Defective Pieces: Understanding the Problem

This problem asks us to calculate the mean (average) value and variance (spread) of the number of defective pieces produced by a machine in a day. We are given the probabilities associated with producing 0, 1, or 2 defective pieces.

Let \(X\) be the random variable representing the number of defective pieces produced in a day. The possible values for \(X\) and their corresponding probabilities are:

• When 0 defective pieces are produced, the probability \(P(X=0) = \frac{1}{6}\).
• When 1 defective piece is produced, the probability \(P(X=1) = \frac{2}{3}\).
• When 2 defective pieces are produced, the probability \(P(X=2) = \frac{1}{6}\).

We can organize this information in a table:

Mean Value Calculation for Defective Pieces

The mean value, also known as the expected value, of a discrete random variable \(X\) is denoted as \(E(X)\). It is calculated by summing the product of each possible value of the random variable and its corresponding probability.

The formula for the mean \(E(X)\) is:

\[E(X) = \sum x_i P(X=x_i)\]

Let's calculate the mean using the given values:

\[E(X) = (0 \times \frac{1}{6}) + (1 \times \frac{2}{3}) + (2 \times \frac{1}{6})\] \[E(X) = 0 + \frac{2}{3} + \frac{2}{6}\]

To add these fractions, we can simplify \(\frac{2}{6}\) to \(\frac{1}{3}\):

\[E(X) = 0 + \frac{2}{3} + \frac{1}{3}\] \[E(X) = \frac{2+1}{3}\] \[E(X) = \frac{3}{3}\] \[E(X) = 1\]

So, the mean value of the number of defective pieces produced by the machine in a day is 1.

Variance Calculation for Defective Pieces

The variance measures how spread out the values of the random variable are from the mean. For a discrete random variable, the variance is denoted as \(Var(X)\) or \(\sigma^2\).

The formula for the variance \(Var(X)\) is:

\[Var(X) = E(X^2) - (E(X))^2\]

First, we need to calculate \(E(X^2)\), which is the expected value of \(X\) squared.

\[E(X^2) = \sum x_i^2 P(X=x_i)\]

Let's calculate \(E(X^2)\):

\[E(X^2) = (0^2 \times \frac{1}{6}) + (1^2 \times \frac{2}{3}) + (2^2 \times \frac{1}{6})\] \[E(X^2) = (0 \times \frac{1}{6}) + (1 \times \frac{2}{3}) + (4 \times \frac{1}{6})\] \[E(X^2) = 0 + \frac{2}{3} + \frac{4}{6}\]

Now, simplify \(\frac{4}{6}\) to \(\frac{2}{3}\):

\[E(X^2) = 0 + \frac{2}{3} + \frac{2}{3}\] \[E(X^2) = \frac{2+2}{3}\] \[E(X^2) = \frac{4}{3}\]

Now we have \(E(X^2) = \frac{4}{3}\) and we previously calculated \(E(X) = 1\). We can substitute these values into the variance formula:

\[Var(X) = E(X^2) - (E(X))^2\] \[Var(X) = \frac{4}{3} - (1)^2\] \[Var(X) = \frac{4}{3} - 1\]

To subtract, express 1 as \(\frac{3}{3}\):

\[Var(X) = \frac{4}{3} - \frac{3}{3}\] \[Var(X) = \frac{4-3}{3}\] \[Var(X) = \frac{1}{3}\]

Thus, the variance of the number of defective pieces produced by the machine in a day is \(\frac{1}{3}\).

Conclusion

Based on our calculations, the mean value of the number of defective pieces is 1, and the variance is \(\frac{1}{3}\).

Therefore, the correct option is the one stating 1 and \(\frac{1}{3}\).