Approximate probabilities of a binomial random variable using the normal distribution

Question

A fabrication process is making 8% broken units. Everyday we take 200 units to check how many unites are broken (X). With the normal law as an approximation of the binomial law, find : $$ P (X \le 16)$$ $$P(X=150)$$ $$P(12 \le X \le 20)$$ $$P(X=14)$$

I have a random variable that obey a binomial law with :

$$n = 200 $$ $$p = 0.08$$

I want to approximate it with a normal law. I get :

$$\mu = 200 * 0.08 = 16 $$ $$\sigma^2 = 200 * 0.08( 1 - 0.08) = 14.72$$

I want to find

$$P(X=15) = ?$$

I try this :

$$ 1 - P( X <= 14) $$ $$ 1 - \phi (\frac{14 - 16}{14.72}) $$ $$ 1 - \phi(-0.14) $$ $$ 1 - ( 1 - \phi(0.14)) $$ $$ \phi(0.14) $$ $$ 0.55670$$

Which is not the result I should find.

What am I doing wrong?

Thank you.

Answer 1

I believe the original distribution is $X \sim Binom(n = 20, p = .08).$ Using R (in which the PDF is dbinom and the CDF is 'pbinom`), I get exact probabilities as follows:

 n = 200;  p = .08
 pbinom(16, n, p)
 ## 0.5660199
 dbinom(150, n, p)
 ## 2.040844e-119
 diff(pbinom(c(11, 30), n, p))
 ## 0.8828718
 sum(dbinom(12:30, n, p))
 ## 0.8828718
 dbinom(14, n, p)
 ## 0.09541156

The respective, normal approximations, with continuity corrections, are:

 mu = n*p;  sg = sqrt(n*p*(1-p))
 pnorm(16.5, mu, sg)
 ## 0.551844
 diff(pnorm(c(149.5, 150.5), mu, sg))
 ## 0
 diff(pnorm(c(11.5, 30.5), mu, sg))
 ## 0.8795021
 diff(pnorm(c(13.5, 14.5), mu, sg))
 ## 0.09058454

This particular binomial distribution, even though skewed, should be reasonably well approximated by the normal distribution. You can expect almost two place accuracy. For the continuity correction, see the note by @AndreNicolas, and the first figure below.

If you want to use standard normal tables, you need to standardize first, but R software does not require you to standardize in these problems. Results from tables of standard normal $Z$ will not be quite the same because of rounding errors. Specifically, $$P(X=14) = P(13.5 \le X \le 14.5) \approx P\left(\frac{13.5-16}{3.84} < Z < \frac{14.5-16}{3.84}\right)\\ = \Phi(-0.39) - \Phi(-0.65) = 0.1016.$$

Approximations based on $Pois(np)$ are as follows.

 ppois(16, mu)
 ## 0.5659624
 dpois(150, mu)
 ## 8.173203e-90
 diff(ppois(c(11, 30), mu))
 ## 0.87244
 sum(dpois(12:30, mu))
 ## 0.87244
 dpois(14, mu)
 ## 0.09301644

The figure below shows the exact binomial probabilities (bars), approximating Poisson probabilities (dots), and the normal density curve with mean and variance matching those of the binomial distribution. Vertical dotted lines show the area under the normal curve that approximates $P(12 \le X \le 20).$

Note: For a binomial distribution with larger $n$ and smaller $p$, such that the mean remains $\mu = np = 16,$ the Poisson approximation would improve noticeably; the normal perhaps not changed by much.