Right Wing Nation

Class today, so while I’m gone, here are some “test your knowledge” quiz questions I have given my students. Each is easily answered with no more than simple descriptive statistics, but tests the student’s knowledge of the concepts (as opposed to whether the student can calculate an arithmetic mean or standard deviation). A student sitting in front of Excel, SPSS, or SAS should be able to answer these three questions in three minutes.

Answers when I get back this afternoon.

1. A tire manufacturer produces a particular model tire whose tread wear life is normally distributed with a mean of 39,000 miles and a standard deviation of 5,300 miles. The manufacturer wishes to provide a guaranteed tread life for this model which would be exceeded by 98% of all tires. What tread life would meet this requirement?
2. The mechanical process which fills 10-lb bags of dog food is subject to random fluctuations in the amount placed in each bag. The amount placed in each bag is approximately normally distributed with a mean of 170 ounces and a standard deviation of 4.3 ounces. Determine an interval centered on the mean such that the weight of the contents of 99% of the bags will fall within that interval.
3. The scores on an exam are approximately normally distributed with a mean of 75 and a standard deviation of 10. If the professor wants 10% of the class to receive As, then what is the minimum score a student can get and receive an A on the exam?

Answers.

All of these are critical value problems, where we calculate either the probability of a value being less than (or greater than) or equal to a critical value, or calculate the critical value based on the probability. The only difference between the first and second problem is that the first problem is one-tailed (we need to find the critical value for which all values will be greater than or equal to the top 98% tail) and the second problem is two-tailed (we need to find the critical value for which all all values will fall between the two tails).

1. A tire manufacturer produces a particular model tire whose tread wear life is normally distributed with a mean of 39,000 miles and a standard deviation of 5,300 miles. The manufacturer wishes to provide a guaranteed tread life for this model which would be exceeded by 98% of all tires. What tread life would meet this requirement?

The first step is always the same for every problem: Enter the data given, then the data we can deduce. We are given the mean, the standard deviation, and the probability of the area of the curve less than the critical value (98%).

Because we know that P(right area) is 98%, we can subtract it from 1 to get the P(left area).

We can use Excel’s NORMSINV function to take the probability and return a z-score, the critical value in standard deviations.

So the critical value is 2.05375 standard deviations below the mean. Now, let’s convert that to a number.

First, we multiply the z(left area) by the standard deviation. That converts the value to a number. Then, all we have to do is add the critical value to the mean to get the answer (we add because the number is negative).

Answer: A tread life of 28,115.14 miles would be exceeded by 98% of all tires.