Q. 8.14 - A First Course in Probability

Question

A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2000 hours is at least 0.95?

Step-by-Step Solution

Answer

$n\ge 23$

Given information

$𝜇=100 \mathrm{and} 𝜎=30$

We need to find n.

Let p be the probability they will last for atleast 2000 hours.

Fomulation

$p=P\left(\underset{i=1}{\overset{n}{\sum X_i}}\ge 2000\right)\approx P\left(Z\ge \frac{2000-100n}{30\sqrt{n}}\right)$

where Z  is a standard normal random variable.

Calculating n

We know that

$$\begin{gathered} P(Z>-1.64)=0.95 \\ \Rightarrow \frac{2000-100n}{30\sqrt{n}}\le -1.64 \\ \Rightarrow 2000-100n\le -49.2\sqrt{n} \end{gathered}$$

Upon using calculator, we get

$n\ge 23.$