Question

An optical inspection system is used to distinguish among different part types. The probability of a correct classification of any part is 0.97. Suppose that three parts are inspected and that the classifications are independent. Let the random variable X denote the number of parts that are correctly classified. Determine the mean and variance of X. Round your answers to three decimal places (e.g. 98.765). Mean = Variance =

Answer 1

Answer: Mean = 2.91 and Variance = 0.087

Step-by-step explanation:

Given : The probability of a correct classification of any part p= 0.97.

The number of parts are inspected : n=3

Let the random variable X denote the number of parts that are correctly classified.

For binomial experiment , the mean and variance is given by :-

$\mu=np\\\\\sigma^2=np(1-p)$

Then, the  the mean and variance of X will be :-

$\mu=3(0.97)=2.91\\\\\sigma^2=np(1-p)=3(0.97)(0.03)=0.0873\approx0.087\ \ [ \text{Rounded to three decimal places}]$