Question

If the random variable x is given by x: waiting time in minutes for bus and probability distribution function (p.d.f.) of x is given by
 $ f(x)=\left\{ \begin{align}
  & \dfrac{1}{5},0\le x\le 5 \\
 & 0\text{ otherwise} \\
\end{align} \right\} $
Then the probability of waiting time not more than 4 should be equal to
a.0.3
b.0.8
c.0.2
d.0.5

Answer 1

Hint: In this question, we are given the probability distribution of x. We are asked to find the probability of waiting time not more than 4, therefore, the required probability should be the integral of the probability between 0 and 4. Using the given distribution, we can integrate the distribution function between the required intervals to obtain the answer to this given question.

Complete step-by-step answer:
The given random variable is x and its probability distribution is given by
 $ f(x)=\left\{ \begin{align}
  & \dfrac{1}{5},0\le x\le 5 \\
 & 0\text{ otherwise} \\
\end{align} \right\}..............................(1.1) $
We are asked to find the probability of waiting time not more than 4. This, corresponds to the probability that the waiting time of the bus is between 0 and 4 with the endpoints 0 and 4 included……(1.2)
Now, we know that the formula for finding the probability of a random variable z following a probability distribution function f(z) to lie in the interval a to b is given by
 $ P(a\le z\le b)=\int_{a}^{b}{f(z)dz...............................(1.3)} $
Therefore, using (1.3) and (1.2), we can obtain the probability that the waiting time of the bus is not more than 4 as
 $ P(0\le x\le 4)=\int_{0}^{4}{f(x)dx}...................(1.4) $
The integral of cdx, where c is a constant, over a given range a to b is given by
 $ \int_{a}^{b}{cdx}=c\int_{a}^{b}{dx=}c\left( b-a \right)............................(1.5) $
However, in the interval 0 to 4, the value of f(x) should be $ \dfrac{1}{5} $ from equation (1.1). Thus, using it in equation (1.4) and using equation (1.5), we obtain
 $ \begin{align}
  & P(0\le x\le 4)=\int_{0}^{4}{f(x)dx} \\
 & =\int_{0}^{4}{\dfrac{1}{5}\times dx}=\dfrac{1}{5}\int_{0}^{4}{dx} \\
 & =\dfrac{1}{5}\times (4-0)=\dfrac{4}{5} \\
\end{align} $
which is the required probability. Thus, the answer to the given question should be $ \dfrac{4}{5} $ .
hence option (b) which is equal to 0.8 is the right answer.

Note: We should note that equation (1.5) holds only if c is a constant. However, if c is not a constant and is a function of x, then we have to evaluate the integral using other methods appropriate for the function for c. For example, if c=x, then $ \int_{a}^{b}{cdx}=\int_{a}^{b}{xdx=\left( \dfrac{{{b}^{2}}}{2}-\dfrac{{{a}^{2}}}{2} \right)} $ .