Question

find the probability that at most 5 defective bolts will be found in a box of 200 bulbs if it is known that 2percent of such bolts are expected to be defective (you may take the distribution to be poisson.)

Answer 1

The poisson distribution is given by :

f(o, x) = $\frac{o^xe^-^o}{x!}$

Where : o = the mean.

The mean in our case is given as :

2/100 × 200 = 4

o = 4

We want the probability that x ≤ 5

P(x =  5) = $\frac{4^5e^-^4}{5!}$

We solve this as follows:

=  $\frac{1024 * 0.018316}{120}$

= 0.1563

We do this for 1 up to 4

P(x = 1) = $\frac{4^1e^-^4}{1!}$ = 0.07326

P(x = 2) = $\frac{4^2e^-^4}{2!}$ = 0.12821

P(x = 3) = $\frac{4^3e^-^4}{3!}$ = 0.19537

P(x =4) = $\frac{4^4e^-^4}{4!}$ = 0.19537

P(x≤5) = P(x = 1) + P(x = 2) + P(x =3) + P(x = 4) + P(x = 5)

= 0.07326 + 0.12821 + 0.19537 + 0.19537 + 0.1563 = 0.74851