Q. 5.24 - A First Course in Probability

Question

The lifetimes of interactive computer chips produced

by a certain semiconductor manufacturer are normally distributed with parameters$\mu = 1.4 \times {10}^6$ hours and $\sigma = 3 \times {10}^5$ hours. What is the approximate probability that a batch of $100$ chips will contain at least $20$ whose lifetimes are less than $1.8 \times {10}^6$?

Step-by-Step Solution

Answer

The required probability is $1$.

Step 1. Given information.

Here, it is given that the lifetimes of interactive computer chips are normally distributed with parameters

$$\begin{gathered} \mu = 1.4 \times {10}^6 \text{hours} \\ \sigma = 3\times {10}^5 \text{hours} \end{gathered}$$

Step 2. Find the life time of an interactive computer chip.

Let $X$ denote the life time of an interactive computer chip.

$$\begin{gathered} P\left(X\le 1.8\times {10}^6\right)=P\left(Z\le \frac{1.8\times {10}^6-1.4\times {10}^6}{3\times {10}^5}\right) \\ =P\left(Z\le 1.333\right) \\ =0.9082 \end{gathered}$$

Step 3. Find the probability that a batch of 100 chips will contain at least 20 such chips whose life times are less than 1 . 8 × 10 6 .

$$\begin{gathered} p=0.9082 \\ \mu =np=100\times 0.9082=90.82 \\ \sigma =\sqrt{npq}=\sqrt{90.82(1-0.9082)}=2.89 \end{gathered}$$

$$\begin{gathered} P\left(Y\ge 19.5\right)=P\left(z\ge \frac{19.5-90.82}{2.89}\right) \\ =P\left(z\ge -24.678\right) \\ \cong 1 \end{gathered}$$

Therefore, the approximate probability that a batch of $100$chips will contain at least 20 such chips whose life times are less than $1.8\times {10}^6$ is $1$.