2. A cumulative distribution function of a discrete random variable $$ X $$ is given as $$ F(X)=\left\{\begin{array}{lc} 0, & x \leq 0 \ \frac{k}{8}, & 1 \leq x

Question

2. A cumulative distribution function of a discrete random variable $$ X $$ is given as $$ F(X)=\left\{\begin{array}{lc} 0, & x \leq 0 \ \frac{k}{8}, & 1 \leq x<2 \ \frac{3 k}{8}, & 2 \leq x<3 \ \frac{k+5}{8}, & 3 \leq x<4 \ \frac{5 k}{10}, & x \geq 4 \end{array}\right. $$ where $$ k $$ is a constant. (a) Show that $$ k=2 $$. [2 marks] (b) Find the median of the distribution. [2 marks] (c) Find the probability distribution function of X . Hence, find the mode. [3 marks] (d) Find the expectation of X . [3 marks] (e) $$ \operatorname{Var}(3-X) $$ [4 marks]

UpStudy AI · Accepted Answer

**(a) Determine $$ k $$:**  
A cumulative distribution function (CDF) must satisfy $$ F(x)\to 1 $$ as $$ x\to\infty $$. Here, for $$ x\ge4 $$ the CDF is given as  
$$
F(x)=\frac{5k}{10}=\frac{k}{2}.
$$
Thus, we require  
$$
\frac{k}{2}=1,
$$
which implies  
$$
k=2.
$$

---

**(b) Find the Median:**  
With $$ k=2 $$, the CDF becomes:
- For $$ 1\le x<2 $$: $$ F(x)=\frac{2}{8}=\frac{1}{4} $$.
- For $$ 2\le x<3 $$: $$ F(x)=\frac{3\cdot 2}{8}=\frac{6}{8}=\frac{3}{4} $$.
- For $$ 3\le x<4 $$: $$ F(x)=\frac{2+5}{8}=\frac{7}{8} $$.
- For $$ x\ge4 $$: $$ F(x)=\frac{2}{2}=1 $$.

Since $$ X $$ is discrete the probability masses occur at the jump points:
- At $$ x=1 $$:
  $$
  P(X=1)=F(1)-\lim_{x\to 1^-}F(x)=\frac{1}{4}-0=\frac{1}{4}.
  $$
- At $$ x=2 $$:
  $$
  P(X=2)=F(2)-F(2^-)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}.
  $$
- At $$ x=3 $$:
  $$
  P(X=3)=F(3)-F(3^-)=\frac{7}{8}-\frac{3}{4}=\frac{7}{8}-\frac{6}{8}=\frac{1}{8}.
  $$
- At $$ x=4 $$:
  $$
  P(X=4)=1-\frac{7}{8}=\frac{1}{8}.
  $$

The median $$ m $$ is defined as the smallest value such that $$ F(m) \ge \frac{1}{2} $$. From the cumulative sums:
- $$ F(1)=\frac{1}{4} < \frac{1}{2} $$,
- $$ F(2)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \ge \frac{1}{2} $$.

Thus, the median is  
$$
m=2.
$$

---

**(c) The Probability Distribution Function and the Mode:**  
From the differences computed:
- $$ P(X=1)=\frac{1}{4} $$,
- $$ P(X=2)=\frac{1}{2} $$,
- $$ P(X=3)=\frac{1}{8} $$,
- $$ P(X=4)=\frac{1}{8} $$.

The mode is the value of $$ X $$ with the highest probability. Since  
$$
P(X=2)=\frac{1}{2}
$$
is the largest, the mode is  
$$
\text{mode}=2.
$$

---

**(d) Find the Expectation $$ E[X] $$:**  
We compute  
$$
E[X]=\sum_x x\,P(X=x)= 1\cdot \frac{1}{4}+2\cdot \frac{1}{2}+3\cdot \frac{1}{8}+4\cdot \frac{1}{8}.
$$
Calculating each term:
- $$ 1\cdot \frac{1}{4}=\frac{1}{4} $$,
- $$ 2\cdot \frac{1}{2}=1 $$,
- $$ 3\cdot \frac{1}{8}=\frac{3}{8} $$,
- $$ 4\cdot \frac{1}{8}=\frac{4}{8}=\frac{1}{2} $$.

Summing, convert to a common denominator (8):
$$
E[X]=\frac{2}{8}+\frac{8}{8}+\frac{3}{8}+\frac{4}{8}=\frac{2+8+3+4}{8}=\frac{17}{8}.
$$

---

**(e) Find $$ \operatorname{Var}(3-X) $$:**  
For any linear transformation $$ Y = aX + b $$,  
$$
\operatorname{Var}(Y)=a^2\operatorname{Var}(X).
$$
Here, $$ 3-X $$ has $$ a=-1 $$ and $$ b=3 $$, so
$$
\operatorname{Var}(3-X)=(-1)^2\operatorname{Var}(X)=\operatorname{Var}(X).
$$
To find $$ \operatorname{Var}(X) $$, we need $$ E[X^2] $$:
$$
E[X^2]=1^2\cdot\frac{1}{4}+2^2\cdot\frac{1}{2}+3^2\cdot\frac{1}{8}+4^2\cdot\frac{1}{8}.
$$
Calculating:
- $$ 1^2\cdot\frac{1}{4}=\frac{1}{4} $$,
- $$ 2^2\cdot\frac{1}{2}=4\cdot\frac{1}{2}=2 $$,
- $$ 3^2\cdot\frac{1}{8}=9\cdot\frac{1}{8}=\frac{9}{8} $$,
- $$ 4^2\cdot\frac{1}{8}=16\cdot\frac{1}{8}=2 $$.

Thus,
$$
E[X^2]=\frac{1}{4}+2+\frac{9}{8}+2.
$$
Express in eighths:
$$
\frac{1}{4}=\frac{2}{8},\quad 2=\frac{16}{8},
$$
so
$$
E[X^2]=\frac{2}{8}+\frac{16}{8}+\frac{9}{8}+\frac{16}{8}=\frac{43}{8}.
$$
Now, using  
$$
\operatorname{Var}(X)=E[X^2]-(E[X])^2,
$$
and we have $$ E[X]=\frac{17}{8} $$, so
$$
\operatorname{Var}(X)=\frac{43}{8}-\left(\frac{17}{8}\right)^2=\frac{43}{8}-\frac{289}{64}.
$$
Converting $$ \frac{43}{8} = \frac{344}{64} $$, we get
$$
\operatorname{Var}(X)=\frac{344-289}{64}=\frac{55}{64}.
$$
Thus,
$$
\operatorname{Var}(3-X)=\frac{55}{64}.
$$
**(a) $$ k = 2 $$** **(b) Median = 2** **(c) Probability Distribution Function:** - $$ P(X=1) = 0.25 $$ - $$ P(X=2) = 0.5 $$ - $$ P(X=3) = 0.125 $$ - $$ P(X=4) = 0.125 $$ Mode = 2 **(d) Expectation $$ E[X] = 2.125 $$** **(e) $$ \operatorname{Var}(3-X) = 0.8594 $$**

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