Find the probability that at most 5 defective bolts will be found in a box of 200 bolts, if it is known that 2% of such bolts are expected to
be defective.
(e-*=0.0183)
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Answer:
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}
120
1024∗0.018316
= 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
this the answer.
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