1. The continuous random variable $$ X $$ has cumulative probability density function $$ F(x) $$ is given by $$ F(X)=\left\{\begin{array}{cc} 0, & x \leq 2 \ \frac{x^{2}}{6}-\frac{2}{3} x+a & , 2 \leq x \leq 3 \ b+\frac{x}{3}, & 3 \leq x \leq 5 \ 2 x-\frac{x^{2}}{6}+c, & 5 \leq x \leq 6 \ 1, & x \geq 6 \end{array}\right. $$ (a) Find the values of $$ a, b, c $$. [3 marks] (b) Hence, find (i) $$ P(1.5 \leq X \leq 2.5) $$. [ 2 marks] (ii) the third quartile. [3 marks] (iii) Find the probability distribution function of X .

Question

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To solve the problem, we will follow these steps:

### Part (a): Find the values of $$ a, b, c $$

1. **Continuity at $$ x = 2 $$**:
   - The value of $$ F(x) $$ must be continuous at $$ x = 2 $$.
   - From the left, $$ F(2) = 0 $$.
   - From the right, $$ F(2) = \frac{2^2}{6} - \frac{2}{3} \cdot 2 + a $$.

Setting these equal gives:
   $$
   0 = \frac{4}{6} - \frac{4}{3} + a
   $$
   Simplifying:
   $$
   0 = \frac{2}{3} - \frac{4}{3} + a \implies a = \frac{2}{3}
   $$

2. **Continuity at $$ x = 3 $$**:
   - From the left, $$ F(3) = \frac{3^2}{6} - \frac{2}{3} \cdot 3 + a $$.
   - From the right, $$ F(3) = b + \frac{3}{3} = b + 1 $$.

Setting these equal gives:
   $$
   \frac{9}{6} - 2 + \frac{2}{3} = b + 1
   $$
   Simplifying:
   $$
   \frac{3}{2} - 2 + \frac{2}{3} = b + 1
   $$
   Converting to a common denominator (6):
   $$
   \frac{9}{6} - \frac{12}{6} + \frac{4}{6} = b + 1 \implies \frac{1}{6} = b + 1 \implies b = -\frac{5}{6}
   $$

3. **Continuity at $$ x = 5 $$**:
   - From the left, $$ F(5) = b + \frac{5}{3} $$.
   - From the right, $$ F(5) = 2 \cdot 5 - \frac{5^2}{6} + c $$.

Setting these equal gives:
   $$
   -\frac{5}{6} + \frac{5}{3} = 10 - \frac{25}{6} + c
   $$
   Simplifying:
   $$
   -\frac{5}{6} + \frac{10}{6} = 10 - \frac{25}{6} + c \implies \frac{5}{6} = 10 - \frac{25}{6} + c
   $$
   Converting $$ 10 $$ to sixths:
   $$
   \frac{5}{6} = \frac{60}{6} - \frac{25}{6} + c \implies \frac{5}{6} = \frac{35}{6} + c \implies c = -\frac{30}{6} = -5
   $$

Thus, we have:
- $$ a = \frac{2}{3} $$
- $$ b = -\frac{5}{6} $$
- $$ c = -5 $$

### Part (b): Find $$ P(1.5 \leq X \leq 2.5) $$

To find $$ P(1.5 \leq X \leq 2.5) $$, we can use the cumulative distribution function $$ F(x) $$:

$$
P(1.5 \leq X \leq 2.5) = F(2.5) - F(1.5)
$$

1. **Calculate $$ F(1.5) $$**:
   - Since $$ 1.5 < 2 $$, $$ F(1.5) = 0 $$.

2. **Calculate $$ F(2.5) $$**:
   - For $$ 2 \leq x \leq 3 $$:
   $$
   F(2.5) = \frac{(2.5)^2}{6} - \frac{2}{3} \cdot 2.5 + \frac{2}{3}
   $$
   Simplifying:
   $$
   F(2.5) = \frac{6.25}{6} - \frac{5}{6} + \frac{2}{3} = \frac{6.25 - 5 + 4}{6} = \frac{5.25}{6} = \frac{21}{24}
   $$

Thus:
$$
P(1.5 \leq X \leq 2.5) = F(2.5) - F(1.5) = \frac{21}{24} - 0 = \frac{21}{24}
$$

### Part (c): Find the third quartile

The third quartile $$ Q_3 $$ is the value of $$ x $$ such that $$ F(x) = 0.75 $$.

1. **For $$ 3 \leq x \leq 5 $$**:
   $$
   F(x) = -\frac{5}{6} + \frac{x}{3}
   $$
   Setting this equal to $$ 0.75 $$:
   $$
   -\frac{5}{6} + \frac{x}{3} = 0.75
   $$
   Converting $$ 0.75 $$ to sixths:
   $$
   -\frac{5}{6} + \frac{x}{3} = \frac{9}{12} \implies \frac{x}{3} = \frac{9}{12} + \frac{5}{6} = \frac{9}{12} + \frac{10}{12} = \frac{19}{12}
   $$
   Thus:
   $$
   x = 3 \cdot \frac{19}{12} = \frac{57}{12} = 4.75
   $$

### Part (d): Find the probability distribution function of $$ X $$

The probability density function $$ f(x) $$ is the derivative of the cumulative distribution function $$ F(x) $$:

1. For $$ x < 2 $$:
   $$
   f(x) = 0
   $$

2. For $$ 2 \leq x \leq 3 $$:
   $$
   f(x) = \frac{d}{dx}\left(\frac{x^2}{6} - \frac{2}{3}x + \frac{2}{3}\right) = \frac{x}{3} - \frac{2}{3}
   $$

3. For $$ 3 \leq x \leq 5 $$:
   $$
   f(x) = \frac{d}{dx}\left(-
**Part (a): Values of $$ a, b, c $$** - $$ a = \frac{2}{3} $$ - $$ b = -\frac{5}{6} $$ - $$ c = -5 $$ **Part (b): Probability $$ P(1.5 \leq X \leq 2.5) $$** \[ P(1.5 \leq X \leq 2.5) = \frac{21}{24} $$ **Part (c): Third Quartile $$ Q_3 $$** $$ Q_3 = 4.75 $$ **Part (d): Probability Distribution Function of $$ X $$** $$ f(x) = \begin{cases} 0 & \text{if } x < 2 \ \frac{x}{3} - \frac{2}{3} & \text{if } 2 \leq x \leq 3 \ \frac{1}{3} & \text{if } 3 < x \leq 5 \ -\frac{x}{6} + \frac{1}{6} & \text{if } 5 < x \leq 6 \ 0 & \text{if } x > 6 \end{cases} $$

The continuous random variable has cumulative probability density function is given by

(a) Find the values of .
[3 marks]
(b) Hence, find
(i) .
[ 2 marks]
(ii) the third quartile.
[3 marks]
(iii) Find the probability distribution function of X .

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(a) Find the values of .

(b) Hence, find

(i) .

(ii) The third quartile.

(iii) Find the probability distribution function of .

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The continuous random variable has cumulative probability density function is given by (a) Find the values of . [3 marks] (b) Hence, find (i) . [ 2 marks] (ii) the third quartile. [3 marks] (iii) Find the probability distribution function of X .