Measures of location and dispersion — mean, coding, and standard deviation

A data set $X$ is coded as $Y = (X - 40) / 4$. Given $\text{Var}(Y) = 5$, what is $\text{Var}(X)$?

• $5 \times 4^2 = 80$

  Correct. Here the coding rule divides by 4 to produce $Y$, so decoding multiplies by $4^2 = 16$ to recover $\text{Var}(X)$ — the same squared-constant, reversed-operation rule as before, just written with the constant as a divisor instead of a multiplier.

• B$5 \times 4 = 20$

  The direction (multiplying, to undo a division) is right, but the constant isn't squared — this is the forgot-to-square error, transplanted onto the divisor form of the coding rule instead of the multiplier form.

• C$5 \div 4^2 = 0.3125$

  The constant is squared correctly, but this divides instead of multiplying — since coding DIVIDED to reach $Y$, decoding has to MULTIPLY to reverse it, not divide again.

• D$5 \div 4 = 1.25$

  Both errors together: no square, and the wrong direction. $\text{Var}(X)$ has to be LARGER than $\text{Var}(Y)$ here, since coding divided the data down — this answer moves it smaller instead.

Traps tested: Coding decode forgot to square constant · Coding decode multiplied instead of divided

A student's working for the standard deviation of a set of exam scores (out of 100) produces a final answer of 236. What should immediately raise suspicion, beyond re-checking the algebra?

• 236 is far larger than any plausible spread for scores confined to a 0–100 range — and a real WST01 mark scheme states plainly that manifestly implausible answers are never credited, even under a follow-through rule that would otherwise excuse an earlier slip

  Correct. Sanity-checking a final answer against what's actually being measured catches this kind of error before it's ever submitted — and it's backed by the mark scheme's own general marking guidance, which states "manifestly absurd answers should never be awarded A marks," specifically so a chain of otherwise-followed-through working can't rescue a result that plainly can't be right.

• BNothing — standard deviations can be any size, so 236 is just as plausible as any other number without more context

  The DATA'S OWN RANGE is real context: scores out of 100 cannot possibly vary by more than 100 from each other, so a spread of 236 is a contradiction with the scale of the data itself, not merely an unfamiliar-looking number.

• C236 must be wrong because standard deviations are always whole numbers

  Standard deviations are frequently NOT whole numbers — this isn't a real property of the statistic at all, and reaching for it here would flag plenty of perfectly correct answers as suspicious for no reason.

• D236 must be wrong because a standard deviation can never exceed the mean

  This also isn't a general property of standard deviation — a data set can have a standard deviation larger than its mean, particularly one with a long tail. The real problem with 236 here is its size relative to the DATA'S OWN maximum possible range, not its size relative to the mean specifically.

Traps tested: No plausibility check on final answer · Invented property of standard deviation

A frequency table has values 10, 20, 30, 40 with frequencies 3, 5, 7, 5 ($\Sigma f = 20$). What is the mean?

• $\Sigma fx / \Sigma f = 540 / 20 = 27$

  Correct. $\Sigma fx = 3(10)+5(20)+7(30)+5(40) = 30+100+210+200 = 540$, and $\Sigma f = 20$.

• B$540 / 4 = 135$ — dividing by the number of distinct values in the table

  The number of distinct rows (4) plays the same wrong role here as "the number of classes" did in the grouped-table trap above — it isn't the total count of data values, which is what a mean's denominator always has to be.

• C$(10+20+30+40)/20 = 5$ — the plain sum of the four values, divided by $\Sigma f$

  This ignores the frequencies in the NUMERATOR entirely, treating each value as if it occurred exactly once regardless of how many times it actually appears — the value 30 occurred 7 times and 10 occurred only 3, and a mean that doesn't weight by frequency isn't measuring the actual data set at all.

• D30, the value with the highest frequency

  This is the MODE (the most common value), a different statistic entirely — reporting it as the mean confuses "most frequent" with "average," which don't have to be anywhere near each other.

Traps tested: Mean denominator is class count not total frequency · Mean numerator ignores frequency weighting · Mode picked instead of mean

A grouped frequency table has $n = 39$. One student finds $Q_1$'s position using $n/4 = 9.75$; another uses $(n+1)/4 = 10$. Both interpolate correctly from their own position within whichever class it falls into. What's true of their two final answers?

• They may come out slightly different — a real WST01 mark scheme is confirmed to tolerate either convention, but the two positions aren't guaranteed to interpolate to the exact same value

  Correct. $9.75$ and $10$ are close but not identical positions — in this lesson's own masses table, the equivalent pair of positions ($15$ vs $15.25$) produced $Q_1 = 20$ against $Q_1 = 20.125$: close, genuinely accepted either way, but not the same number.

• BThey must be identical, since n/4 and (n+1)/4 are just two ways of writing the same rule

  They're two DIFFERENT rules that usually land close together, not the same rule written two ways — for large $n$ the difference between $n/4$ and $(n+1)/4$ shrinks toward nothing, but it's never exactly zero, and at a class boundary it can matter.

• CStudent B's answer (using n+1) is definitely correct, since n+1 is the more standard convention

  Neither convention is "more standard" in a way that makes the other wrong — a real mark scheme accepts either. If anything, the confirmed evidence points the other way: "those that worked with n rather than n + 1 were usually more successful" on a real series.

• DStudent A's answer (using n) is definitely correct, since only n appears in the formula printed in the booklet

  Neither the interpolation formula itself nor either convention for the quartile's position is printed in the formula booklet at all — both have to be remembered, which is exactly why this lesson's teach block on standard deviation makes the same point about mean and SD.

Traps tested: N and n plus 1 treated as identical · One interpolation convention assumed superior · Nonexistent booklet formula invoked

Group C has 20 data values, mean 45, standard deviation 3. To combine it with another group, you need $\Sigma x^2$ for Group C. Which expression is correct?

• $20 \times (3^2 + 45^2) = 20 \times 2034 = 40680$

  Correct. $\Sigma x^2 = n(\text{SD}^2 + \text{mean}^2)$, with the SD squared before it's added to the squared mean.

• B$20 \times (3 + 45^2) = 20 \times 2028 = 40560$

  The standard deviation (3) is used un-squared — the exact, confirmed real error on this step: candidates who "forgot to square the standard deviation... when finding Σy²" on a genuine WST01 question. 40560 is close enough to the correct 40680 that this slip is easy to miss without checking the intermediate working.

• C$20 \times (3^2 + 45) = 20 \times 54 = 1080$

  This squares the SD but leaves the MEAN un-squared instead — a different mistake in the same formula, and one that produces a far more obviously wrong (tiny) result.

• D$3^2 + 45^2 = 2034$ — without multiplying by $n$ at all

  $\Sigma x^2$ is a total across all 20 values, not a per-value average — leaving out the $\times n$ step produces a number 20 times too small to be a genuine sum of squares for this group.

Traps tested: Forgot to square sd recovering sum of squares · Mean not squared recovering sum of squares · Sum of squares not scaled by n