Q.(a) The amount of bread (in hundreds of pounds) $x$ that a certain bakery is able to sell in a day is found to be a numerical valued random phenomenon, with a probability function specified by the probability density function $f(x)$ is given by, $f(x) = \begin{cases} Ax, & \text{for } 0 \le x < 10 \\ A(20 - x), & \text{for } 10 \le x < 20 \\ 0, & \text{otherwise} \end{cases}$

(a) Total area $=1$ gives $A = 0.01$; the areas give $0.5$, $0.5$, $0.75$. (b) Reducing the system leaves $(a-3)z = b-10$, which classifies the three cases.

Part (a) — Triangular density

$f(x) = \begin{cases} Ax, & 0 \le x < 10 \\ A(20 - x), & 10 \le x < 20 \\ 0, & \text{otherwise.} \end{cases}$

(i) Find $A$ from $\int_{-\infty}^{\infty} f(x)\,dx = 1$:

$\int_0^{10} Ax\,dx + \int_{10}^{20} A(20 - x)\,dx = 1.$

$A\left[\frac{x^2}{2}\right]_0^{10} + A\left[20x - \frac{x^2}{2}\right]_{10}^{20} = A(50) + A(50) = 100A = 1.$

$\Rightarrow A = \frac{1}{100} = 0.01.$

(ii) Probabilities.

1. More than $10$: $P(X > 10) = \int_{10}^{20} A(20 - x)\,dx = 50A = 0.5.$
2. Less than $10$: $P(X < 10) = \int_0^{10} Ax\,dx = 50A = 0.5.$
3. Between $5$ and $15$: $P(5 < X < 15) = \int_5^{10} Ax\,dx + \int_{10}^{15} A(20 - x)\,dx.$ $\int_5^{10} Ax\,dx = A\left[\frac{x^2}{2}\right]_5^{10} = A(50 - 12.5) = 37.5A = 0.375.$ $\int_{10}^{15} A(20 - x)\,dx = A\left[20x - \frac{x^2}{2}\right]_{10}^{15} = A(187.5 - 150) = 37.5A = 0.375.$ $P(5 < X < 15) = 0.375 + 0.375 = 0.75.$

Part (b) — Values of $a$ and $b$

System: $x+y+z=6$, $x+2y+3z=10$, $x+2y+az=b$.

Subtracting equation 2 from equation 3 (they share $x + 2y$):

$(a - 3)z = b - 10.$

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