Help with Statistics!!!

Let x¯, M and σ^2 be respectively the mean mode and variance of n observations x1, x2 ,....., xn and di = (−x1−a) ,i=1,2,.....,n, where a is any number. Statement I : Variance of d1,d1,.....,dn is σ2 Statement II : Mean and mode of d1,d2,.....,dn are −x¯−a and -M-a, respectively

Note by Ashley Shamidha
6 years, 2 months ago

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Comments

Please say whether the statements are correct and relevant to each other. Also prove your answer.

Ashley Shamidha - 6 years, 2 months ago

As for the case of mode, there is nothing difficult to understand. You see, if the observations x1,x2,...,xn{ x }_{ 1 },{ x }_{ 2 },...,{ x }_{ n } have frequencies f1,f2,...,fn{ f }_{ 1 },{ f }_{ 2 },...,{ f }_{ n }, then obviously the observations (−x1−a),(−x2−a),...,(−xn−a)(-{ x }_{ 1 }-a),(-{ x }_{ 2 }-a),...,(-{ x }_{ n }-a) must have the same set of frequencies, and thus if the r-th x value has the highest frequency, then the r-th d value will have the same, i.e., will be the mode. Now, dir=(−xr−a)=−M−a{ d }_{ ir}=(-{ x }_{ r }-a)=-M-a which shows why the mode is (-M-a). Now, for the mean calculation, we have dˉ=∑1=1ndifi∑1=1nfi=∑i=1n(−xi−a)fi∑1=1nfi=−∑i=1nxifi−∑i=1nafi∑1=1nfi=−xˉ∑i=1nfi−a∑i=1nfi∑i=1nfi=−xˉ−a\bar { d } =\frac { \sum _{ 1=1 }^{ n }{ { d }_{ i }{ f }_{ i } } }{ \sum _{ 1=1 }^{ n }{ { f }_{ i } } } =\frac { \sum _{ i=1 }^{ n }{ \left( -{ x }_{ i }-a \right) { f }_{ i } } }{ \sum _{ 1=1 }^{ n }{ { f }_{ i } } } =\frac { -\sum _{ i=1 }^{ n }{ { x }_{ i }{ f }_{ i } } -\sum _{ i=1 }^{ n }{ a{ f }_{ i } } }{ \sum _{ 1=1 }^{ n }{ { f }_{ i } } } =\frac { -\bar { x } \sum _{ i=1 }^{ n }{ { f }_{ i } } -a\sum _{ i=1 }^{ n }{ { f }_{ i } } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } =-\bar { x } -a Similarly we can calculate variance as below:σd2=∑i=1n(di−dˉ)2fi∑i=1nfi=∑i=1n{(−xi−a)−(−xˉ−a)}2fi∑i=1nfi=∑i=1n(xi−xˉ)2fi∑i=1nfi=xd2{ \sigma }_{ d }^{ 2 }=\frac { \sum _{ i=1 }^{ n }{ { \left( { d }_{ i }-\bar { d } \right) }^{ 2 }{ f }_{ i } } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } =\frac { \sum _{ i=1 }^{ n }{ { \left\{ \left( { -x }_{ i }-a \right) -\left( -\bar { x } -a \right) \right\} }^{ 2 }{ f }_{ i } } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } =\frac { \sum _{ i=1 }^{ n }{ { \left( { x }_{ i }-\bar { x } \right) }^{ 2 }{ f }_{ i } } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } ={ x }_{ d }^{ 2 }The basic idea is very simple, in case of mode or mean, we are searching for a central representative value of the whole data set. So, if any change is made to the data variable (scale change and/or origin shifting, here multiplying by -1 and then adding -a respectively), the same change is expected to be reflected in the mean or mode value. But in case of variance, which is the measure of how much the values are dispersed from the central value, no origin shifting is reflected (in simple words, the a's get cancelled during taking the differences). Hope this helped you.

Kuldeep Guha Mazumder - 5 years, 6 months ago
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