Among all the computer chips produced by a certain factory, 6 percent are defective. A sample of 400 chips is selected for inspection.
a) What is the probability that this sample contains between 20 and 25 defective chips (including 20 and 25)?
b) Suppose that each of 40 inspectors collects a sample of 400 chips. What is the probability that at least 8 inspectors will find between 20 and 25 defective chips in their samples?
Solution
Let X be the number of defective chips in the sample of 400 and $p$ be the probability that the computer chip is defective.
Given that $n =400$ and $p=0.06$. Thus $X\sim B(400, 0.06)$.
The distribution of X is approximately Normal with mean = $\mu= n*p = 400 \times 0.06 = 24$ and sd = $\sigma= \sqrt{n*p*(1-p)} = \sqrt{400 \times 0.06 \times (1- 0.06)}=4.75$
a. The probability that this sample contains between 20 and 25 defective chips (including 20 and 25) is
$$ \begin{aligned} P(20\leq X\leq 25) & = P(19.5 < X < 25.5)\\ &\quad \quad (\text{ using continuity correction})\\ &=P(\frac{19.5-24}{4.75} < \frac{X-\mu}{\sigma} < \frac{25.5-24}{4.75})\\ &=P(-0.95 < Z < 0.32)\\ &= P(Z < 0.32)-P(Z < -0.95)\\ & = 0.6239-0.1717\\ & = 0.4522\\ \end{aligned} $$
b. Let $Y$ be the number of inspectors who find between 20 and 25 defective chips in a sample of 400.
The distribution of $Y$ is approximately Normal with mean = $\mu= n*p = 40 \times 0.45 = 18$ and sd = $\sigma= \sqrt{n*p*(1-p)} = \sqrt{40 \times 0.45 \times (1- 0.45)}=3.15$
$$ \begin{aligned} P(Y\geq 8) & = P(Y > 7.5)\\ &\quad \quad (\text{ using continuity correction})\\ & = 1- P(Y < 7.5)\\ &= 1-P\bigg(\frac{Y-\mu}{\sigma} < \frac{7.5- 18}{3.15}\bigg)\\ & =1-P(Z < -3.33)\\ & = 1-0.0004\\ & = 0.9996 \end{aligned} $$
Conclusion
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