Mathematics • Year 10 • Unit 4 • Lesson 20
Statistics & Probability Review — Skill Drill
Rebuild every Unit 4 skill from scratch: classify data, calculate the five-number summary, read a scatter plot, apply the addition / multiplication / conditional probability rules. This is the unit's last warm-up — make sure every skill is rock solid before the master worksheet.
1. I do — fully worked review example
Mixed problem that touches three different Unit 4 skills: five-number summary, addition rule, and conditional probability.
Problem. The test scores of 10 students are: 45, 52, 58, 62, 65, 68, 72, 75, 80, 95. Find (a) the mean, (b) the median, (c) the range, (d) the IQR, and (e) the best measure of centre with justification.
Step 1 — Mean.
Sum = 45+52+58+62+65+68+72+75+80+95 = 672. Mean = 672/10 = 67.2.
Step 2 — Median.
Data is already in order. Middle two (5th and 6th values): 65, 68. Median = (65+68)/2 = 66.5.
Step 3 — Range.
Range = max − min = 95 − 45 = 50.
Step 4 — IQR (interquartile range).
Lower half (5 values): 45, 52, 58, 62, 65 → Q1 = 58 (middle of lower 5).
Upper half (5 values): 68, 72, 75, 80, 95 → Q3 = 75.
IQR = Q3 − Q1 = 75 − 58 = 17.
Step 5 — Best measure of centre.
95 is unusually high compared to the rest (a likely outlier). Mean (67.2) is pulled up; median (66.5) is not.
Reason: Lesson 20 MCQ Q3 — "If a data set has an outlier, the best measure of centre is the median."
Answer: Mean = 67.2; median = 66.5; range = 50; IQR = 17; best measure = median (outlier-resistant).
2. We do — fill in the missing steps
Fill the blanks. 4 marks
Problem. A bag has 3 red, 4 blue and 5 green marbles (12 total). Two marbles are drawn without replacement. Find P(both blue).
Step 1 — Independent or dependent?
Without replacement → the bag changes after the first draw → events are __________________.
Step 2 — First draw.
P(first blue) = ____/12 = 1/3.
Step 3 — Second draw (conditional).
After 1 blue removed: ____ blue remain out of ____ total.
P(second blue | first blue) = ____/11.
Step 4 — Multiply along the branch.
P(both blue) = (4/12) × (____/11) = ______/132 = ______ (simplified fraction).
3. You do — independent practice
Mixed coverage of the whole unit.
Foundation — quick recalls
3.1 Classify each as categorical, discrete numerical or continuous numerical: (a) eye colour, (b) number of siblings, (c) height in cm. 1 mark
3.2 The display that best shows correlation between two numerical variables is the ______________. 1 mark
3.3 For 5 values 2, 5, 7, 9, 11 — find the median. 1 mark
3.4 A fair die is rolled. P(rolling a 6) = ______. 1 mark
Standard — apply unit ideas in combination
3.5 For the data set 10, 12, 14, 14, 15, 17, 18, 20, 22, 25 — find the median, Q1, Q3 and IQR. 3 marks
3.6 P(A) = 0.3, P(B) = 0.4, P(A and B) = 0.12. Find P(A or B) and P(A|B). Are A and B independent? 3 marks
Extension — tree diagram + scatter plot
3.7 A biased coin has P(head) = 0.6. The coin is flipped twice. Draw a tree diagram and find P(at least one head). 3 marks
3.8 A scatter plot of "hours studied" vs "exam score" shows a clear upward pattern with most points close to a straight line. Describe the correlation in one sentence (direction + strength), and explain in one sentence how you would draw the line of best fit. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (3R, 4B, 5G without replacement)
Step 1: dependent.
Step 2: P(first blue) = 4/12 = 1/3.
Step 3: 3 blue remain out of 11 total. P(second blue | first blue) = 3/11.
Step 4: P(both blue) = 4/12 × 3/11 = 12/132 = 1/11.
3.1 — Classifications
(a) Categorical (label). (b) Discrete numerical (whole-number count). (c) Continuous numerical (measured).
3.2 — Correlation display
Scatter plot.
3.3 — Median of 5 values
Middle value = 7.
3.4 — P(6)
P(6) = 1/6.
3.5 — Five-number summary
Median (5th and 6th): (15+17)/2 = 16.
Lower half (10,12,14,14,15) → Q1 = 14.
Upper half (17,18,20,22,25) → Q3 = 20.
IQR = 20 − 14 = 6.
3.6 — Addition, conditional, independence
P(A or B) = 0.3 + 0.4 − 0.12 = 0.58.
P(A|B) = 0.12 / 0.4 = 0.3.
P(A|B) = 0.3 = P(A), so A and B are independent.
3.7 — Biased coin twice
Tree: H (0.6), T (0.4) on each flip. P(no heads) = 0.4 × 0.4 = 0.16. P(at least one head) = 1 − 0.16 = 0.84.
3.8 — Scatter plot description
The correlation is strong positive (clear upward trend, points close to a line). To draw the line of best fit, sketch a single straight line through the middle of the cloud so that roughly the same number of points sit above and below it (it does not need to pass through any actual data point).