20 questions arranged by DBE cognitive level — histograms, frequency polygons, ogives, variance, standard deviation, skewness and outliers. Work each one on paper first, then reveal the memo.
20
practice questions
4
cognitive levels
20
worked memos
100%
independently verified
How to use this bank.
Start at Level 1 and move up — don't jump to Level 4 first.
For an ogive question, always sketch the points before reading anything off.
For variance and standard deviation, write out every deviation before squaring.
Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved from scratch before publication, cross-checked against its own working rather than assumed correct. These are original "Equation Station SA Practice Question" items, written to match the exact CAPS scope taught in the Summary Notes for this grade.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 5 Qs
L4 — Problem Solving 6 Qs
20%
Level 1 | Knowledge
Direct Recall
Reading a modal interval, naming a frequency-polygon coordinate, and converting between variance and standard deviation.
Q1Equation Station SA Practice Question1 mark
Histograms
Reading the Modal Interval
A histogram of daily temperatures (°C) over one month has these frequencies: \(18\le x<21\): 4, \(21\le x<24\): 9, \(24\le x<27\): 15, \(27\le x<30\): 7, \(30\le x<33\): 3. State the modal interval.
Memo
✓ The modal interval is the tallest bar on the histogram — the one with the HIGHEST frequency, not the interval with the biggest numbers.✓ The highest frequency is 15, belonging to \(\boxed{24\le x<27}\)
Q2Equation Station SA Practice Question1 mark
Frequency Polygons
Frequency Polygon Coordinate
A grouped frequency table has an interval \(30\le x<40\). At what \(x\)-value would a frequency polygon plot this interval's point?
Memo
✓ A frequency polygon is always plotted at the interval's MIDPOINT, not either boundary.✓ Midpoint \(=\dfrac{30+40}{2}=\boxed{35}\)
Q3Equation Station SA Practice Question1 mark
Variance & SD
Standard Deviation from Variance
A data set has variance \(=49\). Determine its standard deviation.
Memo
✓ Standard deviation is defined as the square root of variance — this converts the "squared units" of variance back into the original data's own units.✓ \(\sigma=\sqrt{49}=\boxed{7}\)
Q4Equation Station SA Practice Question1 mark
Variance & SD
Variance from Standard Deviation
A data set has standard deviation \(=6\). Determine its variance.
Memo
✓ Reversing the square root: variance is standard deviation SQUARED.✓ \(\sigma^2=6^2=\boxed{36}\)
25%
Level 2 | Routine Procedures
One Established Method
Reading a median and a quartile off an ogive, classifying skewness from given mean/median pairs, and finding a fence value.
Q5Equation Station SA Practice Question3 marks
Ogives
Reading the Median off an Ogive
The ages of 80 people are grouped as follows:
Age
Frequency
\(10\le x<20\)
12
\(20\le x<30\)
20
\(30\le x<40\)
28
\(40\le x<50\)
14
\(50\le x<60\)
6
Determine the position (as a value of \(n\)) and estimate the median from a sketch of the ogive.
Memo
✓ \(n=80\), so the median sits at position \(\dfrac{n}{2}=40\).✓ Cumulative frequencies: \(12,\ 32,\ 60,\ 74,\ 80\) — the running total passes 32 and reaches 60 inside \(30\le x<40\), plotted at the upper boundary 30→12 and 40→60.✓ Interpolating between \((30,32)\) and \((40,60)\) for a target of 40: \(30+\dfrac{40-32}{60-32}\times10=30+\dfrac{8}{28}\times10\approx\boxed{32{,}9}\)
Q6Equation Station SA Practice Question3 marks
Ogives
Reading Q1 off an Ogive
The delivery times (minutes) of 60 orders are grouped as follows:
Time
Frequency
\(0\le x<10\)
5
\(10\le x<20\)
9
\(20\le x<30\)
18
\(30\le x<40\)
16
\(40\le x<50\)
8
\(50\le x<60\)
4
Estimate \(Q_1\) from a sketch of the ogive.
Memo
✓ \(n=60\), so \(Q_1\) sits at position \(\dfrac{n}{4}=15\).✓ Cumulative frequencies: \(5,\ 14,\ 32,\ 48,\ 56,\ 60\) — the running total passes 14 and reaches 32 inside \(20\le x<30\), plotted at \((10,5),(20,14),(30,32)\).✓ Interpolating between \((20,14)\) and \((30,32)\) for a target of 15: \(20+\dfrac{15-14}{32-14}\times10=20+\dfrac{1}{18}\times10\approx\boxed{20{,}6}\)
Q7Equation Station SA Practice Question1 mark
Skewness
Classifying Symmetric Data
A data set has a mean of 45 and a median of 45. Classify its skewness.
