01 · Boundaries
Mean is 12 and SD is 3. Find both boundaries for k = 2.
Hint
The distance is 2×3 = 6.
Worked solution
Lower boundary 6; upper boundary 18. Flag x < 6 or x > 18.
Understand · explore · practise
Outliers using mean and standard deviation
Use mean plus or minus a multiple of standard deviation to investigate unusual values, while keeping the rule, denominator, precision and assumptions explicit.
Before you startMean, standard deviation and inequalities.
01 / Define the distance rule
Unusual according to which definition?
For fixed mean μ, SD σ and k > 0: flag x < μ − kσ or x > μ + kσ.
Equivalently, |x − μ| > kσ. Equality is not flagged by this strict rule.
This is an alternative to an IQR rule. Follow the definition and multiplier supplied in a question; do not assume every use of the word outlier has the same meaning.
Distances measured in SDsExplore
A reference dataset has mean 20 and SD 4. Its summaries stay fixed while you classify x.
Unflagged interval: 12 ≤ x ≤ 28.
x is 2 SDs above the mean.
At the boundary: not flagged by the strict rule.
This is a classification rule. It does not assert a normal distribution or guarantee a particular percentage inside the interval.
02 / Calculate a two-sided interval
Subtract and add the same multiple of SD.
Mean 50, SD 6; flag values more than two SDs from the mean.Worked example
Lower boundary = 50 − 2×6 = 38
Subtract the distance from the mean.
Upper boundary = 50 + 2×6 = 62
Add the same distance.
Flag x < 38 or x > 62
38 and 62 are at, not beyond, the boundaries.
02 · Classify
Using the worked example, classify 37, 38, 62 and 63.
Hint
The interval includes equality.
Worked solution
37 and 63 are flagged. 38 and 62 are not flagged by the strict rule.
03 / Use SD rather than variance
The distance must have the original units.
Watch: count standard deviations from the mean
Pause, replay or seek freely. The notes explain the same idea and stay in view.
If variance is v, use σ = √v in the boundaries.
A variance in minutes² cannot be added directly to a mean in minutes.
03 · Variance supplied
Mean time is 15 minutes and variance is 9 minutes². Find boundaries for k = 3.
Hint
SD = √9 = 3 minutes.
Worked solution
Boundaries are 15 − 9 = 6 and 15 + 9 = 24 minutes. Using 15 ± 3×9 would wrongly substitute variance for SD.
04 · Precision
Mean = 10 and variance = 2. For k = 2, is 12.82 above the upper boundary?
Hint
Use 10 + 2√2 without rounding SD first.
Worked solution
Upper boundary ≈ 12.828427. 12.82 is below it and is not flagged on the high side. Rounding √2 to 1.4 too early would change the conclusion.
04 / Express distance in SD units
A signed distance also tells you the side.
For σ > 0, z = (x − μ)/σ. The strict rule is |z| > k.
This arithmetic does not require the data to be normally distributed.
05 · Signed distance
Mean 20 and SD 4: find the standardised distances for 12 and 29.
Hint
Subtract 20 and divide by 4.
Worked solution
For 12, z = −2: two SDs below the mean. For 29, z = 2.25: 2.25 SDs above the mean. With k = 2, only 29 is flagged.
06 · Zero spread
The reference dataset has SD zero. Can you divide by its SD to standardise a new value?
Hint
The denominator would be zero.
Worked solution
No. For zero spread, all reference observations equal their mean. The unstandardised strict rule with kσ = 0 flags a value different from that mean, but z is undefined.
05 / Identify which SD is given
Different denominators can change a classification.
For observed data, descriptive SD uses Sxx/n. A sample SD using Sxx/(n − 1) has a different value. If a question specifies one, use it consistently; do not switch calculator outputs midway.
07 · Two stated conventions
A reference dataset has n = 5, mean 10 and Sxx = 80. For k = 2, compare upper boundaries using denominator n and denominator n − 1.
Hint
Take the square root after dividing Sxx.
Worked solution
Descriptive SD = √16 = 4, giving upper boundary 18. The alternative SD = √20 gives upper boundary 10 + 2√20 ≈ 18.944. A value 18.5 would be flagged by the first rule but not the second.
08 · Outside value
Why should you not add a candidate to the reference dataset while claiming its mean and SD stayed fixed?
Hint
Adding an observation generally changes both summaries.
Worked solution
Classification against a fixed reference and recalculation after adding data are different operations. State which is intended; recompute summaries if the dataset changes.
06 / Recognise sensitivity to extremes
An extreme observation can move its own reference boundaries.
Mean and SD both use every observation. A very large observation can raise the mean and inflate SD, making a rule based on those same data less sensitive to that observation. This is sometimes called masking.
09 · An extreme that is not flagged
For 0, 0, 0, 0, 10, the descriptive mean is 2 and SD is 4. Does the strict k = 2 rule flag 10?
Hint
The upper boundary is 2 + 2×4.
Worked solution
No. The boundary is 10, so the value lies exactly on it. A visibly extreme observation need not be flagged by every mathematical rule.
10 · Compare definitions
For that dataset, the stated quartile convention gives Q₁ = Q₃ = 0. What does the strict k = 1.5 IQR rule do to 10?
Hint
The IQR and both fences are zero.
Worked solution
It flags 10. The IQR rule and mean/SD rule use different summaries; neither result is an arithmetic contradiction.
07 / Keep conclusions limited
A flag is not a probability statement.
11 · A percentage claim
A student says every dataset has about 95% of observations within two SDs of its mean. Is this a general rule?
Hint
A familiar normal-distribution approximation needs its model.
Worked solution
No. Mean/SD summaries do not by themselves imply a normal distribution. Do not attach the normal percentage to an arbitrary dataset.
12 · What next?
A value is flagged by the supplied rule. What would you check before excluding it?
Hint
Separate classification from data quality and study scope.
Worked solution
Check the original record, measurement units, transcription and eligibility. A genuine eligible unusual observation may be important. Document any justified correction or exclusion.
08 / Name the rule and retain precision
Calculate first; investigate afterwards.
Use SD rather than variance, state the multiplier and denominator convention, keep equality and rounding correct, and avoid inventing probabilities or automatically deleting flagged records.
Section 1 of 8 · Define the distance rule