A machine produces a defective component with a probability of 0.015. The number of defective components in a packed box containing 200 components produced by the machine follows a Poisson distribution. The mean and the variance of the distribution are

The correct option is 3 and 3, respectively.

Step-by-step Explanation:

Based on the problem description and as verified from the formulas shown in the attached image, we can define the parameters of the distribution as follows:
1. The probability of producing a defective component, represented by
$P=0.015$
2. The number of components in a packed box, represented by
$n=200$

For a Poisson distribution that approximates a binomial distribution with a large number of trials $n$ and a small probability $P$, the mean (represented by $\lambda$) is calculated as:

$\mathrm{mean}=\lambda =n\times P$
Substituting the given values:
$\lambda =200\times 0.015=3$

One of the key defining properties of the Poisson distribution is that its mean and variance are equal. Thus, we have:

$\mathrm{variance}={\sigma }^2=\lambda =3$

Consequently, both the mean and the variance of the distribution are 3. This matches the calculation steps shown in the image:
$\mathrm{mean}=\lambda =np=200\times 0.015=3$
and
$\mathrm{variance}={\sigma }^2=\lambda =3$.