Memo
✓ Skewness is judged by comparing the mean and the median — when the two are equal, no value is pulling the mean away from the centre in either direction.✓ Mean \(=\) median \(\Rightarrow\) the data is \(\boxed{\text{symmetric}}\)
Q8Equation Station SA Practice Question2 marks
Skewness
Classifying Positively Skewed Data
A data set has a mean of 52 and a median of 47. Classify its skewness and explain what caused it.
Memo
✓ Mean \((52)>\) median \((47)\), so the mean has been pulled ABOVE the median.✓ This happens when a small number of unusually HIGH values stretch a "tail" toward the right.✓ \(\boxed{\text{The data is positively (right) skewed}}\)
Q9Equation Station SA Practice Question3 marks
Outliers
Calculating a Fence Value
A data set has \(Q_1=18\) and \(Q_3=30\). Using the \(1{,}5\times\text{IQR}\) rule, determine the upper fence.
Memo
✓ First find the IQR: \(\text{IQR}=Q_3-Q_1=30-18=12\)✓ The upper fence marks the point beyond which a value counts as an outlier: \(Q_3+1{,}5\times\text{IQR}=30+1{,}5(12)=30+18=\boxed{48}\)
25%
Level 3 | Complex Procedures
Multi-Step Reasoning
Full manual variance and standard deviation, classifying skewness from raw, unsorted data, and the quartile-distance skewness test.
Q10Equation Station SA Practice Question5 marks
Variance & SD
Manual Variance and SD (n = 7)
Calculate the mean, variance and standard deviation of: \(4,\ 6,\ 9,\ 10,\ 11,\ 14,\ 16\)
Memo
✓ Mean: \(\bar{x}=\dfrac{4+6+9+10+11+14+16}{7}=\dfrac{70}{7}=\boxed{10}\)✓ Deviations from the mean: \(-6,-4,-1,0,1,4,6\); squared: \(36,16,1,0,1,16,36\)✓ Sum of squared deviations \(=36+16+1+0+1+16+36=106\)✓ Variance \(=\dfrac{106}{7}\approx\boxed{15{,}14}\); standard deviation \(=\sqrt{15{,}14}\approx\boxed{3{,}89}\)
Q11Equation Station SA Practice Question5 marks
Variance & SD
Manual Variance and SD (n = 6)
Calculate the mean, variance and standard deviation of: \(14,\ 17,\ 19,\ 21,\ 23,\ 26\)
Memo
✓ Mean: \(\bar{x}=\dfrac{14+17+19+21+23+26}{6}=\dfrac{120}{6}=\boxed{20}\)✓ Deviations from the mean: \(-6,-3,-1,1,3,6\); squared: \(36,9,1,1,9,36\)✓ Sum of squared deviations \(=36+9+1+1+9+36=92\)✓ Variance \(=\dfrac{92}{6}\approx\boxed{15{,}33}\); standard deviation \(=\sqrt{15{,}33}\approx\boxed{3{,}92}\)
✓ Sort first: \(8,20,21,22,23,24,25,26,50\) — 9 values.✓ Mean \(=\dfrac{8+20+21+22+23+24+25+26+50}{9}=\dfrac{219}{9}\approx24{,}33\)✓ Median (5th value) \(=23\)✓ Mean \((24{,}33)>\) median \((23)\): the single high value 50 pulled the mean upward — \(\boxed{\text{positively (right) skewed}}\)
✓ Sort first: \(2,28,29,30,31,32,33,34,36\) — 9 values.✓ Mean \(=\dfrac{2+28+29+30+31+32+33+34+36}{9}=\dfrac{255}{9}\approx28{,}33\)✓ Median (5th value) \(=31\)✓ Mean \((28{,}33)<\) median \((31)\): the single low value 2 pulled the mean downward — \(\boxed{\text{negatively (left) skewed}}\)
Q14Equation Station SA Practice Question3 marks
Skewness
Skewness from the Quartile-Distance Test
A data set has the five-number summary: minimum \(10\), \(Q_1=18\), median \(22\), \(Q_3=25\), maximum \(30\). Using the quartile-distance test (comparing how far the median sits from \(Q_1\) and from \(Q_3\)), classify the skewness of this data.
Memo
✓ Median to \(Q_1\): \(22-18=4\). Median to \(Q_3\): \(25-22=3\).✓ The median sits CLOSER to \(Q_3\) (distance 3) than to \(Q_1\) (distance 4).✓ When the median sits closer to \(Q_3\), the lower side has more room to spread out: \(\boxed{\text{the data is negatively (left) skewed}}\)
30%
Level 4 | Problem Solving
Full Investigations
Complete outlier identification from raw data (odd and even n), a symmetric-data interpretation, connecting skewness to a box-and-whisker shape, reconstructing a mean from an ogive, and solving for two unknowns given the mean and variance.
