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: $$ \begin{array}{lccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20 \end{array} $$ The mean value of \(x\) is \(\mu_{x}=1.2\). Calculate the values of \(\sigma_{x}^{2}\) and \(\sigma_{x}\)

Short answer

Once Steps 1 and 2 are properly implemented, it should be straightforward to find the values of \(\sigma_{x}^{2}\) and \(\sigma_{x}\), which are the variance and standard deviation of the defective tires data.

Step-by-step solution

Calculation of variance

To begin with, we calculate the variance, \(\sigma_{x}^{2}\), using the formula \(\sigma_{x}^{2} = \sum (x - \mu_{x})^{2} \cdot p(x)\). By substituting \(x\) values from 0 to 4 and their corresponding probabilities \(p(x)\), and mean value \(\mu_{x}\) which is given to be 1.2, we calculate each term and sum them up to get the variance.

Calculate the Square Root of Variance

Once we find the variance, the next step is to determine the standard deviation which is the square root of the variance. That is, \(\sigma_{x} = \sqrt{\sigma_{x}^{2}}\). This can be calculated directly once we have \(\sigma_{x}^{2}\).

Interpret the results

The values obtained for \(\sigma_{x}^{2}\) and \(\sigma_{x}\) are the variance and the standard deviation of the number of defective tires on a randomly selected automobile checked. They provide a measure of the dispersion or spread of the defective tires data around the mean \(\mu_{x}\) = 1.2.