Question

Suppose that \( X \) is the waiting time in minutes for a bus and its p.d.f. is given by: \[ f(x) = \frac{1}{5}, \text{for } 0 \leq x \leq 5, \text{and} f(x) = 0, \text{otherwise}. \] Find the probability that: (i) waiting time is between 1 to 3 minutes. (ii) waiting time is more than 4 minutes.

Hint:

For uniform distributions, the probability over any interval is given by the product of the length of the interval and the constant value of the p.d.f.

Answer 1

Step 1: Understand the probability density function (p.d.f.).
The p.d.f. \( f(x) \) is a uniform distribution between 0 and 5 minutes, with a constant value of \( \frac{1}{5} \) over this interval. The total area under the curve must equal 1, which is true for a uniform distribution. \[ \int_0^5 f(x) \, dx = \int_0^5 \frac{1}{5} \, dx = 1 \]

Step 2: Calculate the probability for part (i).
We are asked to find the probability that the waiting time is between 1 and 3 minutes. This is the integral of \( f(x) \) from \( x = 1 \) to \( x = 3 \): \[ P(1 \leq X \leq 3) = \int_1^3 f(x) \, dx = \int_1^3 \frac{1}{5} \, dx \]

Step 3: Solve the integral for part (i).
\[ P(1 \leq X \leq 3) = \frac{1}{5} \times (3 - 1) = \frac{1}{5} \times 2 = \frac{2}{5} \]

Step 4: Calculate the probability for part (ii).
Next, we are asked to find the probability that the waiting time is more than 4 minutes. This is the integral of \( f(x) \) from \( x = 4 \) to \( x = 5 \): \[ P(X > 4) = \int_4^5 f(x) \, dx = \int_4^5 \frac{1}{5} \, dx \]

Step 5: Solve the integral for part (ii).
\[ P(X > 4) = \frac{1}{5} \times (5 - 4) = \frac{1}{5} \times 1 = \frac{1}{5} \] Final Answers: \[ P(1 \leq X \leq 3) = \boxed{\frac{2}{5}}, P(X > 4) = \boxed{\frac{1}{5}} \]