For the following probability distribution:
| X | 1 | 2 | 3 | 4 |
| P(X) | $1/10$ | $3/10$ | $3/10$ | $2/5$ |
E(X2) is equal to 10.
Explanation:
We know that
E(X2) = $sum_("i" = 1)^"n" "P"_"i""X"_"i"^2$
= $1 xx 1/10 + 4 xx 1/5 + 9 xx 3/10 + 16 xx 2/5$
= $1/10 + 4/5 + 27/10 + 32/5$
= $28/10 + 36/5$
= $100/10$
= 10