Question

Each item produced by a certain manufacturer is, independently, of acceptable quality with probability $.95$. Approximate the probability that at most $10$of the next $150$items produced are unacceptable.

Short answer

The required probability is$0.8686$.

Step-by-step solution

Step 1. Given Information.

Here, it is given that - The product will be acceptable if the probability is $0.95$.

Items produced = $150$

Step 2. Check conditions for normal approximation to binomial distribution.

Let, number of unacceptable items be $X$.

Let us assume that $X$ follows the binomial distribution, where

$n=150,\mathrm{and}p=1-0.95=0.05$

$$\begin{gathered} np>5 \\ \Rightarrow 150(0.05)>5 \\ \Rightarrow 7.5>5 \end{gathered}$$

$$\begin{gathered} n(1-p)>5 \\ \Rightarrow 150(1-0.05)>5 \\ \Rightarrow 142.5>5 \end{gathered}$$

Therefore, the conditions for normal approximation are satisfied.

Step 3. Calculate mean and standard deviation.

Suppose $X$is a binomial random variable with parameters $n,\text{and}p$, then

$Z=\frac{X-np}{\sqrt{np(1-p)}}~N(0,1)$

$$\begin{gathered} \mu =np \\ =150\times 0.05 \\ =7.5 \end{gathered}$$

$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ =\sqrt{150(0.05)(1-0.05)} \\ =\sqrt{7.125} \\ =2.6693 \end{gathered}$$

Step 4. Find the probability that at most 10 of the next 150 items produced are unacceptable.

$$\begin{gathered} P\left(X\le 10\right)=P\left(X<10.5\right) \\ =P\left(\frac{X-np}{\sqrt{np(1-p)}}<\frac{10.5-np}{\sqrt{np(1-p)}}\right) \\ =P\left(\frac{X-7.5}{2.6693}<\frac{10.5-7.5}{2.6693}\right) \\ =P(Z<1.12) \\ =\phi (1.12) \\ =0.8686 \end{gathered}$$

Therefore, the probability that at most $10$ of the next $150$ items produced are unacceptable is$0.8686$.