Question

In a factory, 100 bulbs are in use. The table lists the cumulative probability of the failure of a bulb for various durations. \[ \begin{array}{|c|c|} \hline \text{Duration (month)} & \text{Cumulative probability} \\ \hline 1 & 0.10 \\ 2 & 0.25 \\ 3 & 0.47 \\ 4 & 0.68 \\ 5 & 1.00 \\ \hline \end{array} \] The factory follows the individual replacement policy. If the cost of replacing a bulb is ₹300, then the expected cost (in ₹) of replacement per month is \(\underline{\hspace{2cm}}\). [round off to nearest integer]

Hint:

For individual replacement, compute expected lifetime from the failure distribution and divide total units by expected life to get expected failures per month.

Answer 1

Correct Answer: 8570 - 8701

The probability of failure in each month is obtained by differences: \[ \begin{aligned} P_1 &= 0.10 \\ P_2 &= 0.25 - 0.10 = 0.15 \\ P_3 &= 0.47 - 0.25 = 0.22 \\ P_4 &= 0.68 - 0.47 = 0.21 \\ P_5 &= 1.00 - 0.68 = 0.32 \\ \end{aligned} \] The expected life of a bulb is: \[ E(L) = \sum tP(t) \] \[ E(L) = 1(0.10) + 2(0.15) + 3(0.22) + 4(0.21) + 5(0.32) \] \[ E(L) = 0.10 + 0.30 + 0.66 + 0.84 + 1.60 = 3.50\ \text{months} \] For 100 bulbs, monthly failures = \[ \frac{100}{E(L)} = \frac{100}{3.5} \approx 28.57 \] Cost per replacement = ₹ 300 Thus monthly replacement cost: \[ 300 \times 28.57 = 8571 \] The expected cost lies in the range: \[ \boxed{8570\text{ to }8701} \]