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IF the variance of 10 natural numbers 1, 1, 1……… 1, k less than 10, then the maximum possible value of k is…………………

0 2 Views | Posted a month ago
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  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    σ 2 = ∑ x 2 n − ( ∑ x n ) 2 = 9 + k 2 1 0 − ( 9 + k 1 0 ) 2 < 1 0

    ⇒ k < 1 0 1 0 3 + 1 ⇒ k ≤ 1 1        

    ∴ maximum value of k is 11

Similar Questions for you

The number of values of a ∈ N  such that the variance of 3, 7, 12, a, 43 -a is a natural number is:
A
alok kumar singh

M e a n = 3 + 1 2 + 7 + a + ( 4 3 − a ) 5 = 1 3  

Variance =  3 2 + 1 2 2 + 7 2 + a 2 + ( 4 3 − a ) 2 5 − ( 1 3 ) 2  

⇒ 2 a 2 − a + 1 5 → N a t u r a l     n u m b e r  

Let 2a2 – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not  4 λ     o r     4 λ + 1     f r o m .  

As each square form is  4 λ     o r     4 λ + 1  

Mean and variance of 10 positive observations is 7 and 8. If we add 5 to each number then sum of mean and variance of new data is
R
Raj Pandey

Clearly mean will increase by 5 units and variance will be un changed so the sum would be 7 + 8 + 5 = 20 .7+8+5=20

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