Question

Two machines in a factory are supposed to work at the same speed to pass inspection. On five random​ days, the number of items built by each machine is recorded in the table. The inspector believes that the machines should not pass inspection because the mean speed of Machine X is much faster than the mean speed of Machine Y. Which measure of center and variability should be used to compare the performances of each​ machine? Then explain why the inspector is correct or incorrect with his decision.
Number of Items Built
Machine X
19
14
20
15
17
Machine Y
20
3
16
18
17
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Part 1
Which measure of center and variability should be used to compare the performances of each​ machine? Select all that apply.
A.
Mean
B.
Median
C.
Range
D.
Interquartile range

Answer 1

Which to use: B (median) and D (interquartile range).

Why: Machine Y has an unusually low value (3) that strongly skews the mean and inflates the range. Median and IQR are robust to such outliers and better summarize center and spread here.

Quick calculations:
- Means: X = 85/5 = 17.0, Y = 74/5 = 14.8 (difference 2.2, pulled down by the 3).
- Medians: X = 17, Y = 17 (equal).
- IQRs: X = 19.5 − 14.5 = 5.0, Y = 19 − 9.5 = 9.5 (Y shows greater spread).

Conclusion: Using robust measures (median, IQR) the machines have the same typical output (median 17), so the inspector is incorrect to reject them based on the mean being “much faster” for X — that mean difference is driven by an outlier and is not reflected in the medians. Note the sample is small, so any firm decision should be made with more data.