Mathematics • Year 9 • Unit 4 • Lesson 19
Measures of Centre and Probability Rules
Build fluency with mean / median / mode / IQR, and with the lesson's three probability rules: complement, addition, multiplication. One worked example, one faded example, then eight graduated problems.
1. I do — fully worked example
A data summary plus a probability calculation, side by side.
Problem. The data set is: 2, 5, 7, 8, 10, 15, 20.
(a) Find the mean, median, mode, range and IQR.
(b) Use the result of (a) to comment on the skew.
Step 1 — Sort and count.
Already sorted. n = 7 values.
Step 2 — Mean.
Sum = 2+5+7+8+10+15+20 = 67. Mean = 67 ÷ 7 ≈ 9.57.
Step 3 — Median.
Middle value of 7 is the 4th value = 8.
Step 4 — Mode, range.
No value repeats → no mode. Range = 20 − 2 = 18.
Step 5 — IQR.
Lower half (below the median): 2, 5, 7. Q1 = 5.
Upper half (above the median): 10, 15, 20. Q3 = 15.
IQR = Q3 − Q1 = 15 − 5 = 10.
Step 6 — Skew.
Mean (9.57) > median (8), so the data is RIGHT-SKEWED (tail to the right).
Answers: mean ≈ 9.57, median = 8, no mode, range = 18, IQR = 10; data is right-skewed.
2. We do — fill in the missing steps
A two-event probability problem. Fill the blanks. 4 marks
Problem. In a class of 40, 25 students passed Maths, 20 passed Science, and 12 passed both. Find (a) P(passed Maths or Science) and (b) P(passed neither).
Step 1 — Identify the events.
P(Maths) = ____ ÷ 40 = ______
P(Science) = ____ ÷ 40 = ______
P(Maths ∩ Science) = ____ ÷ 40 = ______
Step 2 — Apply the addition rule.
P(M ∪ S) = P(M) + P(S) − P(M ∩ S) = ____ + ____ − ____ = ______
Step 3 — Apply the complement rule for "neither".
P(neither) = 1 − P(M ∪ S) = 1 − ______ = ______
Step 4 — Sanity check. Number who passed at least one = ____ ÷ 40. Number who passed neither = 40 − ____ = ____. Match your fraction? ____
3. You do — independent practice
Foundation = one rule. Standard = combine two. Extension = multi-step or judgement.
Foundation — single calculation
3.1 Find the mean of: 4, 6, 8, 10, 12. 1 mark
3.2 Find the median of: 3, 6, 7, 9, 11, 14. (Six values — average the middle two.) 1 mark
3.3 P(A) = 0.4. Find P(not A). 1 mark
3.4 Events A and B are mutually exclusive. P(A) = 0.3, P(B) = 0.5. Find P(A ∪ B). 1 mark
Standard — combine two ideas
3.5 A box plot has min = 5, Q1 = 10, median = 15, Q3 = 20, max = 30. (a) Find the IQR. (b) Find the range. (c) Comment on the skew (Q3 − median vs median − Q1). 2 marks
3.6 A bag has 3 red, 4 blue and 3 green marbles. Two are drawn without replacement. Find P(both blue). 2 marks
Extension — push your thinking
3.7 A card is drawn from a standard 52-card deck and a fair coin is tossed. (a) Find P(ace AND heads). State whether these events are independent and why. (b) Find P(ace OR heads). 3 marks
3.8 A data set is: 4, 6, 8, 10, 12, 100. (a) Find the mean and median. (b) Which measure of centre is more appropriate, and why? (c) What kind of value is the 100, and what skew does it create? 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (Maths/Science class)
Step 1: P(M) = 25/40 = 0.625; P(S) = 20/40 = 0.5; P(M ∩ S) = 12/40 = 0.3.
Step 2: P(M ∪ S) = 0.625 + 0.5 − 0.3 = 0.825 (or 33/40).
Step 3: P(neither) = 1 − 0.825 = 0.175 (or 7/40).
Step 4: 33 passed at least one; 40 − 33 = 7 passed neither; 7/40 = 0.175. ✓
3.1 — Mean of 4,6,8,10,12
Sum = 40. Mean = 40 ÷ 5 = 8.
3.2 — Median of 3,6,7,9,11,14
Middle two are 7 and 9. Median = (7 + 9) ÷ 2 = 8.
3.3 — Complement
P(not A) = 1 − 0.4 = 0.6.
3.4 — Mutually exclusive A and B
P(A ∪ B) = P(A) + P(B) = 0.3 + 0.5 = 0.8. (No overlap to subtract.)
3.5 — Box plot
(a) IQR = 20 − 10 = 10.
(b) Range = 30 − 5 = 25.
(c) Q3 − median = 20 − 15 = 5; median − Q1 = 15 − 10 = 5. Symmetric box → roughly symmetric distribution (no clear skew from the quartiles).
3.6 — Two blues without replacement
Total marbles = 10. P(first blue) = 4/10. After taking one blue, 3 blue out of 9 remain: P(second blue | first blue) = 3/9.
P(both blue) = (4/10) × (3/9) = 12/90 = 2/15 ≈ 0.133.
3.7 — Card + coin
(a) Events independent because tossing a coin doesn't affect drawing a card.
P(ace) = 4/52 = 1/13. P(heads) = 1/2.
P(ace AND heads) = (1/13) × (1/2) = 1/26 ≈ 0.038.
(b) P(ace OR heads) = P(ace) + P(heads) − P(both) = 1/13 + 1/2 − 1/26 = 2/26 + 13/26 − 1/26 = 14/26 = 7/13 ≈ 0.538.
3.8 — Outlier in data
Sum = 4 + 6 + 8 + 10 + 12 + 100 = 140. (a) Mean = 140 ÷ 6 ≈ 23.3. Median = (8 + 10) ÷ 2 = 9.
(b) Median is more appropriate — the 100 is an outlier that drags the mean up to 23.3, which doesn't represent the centre of the data well.
(c) 100 is an outlier (high value); it creates a right skew (the tail goes to the right / the high end).