Quiz 1 Results

Average: 4.3

SD: 3

Median: 4

Min: 0

Max: 10

Below is a histogram of estimated annual rainfall amounts (inches) at a location in the Pacific Northwest for 2,129 years (including the present). The bins are marked by the mid-point, so you can see that each bin is 1.0 inch wide. The table on the next page gives the height of each bin.

1) Approximately what is the median amount of annual rainfall? Explain.

As the table shows, 40.25% of the observations are below 13 inches, and 61.06% are below 14 inches. The median is the number that has 50% of the observations above and below it, so the median must be in the 13-14 bin. This is the bin marked 13.5. We have no way of knowing precisely which number is the sample median because we cannot see the original list of observations. However, it must be near 13.5, and must be more than or equal to 13, and less than or equal to 14.

Many of you referred to a "sorted" list, but note that the original list is not here, and so we can't sort and we can't know exactly what the median is.

It is helpful to make a mark where the median occurs. Many of you would have then noticed that your answer does not appear to be where the histogram is cut in half, as it should.

The symmetry of the histogram is somewhat irrelevant to finding the median. Some of you correctly pointed out that the distribution is right-skewed, but that tells us only that the sample mean will be greater than the sample median. It does not tell us where the median is located. (Some people said something like "the distribution is symmetric and therefore the median is in the center." But the median is always in the center and this begs the question "what do you mean by 'center'?")

2) Suppose we combine the last six bins into one long bin. This bin will go from 17 inches to 24 inches. What will the height of this bin be? (Hint, we are constructing a relative frequency histogram in which the units will be percent per inch.) Show your work.

The area of a bin is proportional to the proportion of observations. Area = height * Width

therefore height = area/width.

Area here must be (49+25+9+5+4+1)/2129 = 93/2129 = 0.0437

Width = 24 - 17 = 7 inches.

height = .0437/7 = 0.0062 percent per inch