Poisson approximation to the binomial distribution calculator

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?

x4.24.44.64.85.05.200.0150.0120.0100.0080.0070.00610.0780.0330.0560.0480.0400.03420.2100.1850.1630.1430.1250.10930.3950.3590.3260.2940.2650.23840.5900.5510.5130.4760.4400.40650.7530.7200.6860.6510.6160.58160.8670.8440.8180.7910.7620.73270.9360.9210.9050.8870.8670.84580.9720.9640.9550.9440.9320.91890.9890.9850.9800.9750.9680.960100.9960.9940.9920.9900.9860.982110.9990.9980.9970.9960.9950.993121.0000.9990.9990.9990.9980.997131.0001.0001.0001.0000.9990.999141.0001.0001.0001.0001.0001.000151.0001.0001.0001.0001.0001.000161.0001.0001.0001.0001.0001.000

Answer

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 \[p\le 0.10\].

The Poisson Probability Calculator can calculate the probability of an event occurring in a given time interval. Before using the calculator, you must know the average number of times the event occurs in the time interval. The symbol for this average is $ \lambda $, the greek letter lambda. You also need to know the desired number of times the event is to occur, symbolized by x.

If you’d like to construct a complete probability distribution based on a value for $ \lambda $ and x, then go ahead and take a look at the Poisson Distribution Calculator. It will calculate all the poisson probabilities from 0 to x.

If using a calculator, you can enter $ \lambda = 4.8 $ and $ x = 6 $ into a poisson probability distribution function [PDF]. If doing this by hand, apply the poisson probability formula: $$ P[x] = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!} $$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $ x = 6 $ and $ \lambda = 4.8 $, we have $$ P[6] = \frac{{e^{-4.8}} \cdot {4.8^6}}{6!} $$ Evaluating the expression, we have $$ P[6] = 0.13979814691511 $$

Solution

Answer:

Solution:

How to calculate Poisson approximation to binomial distribution?

Can a binomial distribution be a Poisson distribution?

What is the relationship between binomial distribution and Poisson distribution?

Does binomial converge Poisson?

Bài Viết Liên Quan

Toplist mới

Bài mới nhất

Chủ Đề