The class mark of the class \(12.5\)–\(17.5\) is
- (a)\(15\)
- (b)\(5\)
- (c)\(12.5\)
- (d)\(17.5\)
Show answer
Answer: (a) \(15\)
Why: Class mark = (lower limit + upper limit)/2.
\(\dfrac{12.5 + 17.5}{2} = \dfrac{30}{2} = 15\).
Statistics Class 10: MCQ and case study questions
15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
- 15 MCQs (1 mark each)
- 2 case studies (4 marks each)
- No calculator needed
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Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
A grouped distribution has mean \(30\) and mode \(24\). From \(3\,\text{Median} = \text{Mode} + 2\,\text{Mean}\), its median is
- (a)\(27\)
- (b)\(28\)
- (c)\(26\)
- (d)\(29\)
Show answer
Answer: (b) \(28\)
Why: 3 Median = Mode + 2 Mean = 24 + 60.
\(3\,\text{Median} = 24 + 2(30) = 84 \Rightarrow \text{Median} = 28\).
Using the step-deviation method with assumed mean \(a = 25\), class width \(h = 5\), \(\sum f_iu_i = -10\) and \(\sum f_i = 50\), the mean is
- (a)\(26\)
- (b)\(25\)
- (c)\(24\)
- (d)\(23\)
Show answer
Answer: (c) \(24\)
Why: Mean = a + h × (Σfu/Σf) = 25 + 5 × (−1/5).
\(\bar x = 25 + 5 \times \dfrac{-10}{50} = 25 - 1 = 24\).
For the distribution below, the median class is
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 9 | 6 |
- (a)\(10\)–\(20\)
- (b)\(20\)–\(30\)
- (c)\(30\)–\(40\)
- (d)\(0\)–\(10\)
Show answer
Answer: (b) \(20\)–\(30\)
Why: n = 40, n/2 = 20; cumulative frequencies 5, 13, 25, … first reach 20 in 20–30.
\(n = 40\), \(\dfrac n2 = 20\). Cumulative frequencies: \(5, 13, 25, 34, 40\). The first one \(\ge 20\) is \(25\), in the class \(20\)–\(30\).
The mode of the following data is
| Class | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 4 | 6 | 15 | 9 | 4 |
- (a)\(34\)
- (b)\(35\)
- (c)\(30\)
- (d)\(36\)
Show answer
Answer: (d) \(36\)
Why: Modal class 30–40: Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h.
Modal class \(30\)–\(40\): \(l = 30\), \(f_1 = 15\), \(f_0 = 6\), \(f_2 = 9\), \(h = 10\). Mode \(= 30 + \dfrac{15 - 6}{30 - 6 - 9} \times 10 = 30 + \dfrac{9}{15} \times 10 = 36\).
The mean of the data with classes \(0\)–\(10\), \(10\)–\(20\), \(20\)–\(30\) and frequencies \(4\), \(10\), \(6\) is
- (a)\(15\)
- (b)\(16\)
- (c)\(17\)
- (d)\(16.5\)
Show answer
Answer: (b) \(16\)
Why: Use class marks 5, 15, 25: Σfx = 20 + 150 + 150 = 320 over Σf = 20.
\(\bar x = \dfrac{4(5) + 10(15) + 6(25)}{20} = \dfrac{320}{20} = 16\).
In the formula \(\text{Median} = l + \left(\dfrac{\frac n2 - cf}{f}\right) \times h\), the symbol \(cf\) stands for the cumulative frequency
- (a)of the median class
- (b)of the class after the median class
- (c)of the class before the median class
- (d)of the whole distribution
Show answer
Answer: (c) of the class before the median class
Why: cf counts all observations below the median class.
\(cf\) is the cumulative frequency of the class preceding the median class: the number of observations below \(l\).
The median of the distribution below is
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 9 | 6 |
- (a)\(\dfrac{155}{6}\)
- (b)\(25\)
- (c)\(27\)
- (d)\(\dfrac{145}{6}\)
Show answer
Answer: (a) \(\dfrac{155}{6}\)
Why: Median class 20–30 with l = 20, cf = 13, f = 12, h = 10.
\(\text{Median} = 20 + \dfrac{20 - 13}{12} \times 10 = 20 + \dfrac{70}{12} = \dfrac{155}{6}\) (about \(25.8\)).
If every observation of a data set is multiplied by \(3\), then the mean
- (a)is unchanged
- (b)increases by \(3\)
- (c)is multiplied by \(3\)
- (d)is divided by \(3\)
Show answer
Answer: (c) is multiplied by \(3\)
Why: Σ(3x)/n = 3 × Σx/n.
\(\dfrac{\sum f_i(3x_i)}{\sum f_i} = 3 \times \dfrac{\sum f_ix_i}{\sum f_i}\): the mean is multiplied by \(3\).
In the assumed-mean method, \(a = 50\), \(\sum f_id_i = -40\) and \(\sum f_i = 20\). The mean is
- (a)\(52\)
- (b)\(48\)
- (c)\(50\)
- (d)\(46\)
Show answer
Answer: (b) \(48\)
Why: Mean = a + Σfd/Σf = 50 − 2.
\(\bar x = 50 + \dfrac{-40}{20} = 48\).
