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