Now, the values of X are
X= 2,3,4,5......., 12
We note the values of X and their corresponding probablities without the factor $\frac{1}{36}$
X |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
P(x) |
1 |
2 |
3 |
4 |
5 |
6 |
5 |
4 |
3 |
2 |
1 |
Now, V(X) = E(X2) - E2(X) ..........(1)
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We calculate the mean i.e E(X)
Now, just look at P(X) values from 1 to 6. X values for them are one greater.
So, this is $\sum_{1}^{6}$n(n+1)
Now, for the rest of the values, if P(X)=n then X=13-n
So, this is $\sum_{1}^{5}$n(13-n)
So, E(x)
= $\sum_{1}^{6}$n(n+1) + $\sum_{1}^{5}$n(13-n)
= 7
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We also calculate E(X2)
E(X2)
= $\sum_{1}^{6}$n(n+1)2 + $\sum_{1}^{5}$n(13-n)2
= $\frac{1974}{36}$
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Now, from (1), we get
V(X)
= $\frac{1974}{36}$ - 72
= $\frac{35}{6}$
So, option A should be the correct answer.