2 years ago

Help me quickly 1/2 -1/2 2 -2

1 answer:

7 0

Answer:

$\frac{-1}{2}$

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charle [14.2K]

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:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let <em>X</em> = 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, $(2\times4)=8$ .

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.

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