Answer:
The probability that there will be a total of 7 defects on four units is 0.14.
Step-by-step explanation:
A Poisson distribution describes the probability distribution of number of success in a specified time interval.
The probability distribution function for a Poisson distribution is:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...[/tex]
Let X = number of defects in a unit produced.
It is provided that there are, on average, 2 defects per unit produced.
Then in 4 units the number of defects is, [tex](2\times4)=8[/tex].
Compute the probability of exactly 7 defects in 4 units as follows:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14[/tex]
Thus, the probability of exactly 7 defects in 4 units is 0.14.
