Question

Let $n \geqslant 3$. A list of numbers ${x_1},{x_2},.....{x_n}$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers ${y_1},{y_2},.....{y_n}$ is made as follows : ${y_1} = \dfrac{{{x_1} + {x_2}}}{2}$, ${y_2} = \dfrac{{{x_1} + {x_2}}}{2}$ and ${y_j} = {x_j}$for $j = 3,4,....n$.
The mean and the standard deviation of the list are $\widehat \mu$ and $\widehat \sigma$. Then which of the following is necessarily true?
A. $\mu = \widehat \sigma$ and $\sigma \leqslant \widehat \sigma$
B. $\mu = \widehat \mu$ and $\sigma \geqslant \widehat \sigma$
C. $\sigma = \widehat \sigma$
D. $\mu \ne \widehat \mu$

Answer 1

Given two lists of numbers or observations with different means and different standard deviations. To solve this problem we should be familiar with what mean and standard deviation are. Mean and standard deviation are the statistical values in statistics and probability theory. Mean is just the average of the observations, which is given by the ratio of the sum of observations to the value of total no. of observations, while the standard deviation is a measure of how the numbers spread out.

Complete step-by-step solution:
As already discussed, the mean is given by the average of the observation which is the ratio of the sum of all the observations to the total no. of observations.
$\mu = \dfrac{{\sum\limits_i {{x_i}} }}{n}$
The standard deviation is a measure of the amount of variation or dispersion of the observations. If the standard deviation is low, it means that the values are close to the mean. But if the standard deviation is high this indicates that the values are very far from mean and are spread out over a wide range.
$\sigma = \sqrt {\dfrac{{\sum\limits_i {{x_i}^2} }}{n} - {\mu ^2}}$, is the standard deviation. If squared on both sides, is given below :
${\sigma ^2} = \dfrac{{\sum\limits_i {{x_i}^2} }}{n} - {\mu ^2}$
Consider the first list of observations ${x_1},{x_2},....{x_n}$:
Given the mean of this list is $\mu$ which is given by:
$\Rightarrow \mu = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}$
Standard deviation is given by:
$\Rightarrow {\sigma ^2} = \dfrac{{\sum\limits_i {{x_i}^2} }}{n} - {\mu ^2}$
$\Rightarrow {\sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
Now consider the new list of observations ${y_1},{y_2},....{y_n}$:
Given the mean of this list is $\widehat \mu$ which is given by:
$\Rightarrow \widehat \mu = \dfrac{{{y_1} + {y_2} + {y_3} + ...... + {y_n}}}{n}$
Standard deviation is given by:
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{\sum\limits_i {{y_i}^2} }}{n} - {\widehat \mu ^2}$
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{{y_1}^2 + {y_2}^2 + {y_3} + ...... + {y_n}^2}}{n} - {\widehat \mu ^2}$
Now substitute the given expressions ${y_1} = \dfrac{{{x_1} + {x_2}}}{2}$, ${y_2} = \dfrac{{{x_1} + {x_2}}}{2}$ in new mean expression:
$\Rightarrow \widehat \mu = \dfrac{{\dfrac{{{x_1} + {x_2}}}{2} + \dfrac{{{x_1} + {x_2}}}{2} + {x_3} + ...... + {x_n}}}{n}$ ;
$\because {y_j} = {x_j}$for $j = 3,4,....n$.
$\Rightarrow \widehat \mu = \dfrac{{\dfrac{{2{x_1}}}{2} + \dfrac{{2{x_2}}}{2} + {x_3} + ...... + {x_n}}}{n}$
$\therefore \widehat \mu = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}$
Whereas the mean of the first list of observations is also the same as the new mean.
$\Rightarrow \mu = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}$ and $\widehat \mu = \dfrac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}$
$\therefore \mu = \widehat \mu$
Now calculating the standard deviations of the two lists of observations:
The standard deviation of the first list of observations ${x_1},{x_2},....{x_n}$ is given by :
$\Rightarrow {\sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
Now the standard deviation of the new list of observations ${y_1},{y_2},....{y_n}$is given by :
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{{y_1}^2 + {y_2}^2 + {y_3}^2 + ...... + {y_n}^2}}{n} - {\widehat \mu ^2}$
But just found out that $\widehat \mu = \mu$, hence substituting $\mu$ in place of $\widehat \mu$ , also ${y_1} = \dfrac{{{x_1} + {x_2}}}{2}$, ${y_2} = \dfrac{{{x_1} + {x_2}}}{2}$:
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{{{\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)}^2} + {{\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)}^2} + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
$\because {y_j} = {x_j}$for $j = 3,4,....n$.
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{\dfrac{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}}}{4} + \dfrac{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}}}{4} + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{\dfrac{{2{x_1}^2 + 2{x_2}^2 + 4{x_1}{x_2}}}{4} + + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
$\Rightarrow {\widehat \sigma ^2} = \dfrac{{\dfrac{{{x_1}^2 + {x_2}^2}}{2} + {x_1}{x_2} + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
But the standard deviation of the first list of observation is given by :
$\Rightarrow {\sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}$
Let’s subtract ${\sigma ^2} - {\widehat \sigma ^2}$ which is given by : \Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2} - \left( {\dfrac{{\dfrac{{{x_1}^2 + {x_2}^2}}{2} + {x_1}{x_2} + {x_3}^2 + ...... + {x_n}^2}}{n} - {\mu ^2}} \right)$$$$ \Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + ...... + {x_n}^2}}{n} - \left( {\dfrac{{\dfrac{{{x_1}^2 + {x_2}^2}}{2} + {x_1}{x_2} + {x_3}^2 + ...... + {x_n}^2}}{n}} \right) - {\mu ^2} + {\mu ^2}
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2}}{n} - \dfrac{{\dfrac{{{x_1}^2 + {x_2}^2}}{2} + {x_1}{x_2}}}{n} + \left( {\dfrac{{{x_3}^2 + ...... + {x_n}^2}}{n}} \right) - \left( {\dfrac{{{x_3}^2 + ...... + {x_n}^2}}{n}} \right) + {\mu ^2} - {\mu ^2}$
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2}}{n} - \dfrac{{\dfrac{{{x_1}^2 + {x_2}^2}}{2} + {x_1}{x_2}}}{n}$
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2}}{n} - \dfrac{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}}}{{2n}}$
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{x_1}^2 + {x_2}^2 - 2{x_1}{x_2}}}{{2n}}$
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} = \dfrac{{{{({x_1} - {x_2})}^2}}}{{2n}}$
Where $\dfrac{{{{({x_1} - {x_2})}^2}}}{{2n}} \geqslant 0$
$\Rightarrow {\sigma ^2} - {\widehat \sigma ^2} \geqslant 0$
$\therefore {\sigma ^2} \geqslant {\widehat \sigma ^2}$
$\therefore \mu = \widehat \mu$, ${\sigma } \geqslant {\widehat \sigma}$

Option B is the correct answer.

Note: While solving such kinds of problems be careful while solving the algebraic equations and expressions, because a sign change or a single mistake can ruin the whole process and end up getting a wrong result.