For a distribution, the mean and the median are both \(30\). By the empirical relation, the mode is
- (a)\(30\)
- (b)\(90\)
- (c)\(0\)
- (d)\(60\)
Show answer
Answer: (a) \(30\)
Why: Mode = 3 Median − 2 Mean = 90 − 60.
\(\text{Mode} = 3(30) - 2(30) = 30\).
For the distribution below, the sum of the lower limits of the median class and the modal class is
| Class | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 4 | 6 | 15 | 9 | 4 |
- (a)\(60\)
- (b)\(70\)
- (c)\(50\)
- (d)\(80\)
Show answer
Answer: (a) \(60\)
Why: n = 38, n/2 = 19 falls in 30–40 (cf 25), and 30–40 also has the highest frequency.
Cumulative frequencies: \(4, 10, 25, 34, 38\); \(\dfrac n2 = 19\) lies in \(30\)–\(40\) (median class). Highest frequency \(15\) is also in \(30\)–\(40\) (modal class). Sum \(= 30 + 30 = 60\).
The mean of the distribution below is \(20\). The value of \(p\) is
| Class | 0–10 | 10–20 | 20–30 | 30–40 |
|---|---|---|---|---|
| Frequency | 6 | \(p\) | 10 | 4 |
- (a)\(5\)
- (b)\(6\)
- (c)\(4\)
- (d)\(8\)
Show answer
Answer: (c) \(4\)
Why: Σfx = 420 + 15p must equal 20(20 + p).
Class marks \(5, 15, 25, 35\): \(\sum f_ix_i = 30 + 15p + 250 + 140 = 420 + 15p\), \(\sum f_i = 20 + p\). \(420 + 15p = 400 + 20p \Rightarrow p = 4\).
A cumulative frequency table is used mainly to find the
- (a)mean
- (b)mode
- (c)range
- (d)median
Show answer
Answer: (d) median
Why: The median formula needs the cumulative frequency before the median class.
To locate the median class (where the cumulative frequency first reaches \(\tfrac n2\)) and apply the median formula, we need cumulative frequencies.
For a grouped distribution the modal class is \(60\)–\(80\), with frequency \(20\); the classes just before and after it have frequencies \(10\) and \(14\). The mode is
- (a)\(70\)
- (b)\(75\)
- (c)\(65\)
- (d)\(72.5\)
Show answer
Answer: (d) \(72.5\)
Why: Mode = 60 + (20 − 10)/(40 − 10 − 14) × 20.
\(\text{Mode} = 60 + \dfrac{20 - 10}{2(20) - 10 - 14} \times 20 = 60 + \dfrac{10}{16} \times 20 = 60 + 12.5 = 72.5\).
Case study questions
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
Case study 1: Electricity use in a housing society (4 marks)
The residents' association of a housing society recorded the electricity used last month by \(40\) homes:
| Units used | 100–150 | 150–200 | 200–250 | 250–300 | 300–350 |
|---|---|---|---|---|---|
| Number of homes | 4 | 9 | 14 | 9 | 4 |
(i) Write the modal class. [1 mark]
Show answer
Answer: \(200\)–\(250\)
The highest frequency, \(14\), is in \(200\)–\(250\). A1
(ii) Write the median class. [1 mark]
Show answer
Answer: \(200\)–\(250\)
Cumulative frequencies \(4, 13, 27, 35, 40\); \(\tfrac n2 = 20\) first reached in \(200\)–\(250\). A1
(iii) Find the mode of the data. [2 marks]
Show answer
Answer: \(225\) units
\(\text{Mode} = 200 + \dfrac{14 - 9}{28 - 9 - 9} \times 50\) M1 \(= 200 + 25 = 225\) units. A1
OR Find the mean electricity use, using the step-deviation method. [2 marks]
Show answer
Answer: \(225\) units
\(a = 225\), \(h = 50\), \(u_i = -2, -1, 0, 1, 2\): \(\sum f_iu_i = -8 - 9 + 0 + 9 + 8 = 0\) M1; mean \(= 225 + 50 \times \dfrac{0}{40} = 225\) units. A1
Case study 2: Unit-test marks (4 marks)
A mathematics teacher records the marks (out of \(50\)) scored by the \(40\) students of her class in a unit test:
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Number of students | 3 | 7 | 12 | 10 | 8 |
(i) How many students scored less than \(30\) marks? [1 mark]
Show answer
Answer: \(22\)
\(3 + 7 + 12 = 22\). A1
(ii) Write the median class. [1 mark]
Show answer
Answer: \(20\)–\(30\)
\(\tfrac n2 = 20\); cumulative frequencies \(3, 10, 22, \ldots\) first reach \(20\) in \(20\)–\(30\). A1
(iii) Find the median marks. [2 marks]
Show answer
Answer: \(\dfrac{85}{3}\) (about \(28.3\))
\(\text{Median} = 20 + \dfrac{20 - 10}{12} \times 10\) M1 \(= 20 + \dfrac{25}{3} = \dfrac{85}{3}\). A1
OR Find the mean marks. [2 marks]
Show answer
Answer: \(28.25\)
Class marks \(5, 15, 25, 35, 45\): \(\sum f_ix_i = 15 + 105 + 300 + 350 + 360 = 1130\) M1; mean \(= \dfrac{1130}{40} = 28.25\). A1
Next steps for Statistics
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