The formula for variance and standard deviation are the same as in the case of discrete 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}\\\\\text{or}\;\;\sigma^{2}=\frac{1}{N} \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\frac{\sum_{i=1}^{n} f_{i} x_{i}}{N}\right)^{2}=\frac{1}{N^{2}}\left[N \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\sum_{i=1}^{n} f_{i} x_{i}\right)^{2}\right]

\text{Thus, standard deviation }(\sigma)=\frac{1}{N} \sqrt{N \sum_{i=1}^{n} f_{i} x_{i}^{2}-\left(\sum_{i=1}^{n} f_{i} x_{i}\right)^{2}}

-

\text { Mean }\left(x_{i}-5\right)=\frac{\sum\left(x_{i}-5\right)}{10}=1 \\ \therefore \text{Mean,}\quad \lambda=\left(x_{i}-5+2\right)+2=3

\mu=\operatorname{var} =\left[\left(\mathrm{x}_{1}-5\right)=\frac{\sum\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}}{10}-\frac{\Sigma\left(\mathrm{x}_{\mathrm{i}}-5\right)}{10}\right]=3

Correct Option (3)

Posted by

sudhir.kumar

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