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).