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Given data 60, 60, 44, 58, 68, α, β, 56 has mean 58, variance = 66.2 then find α2 + β2

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  • 1 Answer

  • 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

Similar Questions for you

For the following data table

xi

fi

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

2

4

7

8

6

Find the value of 20 M (where M is median of the data)
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

Let mean and variance of 6 observations a, b, 68, 44, 40, 60 be 55 and 194. If a > b then find a + 3b
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

Let X1, X2,…..X18 be eighteen observations such that ∑ i = 1 1 8 ( X i − α ) = 3 6     a n d     ∑ i = 1 1 8 ( X i − β ) 2 = 9 0 ,  where a and b are distinct real numbers. If the standard deviation of these observations is 1, then the value of   | α − β |     i s . . . . . . . .

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  

 

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

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  

The variance of the data

2001, 2003, 2006, 2007, 2009, 2010 is
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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