$F_X(x) = \begin{cases} 0 & x < t \\ \frac{x-t}{4-t} & t \le x \le 4 \\ 1 & x \ge 4 \end{cases}$
If the median of $X$ is $3$, then what is the value of $t$?
The median ($m$) of a continuous random variable $X$ is the value where the cumulative distribution function (CDF) equals 0.5. Mathematically, this is expressed as $F_X(m) = 0.5$.
We are given that the median of the random variable $X$ is $m = 3$. The CDF is defined as:
$F_X(x) = \frac{x-t}{4-t}$ for $t \le x \le 4$.
Since the median is 3, we know that $F_X(3) = 0.5$. We substitute $x=3$ into the CDF formula:
$F_X(3) = \frac{3-t}{4-t}$
Now, we set the CDF value equal to 0.5 and solve for $t$:
The condition for this part of the CDF is $t \le x \le 4$. Since our calculated $t=2$ and the median $x=3$, the condition $2 \le 3 \le 4$ holds true.
Therefore, the value of $t$ is 2.

The above frequency chart shows the frequency distribution of marks obtained by a set of students in an exam. From the data presented above, which one of the following is CORRECT?