Question

Historical data for a local steel manufacturing company shows that the average number of defects per standard sheet of steel is 2. In addition, the number of defects per unit is distributed according to Poisson distribution.

A batch has just been completed. What is the probability that the first three units manufactured in this batch will contain at least a total of 4 defects?
A.8488
B.7149
C.1512
D.2851

Answer 1

Answer: A.8488

Explanation:

Poisson distribution formula :

`P(X=x)=(e^(-\mu)\mu^x)/(x!)`

, where
`\mu` = mean of the distribution.

Let x be the random variable that represents the number of defects.

Given : The average number of defects per standard sheet of steel is 2.

For 3 sheets , the average number of defects per standard sheet of steel = 2 x 3 = 6

i.e.
`\mu=6`

Now , the probability that the first three units manufactured in this batch will contain at least a total of 4 defects will be :-

`P(X\geq4)=1-P(X<4)\\\\= 1-P(X\leq3)\\\\ 1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))\\\\ =1-((e^(-6)6^0)/(0!)+(e^(-6)6^1)/(1!)+(e^(-6)6^2)/(2!)+\frac {e^(-6)6^3}{3!})\\\\=1-(0.00247875+0.0148725+0.0446175+0.089235)\\\\=1-0.15120375\\\\=0.84879625\approx0.8488`

Hence, the probability that the first three units manufactured in this batch will contain at least a total of 4 defects will be 0.8488.

Thus , the correct option is A .8488.