Cambridge IGCSE Statistics: Mean, median and mode: estimation and choosing an average (0479) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Topic 4, Measures of central tendency (outcomes 4.1 to 4.4), of the Cambridge IGCSE Statistics (0479) syllabus for examination in 2027. Use the study guide for methods and the revision notes for a recap. Other links: the Cambridge IGCSE Statistics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. The number of cars sold by a garage on nine days was 18, 23, 19, 23, 27, 21, 23, 30 and 16. Calculate the mean, and find the median and the mode. [3]

2. The mean of five numbers is 14. Four of the numbers are 11, 16, 9 and 18. Find the fifth number. [3]

3. A class has 15 girls with a mean test mark of 62 and 25 boys with a mean test mark of 54. Calculate the mean mark for all 40 students. [3]

4. The number of people in each of 25 cars passing a checkpoint is shown.

People in car (x)	1	2	3	4	5
Number of cars (f)	2	k	9	7	3

The mean number of people per car is 3.2.

(a) Find the value of k. [3] (b) Find the median number of people per car. [2]

5. The shoe sizes of 30 members of a club are recorded.

Shoe size	5	6	7	8	9
Frequency	4	11	8	5	2

Find the mean, the median and the mode. [5]

6. The number of words in each of 60 sentences from a magazine is summarised below.

Words	1–5	6–10	11–15	16–20	21–25
Frequency	6	15	22	11	6

(a) Write down the modal class. [1] (b) Calculate an estimate of the mean number of words per sentence. [3] (c) Explain why your answer to part (b) is only an estimate. [1]

7. The masses of 80 parcels are recorded. The table shows the number of parcels with mass below each value.

Mass m (kg) below	10	20	30	40	50
Cumulative frequency	4	18	47	71	80

(a) How many parcels have a mass of 30 kg or more but less than 40 kg? [1] (b) Use linear interpolation to estimate the median mass. [3] (c) Describe how you would find the median from a cumulative frequency graph of this data. [2]

8. The times spent by 116 visitors on a website are grouped as shown.

Time t (minutes)	0 ≤ t < 1	1 ≤ t < 3	3 ≤ t < 6	6 ≤ t < 10	10 ≤ t < 20
Frequency	18	30	36	20	12

(a) Which class has the highest frequency? [1] (b) Find the modal class, showing your working. [3] (c) Calculate an estimate of the mean time. [3]

9. Eight houses on one road were sold for the following prices, in thousands of dollars: 118, 121, 124, 124, 127, 130, 133, 790.

(a) Calculate the mean price and find the median price. [2] (b) An estate agent wants to describe the typical sale price on the road. Which average should be used? Give a reason using your answers to part (a). [2] (c) A café owner records the flavour chosen by each customer and wants to decide which flavour to stock most of. Which average should be used, and why? [2]

10. The distances travelled to work by 50 employees are shown.

Distance d (km)	0 ≤ d < 5	5 ≤ d < 10	10 ≤ d < 20	20 or more
Frequency	14	19	12	5

(a) Find the modal class. [2] (b) Estimate the median distance. [3] (c) Explain why the mean cannot be estimated reliably from this table. [1]

11. A helpline records the length of 40 calls, t minutes.

Time t (minutes)	0 ≤ t < 10	10 ≤ t < 20	20 ≤ t < 30	30 ≤ t < 40	40 ≤ t < 50
Frequency	6	p	14	8	2

The estimated mean call length is 22.5 minutes.

(a) Find the value of p. [4] (b) Estimate the median call length. [3] (c) Write down the modal class. [1] (d) One more call is recorded, lasting 95 minutes. Find the new estimated mean and the new estimated median, and state which measure is affected more. [3]

Answers

1. Sum = 18+23+19+23+27+21+23+30+16 = 200, so mean = 200 ÷ 9 = 22.2 [1]. Ordered: 16, 18, 19, 21, 23, 23, 23, 27, 30, so the 5th value gives median = 23 [1]. 23 appears three times, so mode = 23 [1]. [3] Examiner insight: Show the sum before dividing, so a slip in the final digits can still be seen.

