Question

A process yields 10% defective items. If 100 items are randomly selected the process, use normal approximation to find the probability that the number of defectives.

a) exceeds 13 and

b) less than 8.

Answer 1

Let X be the number of defectives in the 100 items.

Use the Normal Approximation to the Binomial Distribution Theorem.

For large 𝑛, 𝑋 has approximately a normal distribution with µ = 𝑛𝑝 and

𝜎

² = 𝑛𝑝(1 − 𝑝) and

$P(X<x)=P(z<\frac{x-0.5-np}{\sqrt{np(1-p)}})$

$P(X\le x)=P(z<\frac{x+0.5-np}{\sqrt{np(1-p)}})$

$P(X>x)=P(z>\frac{x+0.5-np}{\sqrt{np(1-p)}})$

$P(X\ge x)=P(z>\frac{x-0.5-np}{\sqrt{np(1-p)}})$

We have that

𝑝 = 0.1, 𝑛 = 100, 𝑛𝑝 = 100(0.1) = 10, $\sqrt{np(1-p)}=\sqrt{100(0.1)(1-0.1)}=3$

a. $P(X>13)=1-P(X\le13)=1-P(\frac{13+0.5-10}{3})=1-P(Z \le 1.17)=0.5-0.3790=0.121$

b. $P(X<8)=P(X<\frac{8-0.5-10}{3})=P(Z<-0.83)=0.2033$