Probability of defective items: probability approximation

Question

It is known that any item produced by a certain machine will be defective with probability $0.01$, independently of any other item.

What is the probability that a lot of $100$ items will contain more than $2$ defective items? You should use a suitable approximation and state the parameter values.

Answer 1

We know

$$p = .01$$ $$n = 100$$

And our interest is $$x > 2$$

I would use the binomial distribution because the probability of success ( p = .01) is constant and the item is either "defective" or "non-defective". You're interested in:

$$P(x>2)=P (x\geq 2.5)=P (x\geq 3)$$

You can estimate this binomial distribution using the Poisson approximation, because we have very large $n$ and very small $p$, The parameters are: $$n*p = 1$$

For a poison distribution we know that:

$$P(x) =\frac{e^{-\text{np}} \text{np}^x}{x!}$$

Thus,

$$P (x > 2)\approx \frac{\frac{1}{e}*1^x}{x!}$$ $$1-P (x\leq 2)\approx\frac{\frac{1}{e}*1^x}{x!}$$ $$ P (x\leq 2)\approx 1-(1/e\left(\frac{1}{2!}+\frac{1}{1!}+1\right))$$

 1-ppois(2,1)
 = .0803

Much thanks and appreciation to David Quinn for giving me some clarification, I hope I did this correctly for you.

In addition, you can use "PoissonCDF" in a graphine calculator and doing 1 - probability, where parameters Lambda = 1, lower bound = 0 and upper bound = 2. Doing so, 1 - .919699 = .0803 which matches the results produced in the R code above.

Answer 2

Comment: Here are some computations in R: You might as well review several topics for your upcoming exam. You might include in your review any rules of thumb in your text for using various approximations.

Binomial. Let $X \sim Binom(100, .01).$ You seek $P(X > 2) = 1 - P(X \le 2).$

1 - pbinom(2, 100, .01)
## 0.0793732

Poisson. But you were asked to use a 'suitable approximation'. From the binomial the mean is $\mu = np = 100(.01) = 1.$ Let $Y \sim Pois(1).$ You seek $P(Y > 2).$

1 - ppois(2, 1)
## 0.0803014

Normal. The only other common approximation to binomial in elementary courses is the normal. Most of the guidelines for using the normal approximation are violated here. But let's try it. As above, $m = 1.$ Also, $\sigma^2 = np(1-p) = .99,$ so $\sigma = 0.995.$

Let $W \sim Norm(1, .995).$ You seek $P(W > 2) = P(W > 2.5).$ It is best to use the latter--for the 'continuity correction'.

 1 - pnorm(2.5, 1, sqrt(.99))
 ## 0.06583401

If you are instructed to use an approximation for this particular exercise, the Poisson approximation is indeed better than the normal. (But there are instances where the normal is better.) I agree with @DavidQuinn and @Brandon that the Poisson approximation is best and probably what was intended. Also with @Thanissis that, in serious work, one should not use approximations when exact values are readily available.

Here is a figure that compares the three distributions for relevant values. Blue bars show $Binom(100, .01)$ and purple circles the approximating Poisson values. The 'best fitting' normal density is shown as a green curve; the relevant area is to the left of the vertical red dotted line. (Much of that area is to the left of 0.)