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Stuck in the Middle with AVERAGE, MEDIAN, and MODE
Customer
Total Amount Spent Last Year
A
\$300
B
\$90
C
\$2,600
D
\$850
E
\$28,400
F
\$300
The mean (using the AVERAGE function) is \$5,423.33. The median is \$575,
and the mode is \$300. These three amounts aren’t even close! Which one best
represents the typical amount that a customer spent last year?
The issue with this set of data is that one value — \$28,400 — is so much
larger than the other values that it skews the mean. You may be led to
believe that each customer spent about \$5,423. But looking at the real
values, only one single customer spent a lot of money, relatively speaking.
Customers A, B, C, D, and F spent nowhere near \$5,423.33, so how can that
“average” apply to them?
Figure 9-2 shows this situation in which one value is way out of league with
the rest — sometimes called an outlier — which makes the average not too
useful. Figure 9-2 also shows how much the mean changes if the one
spendthrift customer is left out, but if you leave out any other customer, there is
very little change in the mean.
Figure 9-2:
Deciding
what to
do with an
unusual
value.
In Scenario 2, Customer E is left out. The mean and the median are much
closer together — \$968 and \$850, respectively. Either amount reasonably
represents the mid value of what customers spent last year.
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