Question

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of \(120 \mathrm{sec}\) and a standard deviation of \(20 \mathrm{sec}\). The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Short answer

The cutoff time for the fastest 10% of job applicants, who will be given advanced training, is 145.6 seconds.

Step-by-step solution

Identify Given Information

The mean (\( \mu \)) of the distribution is given as 120 seconds, and the standard deviation ( \( \sigma \)) is given as 20 seconds. The fastest 10% of the workers are to be provided with advanced training which means we need to find the cutoff time for the fastest 10% in the distribution. This corresponds to finding the 90th percentile in the distribution since 100% - 10% = 90%.

Use Z-score Formula

The formula to convert an x-value to a z-score is given by the equation \( z = (x - \mu) / \sigma \), where \( z \) is the z-score, \( x \) is the value from our data set, \( \mu \) is the mean and \( \sigma \) is the standard deviation. However, we don't have the x-value (cutoff time), but we do know it is at the 90th percentile of data. From standard normal distribution tables, a z-score of approximately 1.28 corresponds to the 90th percentile.

Solve for the X-Value

Rearrange the equation to solve for the x-value (i.e., the cutoff time): \( x = \mu + z * \sigma \). Substituting the given values into the equation gives \( x = 120 + 1.28 * 20 = 145.6 \) seconds