✓ \(n=11\) (odd), already sorted. Median (6th value) \(=22\).✓ Lower half (excluding the median) \(15,18,19,20,21\Rightarrow Q_1=19\); upper half \(23,24,26,28,52\Rightarrow Q_3=26\).✓ \(\text{IQR}=26-19=7\). Upper fence \(=26+1{,}5(7)=26+10{,}5=36{,}5\).✓ \(52>36{,}5\), so \(\boxed{52\text{ is the only outlier}}\) — every other value sits comfortably below the fence.
✓ \(n=10\) (even), already sorted. Median \(=\dfrac{50+52}{2}=51\).✓ With an even count, the whole data set splits cleanly in half: lower half \(40,44,46,48,50\Rightarrow Q_1=46\); upper half \(52,54,56,58,90\Rightarrow Q_3=56\).✓ \(\text{IQR}=56-46=10\). Lower fence \(=46-1{,}5(10)=31\); upper fence \(=56+1{,}5(10)=71\).✓ \(90>71\), so \(\boxed{90\text{ is the only outlier}}\) — every other value falls between the two fences.
Q17Equation Station SA Practice Question4 marks
Skewness
Confirming Symmetric Data
Nine measurements are: \(5,\ 15,\ 20,\ 23,\ 25,\ 27,\ 30,\ 35,\ 45\). Show that the data is symmetric, and explain what this means about the shape of its distribution.
Memo
✓ Already sorted, 9 values. Mean \(=\dfrac{5+15+20+23+25+27+30+35+45}{9}=\dfrac{225}{9}=25\).✓ Median (5th value) \(=25\).✓ Mean \(=\) median exactly \(=25\), so \(\boxed{\text{the data is symmetric}}\).✓ This means the values are evenly balanced on both sides of the centre — no single unusually high or low value is dragging the mean away from the middle of the data.
Q18Equation Station SA Practice Question4 marks
SkewnessBox-and-Whisker
Connecting Skewness to a Box Plot
A large data set has a mean of \(68{,}5\) and a median of \(72\). (a) Classify the skewness. (b) On a box-and-whisker diagram of this data, would you expect the section below the median or above the median to be longer? Explain.
Memo
✓ (a) Mean \((68{,}5)<\) median \((72)\): a few unusually LOW values have pulled the mean down below the median, so \(\boxed{\text{the data is negatively (left) skewed}}\).✓ (b) Negative skew means the lower half of the data is more spread out than the upper half.✓ \(\boxed{\text{The section below the median (toward the minimum) would be longer}}\), since that is where the unusually low, spread-out values sit.
Q19Equation Station SA Practice Question5 marks
Ogives
Reconstructing the Mean from an Ogive
An ogive for 40 learners' travel times (minutes) shows the following cumulative frequencies at the upper boundary of each 10-minute interval: \(5\) (at \(10\)), \(12\) (at \(20\)), \(25\) (at \(30\)), \(33\) (at \(40\)), \(40\) (at \(50\)). Reconstruct the frequency table and estimate the mean travel time.
Memo
✓ Individual frequencies come from consecutive DIFFERENCES in cumulative frequency: \(5-0=5\); \(12-5=7\); \(25-12=13\); \(33-25=8\); \(40-33=7\).✓ Check: \(5+7+13+8+7=40\), matching the total ✓✓ Midpoints of the 5 intervals: \(5,15,25,35,45\).✓ Estimated mean \(=\dfrac{(5)(5)+(7)(15)+(13)(25)+(8)(35)+(7)(45)}{40}=\dfrac{25+105+325+280+315}{40}=\dfrac{1050}{40}=\boxed{26{,}25\text{ minutes}}\)
Q20Equation Station SA Practice Question6 marks
Variance & SD
Two Unknowns from Mean and Variance
A data set of 6 values has a mean of \(10\) and a variance of \(14\): \(6,\ 9,\ 11,\ 14,\ c,\ d\). Determine the values of \(c\) and \(d\).
Memo
✓ From the mean: total \(=10\times6=60\). The 4 known values sum to \(6+9+11+14=40\), so \(\boxed{c+d=20}\)✓ From the variance: total squared deviation \(=14\times6=84\). The known values' own squared deviations from the mean (10) sum to \((6-10)^2+(9-10)^2+(11-10)^2+(14-10)^2=16+1+1+16=34\), so \((c-10)^2+(d-10)^2=84-34=50\)✓ Substitute \(d=20-c\) into the second equation and expand: \((c-10)^2+(10-c)^2=50\Rightarrow2(c-10)^2=50\Rightarrow(c-10)^2=25\Rightarrow c-10=\pm5\)✓ \(\boxed{c=15,\ d=5}\) (or the other way around)✓ Check: \(6,9,11,14,15,5\) has mean \(\dfrac{60}{6}=10\)✓ and variance \(\dfrac{34+(15-10)^2+(5-10)^2}{6}=\dfrac{34+25+25}{6}=\dfrac{84}{6}=14\)✓