Standard deviation(2)

Probability Level 3

σ = ∑ i = 1 n ( x i − x ˉ ) 2 n \large \sigma = \sqrt{\dfrac{\sum_{i=1}^{n}(x_{i}-\bar{x})^2}{n}}

There is a group of numbers all belong to the set { p , q } \{p,q\} . If all numbers in the group are squared , the standard deviation σ \sigma of the group remains unchanged . Which of the following(s) about the set { p , q } \{p,q\} must satisfy the condition above ?

(1). { p , q } = { 0 , 1 } \{p,q\} = \{0,1\}

(2). { p , q } = { − 1 , 0 } \{p,q\} = \{-1,0\}

(3). { p , q } = { − 1 , 1 } \{p,q\} = \{-1,1\}

(4). { p , q } = { − 2 , 1 } \{p,q\} = \{-2,1\}

(1) and (2) (1) , (2) and (4) (1) only (1) , (2) and (3) (1) and (3)

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Dec 23, 2017

p 1 p_{1} denotes number of p p , p 2 p_{2} denotes number of q q , σ 1 \sigma_{1} denotes original standard deviation and σ 2 \sigma_{2} denotes new standard deviation

For (1) :

( p 1 + p 2 ) σ 1 2 = ( ( 0 ) p 1 + ( 1 ) p 2 p 1 + p 2 − 0 ) 2 + ( ( 0 ) p 1 + ( 1 ) p 2 p 1 + p 2 − 1 ) 2 = ( p 2 p 1 + p 2 ) 2 + ( − p 1 p 1 + p 2 ) 2 ( p 1 + p 2 ) σ 2 2 = ( ( 0 2 ) p 1 + ( 1 2 ) p 2 p 1 + p 2 − 0 2 ) 2 + ( ( 0 2 ) p 1 + ( 1 2 ) p 2 p 1 + p 2 − 1 2 ) 2 = ( p 2 p 1 + p 2 ) 2 + ( − p 1 p 1 + p 2 ) 2 = ( p 1 + p 2 ) σ 1 2 \large (p_{1}+p_{2})\sigma_{1}^2 = (\frac{(0)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-0)^2+(\frac{(0)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-1)^2 = (\frac{p_{2}}{p_{1}+p_{2}})^2+(\frac{-p_{1}}{p_{1}+p_{2}})^2 \\ \large (p_{1}+p_{2})\sigma_{2}^2 = (\frac{(0^2)p_{1}+(1^2)p_{2}}{p_{1}+p_{2}}-0^2)^2+(\frac{(0^2)p_{1}+(1^2)p_{2}}{p_{1}+p_{2}}-1^2)^2 = (\frac{p_{2}}{p_{1}+p_{2}})^2+(\frac{-p_{1}}{p_{1}+p_{2}})^2 = (p_{1}+p_{2})\sigma_{1}^2

∴ \therefore (1) is correct

For (2) :

( p 1 + p 2 ) σ 1 2 = ( ( − 1 ) p 1 + ( 0 ) p 2 p 1 + p 2 − ( − 1 ) ) 2 + ( ( − 1 ) p 1 + ( 0 ) p 2 p 1 + p 2 − 0 ) 2 = ( p 2 p 1 + p 2 ) 2 + ( − p 1 p 1 + p 2 ) 2 ( p 1 + p 2 ) σ 2 2 = ( ( − 1 ) 2 p 1 + ( 0 2 ) p 2 p 1 + p 2 − ( − 1 ) 2 ) 2 + ( ( − 1 ) 2 p 1 + ( 0 2 ) p 2 p 1 + p 2 − 0 2 ) 2 = ( p 2 p 1 + p 2 ) 2 + ( − p 1 p 1 + p 2 ) 2 = ( p 1 + p 2 ) σ 1 2 \large (p_{1}+p_{2})\sigma_{1}^2 = (\frac{(-1)p_{1}+(0)p_{2}}{p_{1}+p_{2}}-(-1))^2+(\frac{(-1)p_{1}+(0)p_{2}}{p_{1}+p_{2}}-0)^2 = (\frac{p_{2}}{p_{1}+p_{2}})^2+(\frac{-p_{1}}{p_{1}+p_{2}})^2 \\ \large (p_{1}+p_{2})\sigma_{2}^2 = (\frac{(-1)^2p_{1}+(0^2)p_{2}}{p_{1}+p_{2}}-(-1)^2)^2+(\frac{(-1)^2p_{1}+(0^2)p_{2}}{p_{1}+p_{2}}-0^2)^2 = (\frac{p_{2}}{p_{1}+p_{2}})^2+(\frac{-p_{1}}{p_{1}+p_{2}})^2 = (p_{1}+p_{2})\sigma_{1}^2

∴ \therefore (2) is correct

For (3) :

( p 1 + p 2 ) σ 1 2 = ( ( − 1 ) p 1 + ( 1 ) p 2 p 1 + p 2 − ( − 1 ) ) 2 + ( ( − 1 ) p 1 + ( 1 ) p 2 p 1 + p 2 − 1 ) 2 = ( 2 p 2 p 1 + p 2 ) 2 + ( − 2 p 1 p 1 + p 2 ) 2 \large (p_{1}+p_{2})\sigma_{1}^2 = (\frac{(-1)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-(-1))^2+(\frac{(-1)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-1)^2 = (\frac{2p_{2}}{p_{1}+p_{2}})^2+(\frac{-2p_{1}}{p_{1}+p_{2}})^2

( p 1 + p 2 ) σ 2 2 = 0 \large (p_{1}+p_{2})\sigma_{2}^2 =0 ( Note that all datas equal 1 1 )

∴ \therefore (3) is wrong

For (4) :

( p 1 + p 2 ) σ 1 2 = ( ( − 2 ) p 1 + ( 1 ) p 2 p 1 + p 2 − ( − 2 ) ) 2 + ( ( − 2 ) p 1 + ( 1 ) p 2 p 1 + p 2 − 1 ) 2 = ( 3 p 2 p 1 + p 2 ) 2 + ( − 3 p 1 p 1 + p 2 ) 2 ( p 1 + p 2 ) σ 2 2 = ( ( − 2 ) 2 p 1 + ( 1 2 ) p 2 p 1 + p 2 − ( − 2 ) 2 ) 2 + ( ( − 2 ) 2 p 1 + ( 1 2 ) p 2 p 1 + p 2 − 1 2 ) 2 = ( 3 p 2 p 1 + p 2 ) 2 + ( 3 p 1 p 1 + p 2 ) 2 = ( p 1 + p 2 ) σ 2 1 \large (p_{1}+p_{2})\sigma_{1}^2 = (\frac{(-2)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-(-2))^2+(\frac{(-2)p_{1}+(1)p_{2}}{p_{1}+p_{2}}-1)^2 = (\frac{3p_{2}}{p_{1}+p_{2}})^2+(\frac{-3p_{1}}{p_{1}+p_{2}})^2 \\ \large (p_{1}+p_{2})\sigma_{2}^2 = (\frac{(-2)^2p_{1}+(1^2)p_{2}}{p_{1}+p_{2}}-(-2)^2)^2+(\frac{(-2)^2p_{1}+(1^2)p_{2}}{p_{1}+p_{2}}-1^2)^2 = (\frac{3p_{2}}{p_{1}+p_{2}})^2+(\frac{3p_{1}}{p_{1}+p_{2}})^2 = (p_{1}+p_{2})\sigma_{2}^1

∴ \therefore (4) is correct

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