Question

The probability distribution of \(x\), the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lrrrrr} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & .54 & .16 & .06 & .04 & .20 \end{array} $$ a. Calculate the mean value of \(x\). b. What is the probability that \(x\) exceeds its mean value?

Short answer

The mean value of \(x\) is 1.06 and the probability that \(x\) exceeds its mean value is 0.30.

Step-by-step solution

Calculate the mean value

The mean or expected value of a discrete random variable \(x\) can be calculated as the sum of the product of each outcome and its respective probability. This can be mathematically represented as \[E(x) = \sum x \cdot P(x)\] Using the given values, we have: \[E(x) = (0 * .54) + (1 * .16) + (2 * .06) + (3 * .04) + (4 * .20)\]

Calculate E(x)

On calculating the above equation, the mean or Expected value, \(E(x)\) is found to be 1.06.

Calculate P(x > E(x))

Now that we have the mean value, we can find the probability that \(x\) exceeds its mean value by summing up the probabilities of all outcomes greater than 1.06. Looking at the distribution, the outcomes greater than 1.06 are 2, 3 and 4. So, \[P(x > E(x)) = P(2) + P(3) + P(4) = .06 + .04 + .20\]

Calculate P(x > E(x))

Adding the respective probabilities, we find that the probability that \(x\) exceeds its mean value is 0.30.