A random variable X has probability density function given by $$ \mathrm{f}(x)=\left\{\begin{array}{ll}k(3-x) & 1 \leqslant x \leqslant 2 \ 0 & \text { otherwise, }\end{array}\right. $$ where k is a constant. a) Show that $$ \mathrm{k}=\frac{2}{3} $$. b) Find the median of X .

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To solve the problem, we will follow the steps outlined:

### Part (a): Show that $$ k = \frac{2}{3} $$

The probability density function (PDF) must satisfy the condition that the total area under the curve is equal to 1. Therefore, we need to calculate the integral of $$ f(x) $$ over its defined range and set it equal to 1.

The PDF is given by:
$$
f(x) = k(3 - x) \quad \text{for } 1 \leq x \leq 2
$$

We will calculate the integral:
$$
\int_{1}^{2} f(x) \, dx = \int_{1}^{2} k(3 - x) \, dx
$$

Calculating the integral:
$$
\int_{1}^{2} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{2}
$$

Now, we will evaluate this expression:
1. At $$ x = 2 $$:
   $$
   3(2) - \frac{(2)^2}{2} = 6 - 2 = 4
   $$
2. At $$ x = 1 $$:
   $$
   3(1) - \frac{(1)^2}{2} = 3 - 0.5 = 2.5
   $$

Now, subtract the two results:
$$
4 - 2.5 = 1.5
$$

Thus, we have:
$$
\int_{1}^{2} k(3 - x) \, dx = k \cdot 1.5
$$

Setting this equal to 1:
$$
k \cdot 1.5 = 1 \implies k = \frac{1}{1.5} = \frac{2}{3}
$$

### Conclusion for Part (a):
We have shown that $$ k = \frac{2}{3} $$.

### Part (b): Find the median of $$ X $$

The median $$ m $$ of a random variable is the value such that:
$$
P(X \leq m) = 0.5
$$

We need to find $$ m $$ such that:
$$
\int_{1}^{m} f(x) \, dx = 0.5
$$

Substituting $$ f(x) $$:
$$
\int_{1}^{m} \frac{2}{3}(3 - x) \, dx = 0.5
$$

Calculating the integral:
$$
\int_{1}^{m} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{m}
$$

Evaluating this:
1. At $$ x = m $$:
   $$
   3m - \frac{m^2}{2}
   $$
2. At $$ x = 1 $$:
   $$
   3(1) - \frac{(1)^2}{2} = 3 - 0.5 = 2.5
   $$

Thus, we have:
$$
\int_{1}^{m} (3 - x) \, dx = \left( 3m - \frac{m^2}{2} \right) - 2.5
$$

Setting this equal to $$ 0.5 $$:
$$
\frac{2}{3} \left( 3m - \frac{m^2}{2} - 2.5 \right) = 0.5
$$

Multiplying both sides by $$ \frac{3}{2} $$:
$$
3m - \frac{m^2}{2} - 2.5 = \frac{3}{4}
$$

Rearranging gives:
$$
3m - \frac{m^2}{2} = 2.5 + \frac{3}{4} = 2.5 + 0.75 = 3.25
$$

Multiplying through by 2 to eliminate the fraction:
$$
6m - m^2 = 6.5
$$

Rearranging gives:
$$
m^2 - 6m + 6.5 = 0
$$

Now we will solve this quadratic equation using the quadratic formula:
$$
m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6.5}}{2 \cdot 1}
$$
$$
= \frac{6 \pm \sqrt{36 - 26}}{2} = \frac{6 \pm \sqrt{10}}{2}
$$
$$
= 3 \pm \frac{\sqrt{10}}{2}
$$

Since $$ m $$ must be in the interval $$[1, 2]$$, we take the smaller root:
$$
m = 3 - \frac{\sqrt{10}}{2}
$$

### Conclusion for Part (b):
The median of $$ X $$ is $$ m = 3 - \frac{\sqrt{10}}{2} $$.
$$ k = \frac{2}{3} $$ and the median of $$ X $$ is $$ 3 - \frac{\sqrt{10}}{2} $$.

A random variable X has probability density function given by \[ \mathrm{f}(x)=\left\{\begin{array}{ll}k(3-x) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise, }\end{array}\right. \] where k is a constant. a) Show that \( \mathrm{k}=\frac{2}{3} \). b) Find the median of X .

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