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Example 12-3
Section* *

Five percent (5%) of Christmas tree light bulbs manufactured by a company are defective. The company's Quality Control Manager is quite concerned and therefore randomly samples 100 bulbs coming off of the assembly line. Let \(X\) denote the number in the sample that are defective. What is the probability that the sample contains at most three defective bulbs?

#### Solution

Can you convince yourself that \(X\) is a binomial random variable? Hmmm.... let's see... there are two possible outcomes (defective or not), the 100 trials of selecting the bulbs from the assembly line can be assumed to be performed in an identical and independent manner, and the probability of getting a defective bulb can be assumed to be constant from trial to trial. So, \(X\) is indeed a binomial random variable. Well, calculating the probability is easy enough then... we just need to use the cumulative binomial table with \(n=100\) and \(p=0.05\).... Oops! The table won't help us here, will it? Even many standard calculators would have trouble calculating the probability using the p.m.f.:

\(P(X\leq 3)=\dbinom{100}{0}(0.05)^0 (0.95)^{100}+\cdots+\dbinom{100}{3}(0.05)^3 (0.95)^{97}\)

Using a statistical software package (Minitab), I was able to use the binomial p.m.f. to determine that:

\(P(X\le 3)=0.0059205+0.0311607+0.0811818+0.1395757=0.25784\)

But, if you recall the way that we derived the Poisson distribution,... we started with the binomial distribution and took the limit as n approached infinity. So, it seems reasonable then that the Poisson p.m.f. would serve as a reasonable approximation to the binomial p.m.f. when your \(n\) is large (and therefore, \(p\) is small). Let's calculate \(P(X\le 3)\) using the Poisson distribution and see how close we get. Well, the probability of success was defined to be:

\(p=\dfrac{\lambda}{n}\)

Therefore, the mean \(\lambda\) is:

\(\lambda=np\)

So, we need to use our Poisson table to find \(P(X\le 3)\) when \(\lambda=100(0.05)=5\). What do you get?

x | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

0 | 0.015 | 0.012 | 0.010 | 0.008 | 0.007 | 0.006 |

1 | 0.078 | 0.033 | 0.056 | 0.048 | 0.040 | 0.034 |

2 | 0.210 | 0.185 | 0.163 | 0.143 | 0.125 | 0.109 |

3 | 0.395 | 0.359 | 0.326 | 0.294 | 0.265 | 0.238 |

4 | 0.590 | 0.551 | 0.513 | 0.476 | 0.440 | 0.406 |

5 | 0.753 | 0.720 | 0.686 | 0.651 | 0.616 | 0.581 |

6 | 0.867 | 0.844 | 0.818 | 0.791 | 0.762 | 0.732 |

7 | 0.936 | 0.921 | 0.905 | 0.887 | 0.867 | 0.845 |

8 | 0.972 | 0.964 | 0.955 | 0.944 | 0.932 | 0.918 |

9 | 0.989 | 0.985 | 0.980 | 0.975 | 0.968 | 0.960 |

10 | 0.996 | 0.994 | 0.992 | 0.990 | 0.986 | 0.982 |

11 | 0.999 | 0.998 | 0.997 | 0.996 | 0.995 | 0.993 |

12 | 1.000 | 0.999 | 0.999 | 0.999 | 0.998 | 0.997 |

13 | 1.000 | 1.000 | 1.000 | 1.000 | 0.999 | 0.999 |

14 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

15 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

16 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

x | 4.2 | 4.4 | 4.6 | 4.8 | 5.0 | 5.2 |

0 | 0.015 | 0.012 | 0.010 | 0.008 | 0.007 | 0.006 |

1 | 0.078 | 0.033 | 0.056 | 0.048 | 0.040 | 0.034 |

2 | 0.210 | 0.185 | 0.163 | 0.143 | 0.125 | 0.109 |

3 | 0.395 | 0.359 | 0.326 | 0.294 | 0.265 | 0.238 |

4 | 0.590 | 0.551 | 0.513 | 0.476 | 0.440 | 0.406 |

5 | 0.753 | 0.720 | 0.686 | 0.651 | 0.616 | 0.581 |

6 | 0.867 | 0.844 | 0.818 | 0.791 | 0.762 | 0.732 |

7 | 0.936 | 0.921 | 0.905 | 0.887 | 0.867 | 0.845 |

8 | 0.972 | 0.964 | 0.955 | 0.944 | 0.932 | 0.918 |

9 | 0.989 | 0.985 | 0.980 | 0.975 | 0.968 | 0.960 |

10 | 0.996 | 0.994 | 0.992 | 0.990 | 0.986 | 0.982 |

11 | 0.999 | 0.998 | 0.997 | 0.996 | 0.995 | 0.993 |

12 | 1.000 | 0.999 | 0.999 | 0.999 | 0.998 | 0.997 |

13 | 1.000 | 1.000 | 1.000 | 1.000 | 0.999 | 0.999 |

14 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

15 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

16 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 | 1.000 |

The cumulative Poisson probability table tells us that finding \(P(X\le 3)=0.265\). That is, if there is a 5% defective rate, then there is a 26.5% chance that the a randomly selected batch of 100 bulbs will contain at most 3 defective bulbs. More importantly, since we have been talking here about using the Poisson distribution to approximate the binomial distribution, we should probably compare our results. When we used the binomial distribution, we deemed \(P(X\le 3)=0.258\), and when we used the Poisson distribution, we deemed \(P(X\le 3)=0.265\). Not too bad of an approximation, eh?

It is important to keep in mind that the Poisson approximation to the binomial distribution works well only when \(n\) is large and \(p\) is small. In general, the approximation works well if \(n\ge 20\) and \(p\le 0.05\), or if \(n\ge 100\) and \(p\le 0.10\).