The probability distribution of a discrete random variable X is given by P(X = x) = { k(x + 1), x = 1, 2, 3, k(x - 3), x = 4, 5, 6, 0, elsewhere. (a). Find the value of the constant k. (b). Find the mean of X. (c). Find the variance of X.

$$\sum P(X=x) = 1$$

Nous listons les probabilités pour chaque valeur de $x$ :
Pour $x=1, 2, 3$: $P(X=x) = k(x+1)$
$P(X=1) = k(1+1) = 2k$
$P(X=2) = k(2+1) = 3k$
$P(X=3) = k(3+1) = 4k$

Pour $x=4, 5, 6$: $P(X=x) = k(x-3)$
$P(X=4) = k(4-3) = k$
$P(X=5) = k(5-3) = 2k$
$P(X=6) = k(6-3) = 3k$

$$2k + 3k + 4k + k + 2k + 3k = 1$$ $$(2+3+4+1+2+3)k = 1$$ $$15k = 1$$ $$k = \frac{1}{15}$$

La valeur de la constante $k$ est $\boxed{\frac{1}{15}}$.

$$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) + 5 \cdot P(X=5) + 6 \cdot P(X=6)$$ $$E(X) = 1(2k) + 2(3k) + 3(4k) + 4(k) + 5(2k) + 6(3k)$$ $$E(X) = 2k + 6k + 12k + 4k + 10k + 18k$$

$$E(X) = (2+6+12+4+10+18)k$$ $$E(X) = 52k$$ $$E(X) = 52 \cdot \frac{1}{15}$$ $$E(X) = \frac{52}{15}$$

La moyenne de $X$ est $\boxed{\frac{52}{15}}$.

Nous devons d'abord calculer $E(X^2)$.

$$E(X^2) = \sum x^2 \cdot P(X=x)$$ $$E(X^2) = 1^2 P(X=1) + 2^2 P(X=2) + 3^2 P(X=3) + 4^2 P(X=4) + 5^2 P(X=5) + 6^2 P(X=6)$$ $$E(X^2) = 1(2k) + 4(3k) + 9(4k) + 16(k) + 25(2k) + 36(3k)$$ $$E(X^2) = 2k + 12k + 36k + 16k + 50k + 108k$$

$$E(X^2) = (2+12+36+16+50+108)k$$ $$E(X^2) = 224k$$ $$E(X^2) = 224 \cdot \frac{1}{15}$$ $$E(X^2) = \frac{224}{15}$$

$$Var(X) = E(X^2) - (E(X))^2$$ $$Var(X) = \frac{224}{15} - \left(\frac{52}{15}\right)^2$$ $$Var(X) = \frac{224}{15} - \frac{2704}{225}$$

Pour soustraire, mettez les fractions sur un dénominateur commun (225).

$$Var(X) = \frac{224 \times 15}{15 \times 15} - \frac{2704}{225}$$ $$Var(X) = \frac{3360}{225} - \frac{2704}{225}$$ $$Var(X) = \frac{3360 - 2704}{225}$$ $$Var(X) = \frac{656}{225}$$

La variance de $X$ est $\boxed{\frac{656}{225}}$.