### Changes of variance, mean and range after adding a value

Question Sample Titled 'Changes of variance, mean and range after adding a value'

Let ${x}_{{1}},{y}_{{1}}$ and ${z}_{{1}}$ be the mean, the range and the variance of a group of numbers ${\left\lbrace{a}_{{1}},{a}_{{2}},{a}_{{3}},\ldots,{a}_{{105}}\right\rbrace}$ respectively. If ${x}_{{2}},{y}_{{2}}$ and ${z}_{{2}}$ are the mean, the range and the variance of the group of numbers ${\left\lbrace{a}_{{1}},{a}_{{2}},{a}_{{3}},\ldots,{a}_{{105}},{x}_{{1}}\right\rbrace}$ respectively. Which of the following must be true?

 I. ${x}_{{1}}={x}_{{2}}$ II. ${y}_{{1}}={y}_{{2}}$ III. ${z}_{{1}}={z}_{{2}}$

A
I and II only
B
I, II and III
C
II and III only
D
I and III only

 (I) By adding the mean of the old set of data into the new set of data, the mean would remain unchanged. Below is the mathematical explanation. ${x}_{{1}}$ $=\dfrac{{1}}{{105}}{\left({a}_{{1}}+{a}_{{2}}+{a}_{{3}}+\ldots+{a}_{{105}}\right)}$ given ${a}_{{1}}+{a}_{{2}}+{a}_{{3}}+\ldots+{a}_{{105}}$ $={105}{x}_{{1}}$ New mean $={x}_{{2}}$  $=\dfrac{{1}}{{106}}{\left({a}_{{1}}+{a}_{{2}}+{a}_{{3}}+\ldots+{a}_{{105}}+{x}_{{1}}\right)}$  $=\dfrac{{1}}{{106}}{\left({105}{x}_{{1}}+{x}_{{1}}\right)}$  $=\dfrac{{1}}{{106}}{\left({106}{x}_{{1}}\right)}$  $={x}_{{1}}$ ∴  ${x}_{{1}}$ $={x}_{{2}}$ (II) As the range is the difference of the largest datum and the smallest datum, adding the mean of the old set of data into the new set of data would not change the range. ∴  ${y}_{{1}}$ $={y}_{{2}}$ (III) Adding the mean of the old set of data into the new set of data would decrease the dispersion of the data by decreasing the standard variance.  As variance is the square of standard deviation ∴   The new variance would decrease as well. Below is the mathematical explanation. ${z}_{{1}}=\dfrac{{1}}{{105}}{\left[{\left({a}_{{1}}-{x}_{{1}}\right)}^{{{2}}}+{\left({a}_{{2}}-{x}_{{1}}\right)}^{{{2}}}+\ldots+{\left({a}_{{105}}-{x}_{{1}}\right)}^{{{2}}}\right]}$ ${z}_{{2}}=\dfrac{{1}}{{106}}{\left[{\left({a}_{{1}}-{x}_{{1}}\right)}^{{{2}}}+{\left({a}_{{2}}-{x}_{{1}}\right)}^{{{2}}}+\ldots+{\left({a}_{{105}}-{x}_{{1}}\right)}^{{{2}}}+{\left({x}_{{1}}-{x}_{{1}}\right)}^{{{2}}}\right]}$ $=\dfrac{{1}}{{106}}{\left[{\left({a}_{{1}}-{x}_{{1}}\right)}^{{{2}}}+{\left({a}_{{2}}-{x}_{{1}}\right)}^{{{2}}}+\ldots+{\left({a}_{{105}}-{x}_{{1}}\right)}^{{{2}}}+{0}\right]}$ $=\dfrac{{1}}{{106}}{\left[{\left({a}_{{1}}-{x}_{{1}}\right)}^{{{2}}}+{\left({a}_{{2}}-{x}_{{1}}\right)}^{{{2}}}+\ldots+{\left({a}_{{105}}-{x}_{{1}}\right)}^{{{2}}}\right]}$ Note that ${z}_{{1}}$ and ${z}_{{2}}$ have the same numerator but a different denominator. ∴  ${z}_{{1}}>{z}_{{2}}$ as ${z}_{{2}}$ has a larger denominator.

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