Permutations And Combinations

Question
CBSEENMA11015510

Let x1, x2, ......, xn be n observations, and let top enclose straight x be their arithmetic mean and σ2 be their variance.
Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4 σ2.
Statement 2: Arithmetic mean of 2x1, 2x2, ......, 2xn is 4top enclose straight x.

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is false

Solution

D.

Statement 1 is true, statement 2 is false

top enclose straight x is the AM and σ2 is the variance of n observations x1, x2, x3......xn
AM of 2x1, 2x2, 2x3, .......2xn
space equals space fraction numerator 2 straight x subscript 1 space plus space 2 straight x subscript 2 space plus 2 straight x subscript 3 space plus.....2 straight x subscript straight n over denominator straight n end fraction

equals space 2 open parentheses fraction numerator straight x subscript 1 plus straight x subscript 2 plus straight x subscript 3 space plus..... plus straight x subscript straight n over denominator straight n end fraction close parentheses space equals space 2 top enclose straight x
Hence, it prove that statement 2 is false.
 

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