A sample of 36 land parcels has mean 48.00 acres, standard deviation 46.44, minimum 7, Q1 18.02, median 32.50, Q3 57.48 and maximum 248. Decide whether the distribution is symmetric or skewed, and use the 1.5 x IQR rule to test for outliers.

Step-by-step solution

1. 1

   (a) Compare the mean with the median. In a symmetric distribution the two nearly coincide. Here

   $\text{mean}-\text{median}=48.00-32.50=15.50$

   which is large compared with the interquartile spread. The mean is pulled above the median, the signature of a right-skewed (positively skewed) distribution.

2. 2

   (a) Confirm with the distances to the extremes. The maximum sits $248-32.50=215.5$ above the median, while the minimum is only $32.50-7=25.5$ below it — a long right tail against a short left one. Note also that the standard deviation $46.44$ is almost as large as the mean, another hallmark of skewed, non-negative data.

3. 3

   (b) Compute the interquartile range. The IQR measures the spread of the middle half of the data and, unlike the standard deviation, is not itself inflated by the extreme values you are testing:

   $IQR=Q_3-Q_1=57.48-18.02=39.46$

4. 4

   (b) Build the two fences. The $1.5\times IQR$ rule places boundaries one and a half interquartile ranges beyond the quartiles:

   $1.5\times IQR=1.5\times39.46=59.19$

   $\text{lower fence}=Q_1-59.19=18.02-59.19=-41.17$
   $\text{upper fence}=Q_3+59.19=57.48+59.19=116.67$

   A negative lower fence is perfectly normal for a quantity that cannot be negative; it simply means no small value can qualify as an outlier.

5. 5

   (b) Test the extreme values against the fences. The minimum $7$ lies comfortably inside the lower fence $-41.17$, so there are no low outliers. The maximum $248$ far exceeds the upper fence $116.67$ — it is more than $4.8$ interquartile ranges above $Q_3$ — so there is at least one high outlier.

6. 6

   (c) Say what the summary cannot tell you. The five-number summary fixes the box ($Q_1$, median, $Q_3$) and shows that $248$ must be plotted as an individual outlier point. What it does not reveal is where the upper whisker stops, because the whisker must end at the largest observation that is still below $116.67$, and that value is not in the table. You would need the individual data values (or at least the largest non-outlying one) — and the same question applies to any other values between $116.67$ and $248$.