Question

An optical inspection system is used to distinguish among different part types. The probability of a correct classification of any part is 0.94. 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.

Answer 1

The mean and variance of X are 2.82 and 0.1692 respectively

Explanation:

The binomial distribution is a discrete probability distribution that tells the percentage in which a result is likely to be obtained between two possibilities when performing n number of tests. In other words, a binomial distribution is a discrete probability distribution that describes the number of successes when conducting n independent experiments on a random variable.

In a binomial distribution, the mean indicates the mean value of a random phenomenon. It is calculated by:

μ=n*p

Where:

• n is the number of trials
• p is the probability of success

This case is a binomial distribution, where n = 3 and p = 0.94. Then:

μ=3*0.94

μ=2.82

The mean of X is 2.82.

Variance is a measure of dispersion that represents the variability of a data series with respect to its mean. It is calculated by:

Var[X]= n*p*q

Where:

• n is the number of trials
• p is the probability of success
• q is the probability of failure

Remembering that the random variable X expresses the number of successes obtained when performing "n" independent Bernouilli tests or tests with probability "p" of success and "(1-p)" of failure, then q = 1-p.

So:

Var[X]= n*p*(1-p)

In this case:

Var[X]=3*0.94*(1-0.94)

Var[X]=3*0.94*0.06

Var[X]=0.1692

The variance of X is 0.1692