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 \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \] and \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \]. Then which of the following is necessarily true?
A.\[\mu = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \] and \[\sigma \leqslant \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \]
B.\[\mu = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \] and \[\sigma \geqslant \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \]
C.\[\sigma = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \]
D.\[\mu \ne \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \]

Answer 1

Hint: First, we will use the formula the mean is the sum of all the observations and dividing it by total number of observations. Then we will use the given conditions in the formula to find the value of \[\mu \] and \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \] to find the required option.

Complete step-by-step answer:
We are given that 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 \[j = 3,4,...,n\].
We also know that the mean and the standard deviation of the list are \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \] and \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\sigma } \].
We know that the formula the mean is the sum of all the observations and dividing it by the total number of observations.
Using the given conditions for \[\mu \], we have
\[ \Rightarrow \mu = \dfrac{{{x_1} + {x_2} + ... + {x_n}}}{n}{\text{ .....eq.(1)}}\]
Using the given conditions for \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \], we have
\[ \Rightarrow \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } = \dfrac{{{y_1} + {y_2} + ... + {y_n}}}{n}\]
Substituting the value of \[{y_1} = \dfrac{{{x_1} + {x_2}}}{2}\], \[{y_2} = \dfrac{{{x_1} + {x_2}}}{2}\] and \[{y_j} = {x_j}\] in the above equation, we get
\[
   \Rightarrow \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } = \dfrac{{\dfrac{{{x_1} + {x_2}}}{2} + \dfrac{{{x_2} + {x_1}}}{2} + {x_3} + ... + {x_n}}}{n} \\
   \Rightarrow \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } = \dfrac{{\dfrac{{{x_1} + {x_2} + {x_2} + {x_1}}}{2} + {x_3} + ... + {x_n}}}{n} \\
   \Rightarrow \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } = \dfrac{{\dfrac{{2{x_1} + 2{x_2}}}{2} + {x_3} + ... + {x_n}}}{n} \\
   \Rightarrow \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } = \dfrac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n} \\
 \]
Using the equation (1) in the above equation, we get
\[ \Rightarrow \mu = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\mu } \]
Hence, option B is correct.

Note:We need to know that average run is the mean. We know that a standard deviation is a measure of how spread out numbers are and the coefficient of variation is a statistical measure of the relative dispersion of a data point in a data series. The key point to solve this question is just to know the formula of coefficient and then substitute value properly.