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There are 6 cards in a bag numbered 1 through 6. Suppose we draw two cards numbered A and B out of the bag(without replacement), what is the variance of A 2B

Sagot :

The Variance of (A +2B) is 12.6724.

What is  Variance ?

Variability is measured by the variance. The average of the squared deviations from the mean is used to compute it. The degree of dispersion in your data collection is indicated by variance. The variance is greater in respect to the mean the more dispersed the data.

Here we have to select two card out of 6 card. now suppose we take 2 card (1,2 ) out of six card (1,2,3,4,5,6) . so here A=1 and B=2 but it is also possible that A=2 and B =1 because it represent same two card . Therefore here order of card matter.

Hence the posible out come of selecting 2 card are (without replacement)

-- (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) -- (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) -- (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) -- (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) -- (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) --

Therefore A+2B of this outcome are

A B A+2B

1 2 5

1 3 7

1 4 9

1 5 11

1 6 13

2 1 4

2 3 8

2 4 10

2 5 12

2 6 14

3 1 5

3 2 7

3 4 11

3 5 13

3 6 15

4 1 6

4 2 8

4 3 10

4 5 14

4 6 16

5 1 7

5 2 9

5 3 11

5 4 13

5 6 17

6 1 8

6 2 10

6 3 12

6 4 14

6 5 16

therefore there variance of A +2B can be computed as follows

x x2

5 25

7 49

9 81

11 121

13 169

4 16

8 64

10 100

12 144

14 196

5 25

7 49

11 121

13 169

15 225

6 36

8 64

10 100

14 196

16 256

7 49

9 81

11 121

13 169

17 289

8 64

10 100

12 144

14 196

16 256

[tex]\Sigma x=315[/tex]       [tex]\sum x^2=3675[/tex]

Therefore variance

[tex]$\begin{aligned}& =\frac{\sum x^2-\frac{\left(\sum x\right)^2}{n}}{n-1} \\& =\frac{3675-\frac{(315)^2}{30}}{30-1} \\& =\frac{3675-3307.5}{29} \\& =12.6724\end{aligned}$[/tex]

The Variance of (A +2B) is 12.6724.

To learn more about Variance visit:https://brainly.com/question/14116780

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