IB Mathematics SL 4.3 Measures of central tendency AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Alex uses a thermometer to record the maximum daily temperature over ten consecutive days. The results, in degrees Celsius (°C), are: 14, 15, 14, 11, 10, 9, 14, 15, 16, 13.
(i) Find the mode. [1]
(ii) Find the mean. [1]
(iii) Find the standard deviation. [2]
▶️ Answer/Explanation
Markscheme
(i)
Data: 14, 15, 14, 11, 10, 9, 14, 15, 16, 13.
Mode: 14 (appears 3 times). A1
[1 mark]
Data: 14, 15, 14, 11, 10, 9, 14, 15, 16, 13.
Mode: 14 (appears 3 times). A1
[1 mark]
(ii)
Sum: \( 14 + 15 + 14 + 11 + 10 + 9 + 14 + 15 + 16 + 13 = 131 \).
Mean: \( \dfrac{131}{10} = 13.1 \). A1
[1 mark]
Sum: \( 14 + 15 + 14 + 11 + 10 + 9 + 14 + 15 + 16 + 13 = 131 \).
Mean: \( \dfrac{131}{10} = 13.1 \). A1
[1 mark]
(iii)
Sample standard deviation: \( \sqrt{\dfrac{\sum (x_i – \bar{x})^2}{n-1}} \), where \( \bar{x} = 13.1 \), \( n = 10 \). M1
Deviations squared: \( (14 – 13.1)^2 = 0.81 \), \( (15 – 13.1)^2 = 3.61 \), \( (14 – 13.1)^2 = 0.81 \), \( (11 – 13.1)^2 = 4.41 \), \( (10 – 13.1)^2 = 9.61 \), \( (9 – 13.1)^2 = 16.81 \), \( (14 – 13.1)^2 = 0.81 \), \( (15 – 13.1)^2 = 3.61 \), \( (16 – 13.1)^2 = 8.41 \), \( (13 – 13.1)^2 = 0.01 \).
Sum: \( 0.81 + 3.61 + 0.81 + 4.41 + 9.61 + 16.81 + 0.81 + 3.61 + 8.41 + 0.01 = 48.9 \).
Variance: \( \dfrac{48.9}{9} \approx 5.4333 \).
Standard deviation: \( \sqrt{5.4333} \approx 2.21133 \approx 2.21 \). A1
[2 marks]
Sample standard deviation: \( \sqrt{\dfrac{\sum (x_i – \bar{x})^2}{n-1}} \), where \( \bar{x} = 13.1 \), \( n = 10 \). M1
Deviations squared: \( (14 – 13.1)^2 = 0.81 \), \( (15 – 13.1)^2 = 3.61 \), \( (14 – 13.1)^2 = 0.81 \), \( (11 – 13.1)^2 = 4.41 \), \( (10 – 13.1)^2 = 9.61 \), \( (9 – 13.1)^2 = 16.81 \), \( (14 – 13.1)^2 = 0.81 \), \( (15 – 13.1)^2 = 3.61 \), \( (16 – 13.1)^2 = 8.41 \), \( (13 – 13.1)^2 = 0.01 \).
Sum: \( 0.81 + 3.61 + 0.81 + 4.41 + 9.61 + 16.81 + 0.81 + 3.61 + 8.41 + 0.01 = 48.9 \).
Variance: \( \dfrac{48.9}{9} \approx 5.4333 \).
Standard deviation: \( \sqrt{5.4333} \approx 2.21133 \approx 2.21 \). A1
[2 marks]
Total Marks: 4
Question
A group of 120 students sat a history exam. The cumulative frequency graph shows the scores obtained by the students.
Diagram not to scale: A cumulative frequency graph showing the scores obtained by 120 students in a history exam.
(a) Determine the median score. [1]
The students were awarded a grade from 1 to 5, depending on the score obtained in the exam. The number of students receiving each grade is shown in the following table.
| Grade | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Number of students | 6 | 13 | 26 | a | b |
(b) Determine an expression for a in terms of b. [2]
(c)(i) The mean grade for these students is 3.65. Determine the number of students who obtained a grade 5. [3]
(c)(ii) Determine the minimum score required for a grade 5. [2]
▶️ Answer/Explanation
Markscheme
(a)
Median (60th student, since \( \frac{120}{2} = 60 \)): From the graph, cumulative frequency of 60 corresponds to a score of 75.
Median score: 75. A1
[1 mark]
Median (60th student, since \( \frac{120}{2} = 60 \)): From the graph, cumulative frequency of 60 corresponds to a score of 75.
Median score: 75. A1
[1 mark]
(b)
Total students: \( 6 + 13 + 26 + a + b = 120 \). M1
Simplify: \( 45 + a + b = 120 \), so \( a = 120 – 45 – b = 75 – b \).
Expression: \( a = 75 – b \). A1
[2 marks]
Total students: \( 6 + 13 + 26 + a + b = 120 \). M1
Simplify: \( 45 + a + b = 120 \), so \( a = 120 – 45 – b = 75 – b \).
Expression: \( a = 75 – b \). A1
[2 marks]
(c)(i)
Mean grade formula: \( \frac{6 \times 1 + 13 \times 2 + 26 \times 3 + (75 – b) \times 4 + b \times 5}{120} = 3.65 \). M1
Numerator: \( 6 + 26 + 78 + (75 – b) \times 4 + b \times 5 = 6 + 26 + 78 + 300 – 4b + 5b = 410 + b \). A1
Solve: \( \frac{410 + b}{120} = 3.65 \), so \( 410 + b = 3.65 \times 120 = 438 \), \( b = 438 – 410 = 28 \).
Number of students with grade 5: 28. A1
[3 marks]
Mean grade formula: \( \frac{6 \times 1 + 13 \times 2 + 26 \times 3 + (75 – b) \times 4 + b \times 5}{120} = 3.65 \). M1
Numerator: \( 6 + 26 + 78 + (75 – b) \times 4 + b \times 5 = 6 + 26 + 78 + 300 – 4b + 5b = 410 + b \). A1
Solve: \( \frac{410 + b}{120} = 3.65 \), so \( 410 + b = 3.65 \times 120 = 438 \), \( b = 438 – 410 = 28 \).
Number of students with grade 5: 28. A1
[3 marks]
(c)(ii)
Students up to grade 4: \( 120 – 28 = 92 \). M1
From the graph, cumulative frequency of 92 corresponds to a score of 84.
Minimum score for grade 5: 84. A1
[2 marks]
Students up to grade 4: \( 120 – 28 = 92 \). M1
From the graph, cumulative frequency of 92 corresponds to a score of 84.
Minimum score for grade 5: 84. A1
[2 marks]
Total Marks: 8