The number of values of a N  such that the variance of 3, 7, 12, a, 43 -a is a natural number is:

Option 1 -

0

Option 2 -

2

Option 3 -

5

Option 4 -

Infinite

0 3 Views | Posted 2 weeks ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 weeks ago
    Correct Option - 4


    Detailed Solution:

    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  

Similar Questions for you

A
alok kumar singh

Variance = x 2 n ( x ¯ ) 2  

6 0 2 + 6 0 2 + 4 4 2 + 5 8 2 + 6 8 2 + α 2 + β 2 + 5 6 2 8 = ( 5 8 ) 2 = 6 6 . 2            

7 2 0 0 + 1 9 3 6 + 3 3 6 4 + 4 6 2 4 + 3 1 3 6 + α 2 + β 2 8 = 3 3 6 4 = 6 6 . 2             

2 5 3 2 . 5 + α 2 + β 2 8 3 3 6 4 = 6 6 . 2            

α2 + β2 = 897.7 × 8

= 7181.6

A
alok kumar singh

xi

fi

c.f.

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

2

4

7

8

6

2

6

13

21

27

N = f = 2 7

( N 2 ) = 2 7 2 = 1 3 . 5

So, we have median lies in the class 12 – 16

I1 = 12, f = 8, h = 4, c.f. = 13

So, here we apply formula

M = I 1 + N 2 c . f . f × h = 1 2 + 1 3 . 5 1 3 8 × 4

= 1 2 + 5 2

M = 2 4 . 5 2 = 1 2 . 2 5

20 M = 20 × 12.25

= 245

A
alok kumar singh

  a + b + 6 8 + 4 4 + 4 0 + 6 0 6 = 5 5

212 + a + b = 330

a + b = 118

x i 2 n ( x ¯ ) 2 = 1 9 4          

a 2 + b 2 + ( 6 8 ) 2 + ( 4 4 ) 2 + ( 4 0 ) 2 + ( 6 0 ) 2 6 = ( 5 5 ) 2 = 1 9 4

= 3219

11760 + a2 + b2 = 19314

a2 + b2 = 19314 – 11760

= 7554

(a + b)2 –2ab = 7554

From here b = 41.795

a + b = 118

a + b + 2b = 118 + 83.59

= 201.59

A
alok kumar singh

Kindly go throuigh the solution

Given   i = 1 1 8 ( x i α ) = 3 6 i . e i = 1 1 8 x i 1 8 α = 3 6 . . . . . . . . . . ( i )

&       i = 1 1 8 ( x i β ) 2 = 9 0 i . e i = 1 1 8 x i 2 2 β x i + 1 8 β 2 = 9 0 . . . . . . . . . . . . . ( i i )      

(i) & (ii)   i = 1 1 8 x i 2 = 9 0 1 8 β + 3 6 β ( α + 2 ) . . . . . . . . . . . . . ( i i i )

Now variance σ 2 = x i 2 n ( x i n ) 2 = 1 given

->(a - b) (a - b + 4) = 0

Since α β s o | α β | = 4  

 

V
Vishal Baghel

Variance of 2001, 2003, 2006, 2007, 2009, 2010 is same as variance of 1, 3, 6, 7, 9, 10

mean =  1 + 3 + 6 + 7 + 9 + 1 0 6 = 6

var (x) =  1 2 + 3 2 + 6 2 + 7 2 + 9 2 + 1 0 2 6 ? 3 6

var (x) = 10

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