The widths of platinum samples manufactured at a factory are normally distributed, with a mean of $1.3 \mathrm{~cm}$ and a standard deviation of $0.5 \mathrm{~cm}$. Find the z-scores that correspond to each of the following widths. Round your answers to the nearest hundredth, if necessary. (a) $2 \mathrm{~cm}$ \[ z= \] (b) $0.2 \mathrm{~cm}$ Activity Details - Task: View this topic

Step 1 ：The z-score is a measure of how many standard deviations an element is from the mean. It is calculated by subtracting the mean from an element and then dividing the result by the standard deviation. In this case, we are given the mean and standard deviation, and we need to find the z-scores for the given widths.

Step 2 ：Given that the mean is \(1.3\) and the standard deviation is \(0.5\), we can calculate the z-scores for the widths \(2\) cm and \(0.2\) cm.

Step 3 ：For the width of \(2\) cm, subtract the mean from the width and divide by the standard deviation: \((2 - 1.3) / 0.5 = 1.4\). So, the z-score for \(2\) cm is \(1.4\).

Step 4 ：For the width of \(0.2\) cm, subtract the mean from the width and divide by the standard deviation: \((0.2 - 1.3) / 0.5 = -2.2\). So, the z-score for \(0.2\) cm is \(-2.2\).

Step 5 ：Final Answer: The z-scores that correspond to the widths are: (a) \(2\) cm: \(\boxed{1.4}\), (b) \(0.2\) cm: \(\boxed{-2.2}\)