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. This means we need to integrate the PDF over its defined range and set the result equal to 1.

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

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

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. Calculate at $$ x = 2 $$:
   $$
   3(2) - \frac{(2)^2}{2} = 6 - 2 = 4
   $$
2. Calculate 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:
$$
P(X \leq m) = 0.5
$$

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

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

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:
$$
m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$ a = 1, b = -6, c = 6.5 $$.

Calculating the discriminant:
$$
b^2 - 4ac = (-6)^2 - 4(1)(6.5) = 36 - 26 = 10
$$

Now substituting into the quadratic formula:
$$
m = \frac{6 \pm \sqrt{10}}{2}
$$

Calculating the two possible values:
$$
m = 3 \pm \frac{\sqrt{10}}{2}
$$

Since $$ m $$ must be between 1 and 2, we take the negative 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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