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 of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Short answer

Therefore, the task completion times that would permit an applicant to qualify for advanced training would be approximately 146 seconds or less.

Step-by-step solution

Understanding the Provided Information

In the problem, we are given that the task completion time follows a normal distribution. The mean (\( \mu \)) = 120 seconds and standard deviation (\( \sigma \)) = 20 seconds. The task is to find out what completion times allow the top 10% of applicants to qualify for advanced training. Since we're looking for the 'fastest' 10%, we will be looking for the 90th percentile.

Standardizing

We have to transform the normal distribution to the standard normal distribution so we can solve the problem. To get the standardized score or z-score for the 90th percentile, we can use the standard normal table or statistical software, where we find that the z-score corresponds approximately to 1.28.

Calculating the Task Completion Time

Now, we use the given mean, standard deviation, and calculated z-value to find out the time that can qualify the top 10%. To do this, we use the formula: \( x = \mu + z * \sigma \), where \( x \) is the task completion time. Inserting the present values, we get \( x = 120 + 1.28*20 \). Solving this, we get \( x \approx 146 \).