2. Total of five = 14 × 5 = 70 [1]. Total of four = 11+16+9+18 = 54 [1]. Fifth number = 70 − 54 = 16 [1]. [3] Examiner insight: “Mean given” questions are about totals. Writing mean × number of values first is the safest way to show the method.

3. Girls’ total = 15 × 62 = 930 [1]. Boys’ total = 25 × 54 = 1350 [1]. Combined mean = (930 + 1350) ÷ 40 = 2280 ÷ 40 = 57 [1]. [3] Examiner insight: Averaging 62 and 54 to get 58 is a common wrong answer, because it ignores the different group sizes.

4. (a) Σf = 21 + k and Σfx = 2 + 2k + 27 + 28 + 15 = 72 + 2k [1]. (72 + 2k) ÷ (21 + k) = 3.2, so 72 + 2k = 67.2 + 3.2k [1]. 4.8 = 1.2k, so k = 4 [1]. [3] (b) n = 25, so the median is the 13th value [1]. Cumulative frequencies are 2, 6, 15, 22, 25, so the 13th value is in the x = 3 column. Median = 3 [1]. [2] Examiner insight: Write Σf and Σfx as expressions in k before forming the equation. Showing each one makes your method easy to follow.

5. Σfx = 20+66+56+40+18 = 200 [1], so mean = 200 ÷ 30 = 6.67 [1]. n = 30, so the median is halfway between the 15th and 16th values [1]. Cumulative frequencies are 4, 15, 23, 28, 30. The 15th value is 6 and the 16th is 7, so median = 6.5 [1]. Highest frequency is 11, so mode = 6 [1]. [5] Examiner insight: The median of a table can fall between two columns. State which two positions you use.

6. (a) The highest frequency is 22, so the modal class is 11–15 [1]. [1] (b) Midpoints are 3, 8, 13, 18 and 23 [1]. Σfx = 18+120+286+198+138 = 760 [1]. Mean ≈ 760 ÷ 60 = 12.7 words [1]. [3] (c) The actual number of words in each sentence is not known, so each sentence is assumed to be at the midpoint of its class [1]. [1] Examiner insight: An explanation of “estimate” needs to mention that values inside a class are unknown or assumed. “It is not accurate” alone is too vague.

7. (a) 71 − 47 = 24 parcels [1]. [1] (b) n/2 = 40, which lies in the class 20 ≤ m < 30, since 18 are below 20 and 47 are below 30 [1]. Median ≈ 20 + (40 − 18) ÷ 29 × 10 [1] = 27.6 kg [1]. [3] (c) Plot cumulative frequency against the upper class boundary and join the points [1]. Read across from 40 (half of 80) on the vertical axis to the curve, then down to read the mass on the horizontal axis[1]. [2] Examiner insight: Mention the upper boundary and half of the total. Reading at the wrong cumulative frequency is a common loss.

8. (a) The class 3 ≤ t < 6 has frequency 36 [1]. [1] (b) The class widths are unequal, so compare frequency density = frequency ÷ width [1]. The densities are 18, 15, 12, 5 and 1.2 [1]. The highest is 18, so the modal class is 0 ≤ t < 1 [1]. [3] (c) Midpoints are 0.5, 2, 4.5, 8 and 15 [1]. Σfx = 9+60+162+160+180 = 571 [1]. Mean ≈ 571 ÷ 116 = 4.92 minutes [1]. [3] Examiner insight: A row of frequency densities is a clear way to justify the modal class. The class with the highest frequency answers a different question.

9. (a) Sum = 1667, so mean = 1667 ÷ 8 = 208.4 thousand dollars (to 1 d.p.) [1]. Ordered, the 4th and 5th values are 124 and 127, so median = 125.5 thousand dollars [1]. [2] (b) Use the median [1]. One sale of 790 pulls the mean to 208.4, which is higher than seven of the eight prices, whereas the median of 125.5 reflects a typical sale [1]. [2] (c) Use the mode [1], because the data is categorical, so the mean and median cannot be calculated, and the mode shows the most popular flavour [1]. [2] Examiner insight: A reason such as “not affected by extreme values” earns more when tied to the figures given, here 790 against the rest.

10. (a) The classes have different widths, so use frequency density. 0–5: 14 ÷ 5 = 2.8; 5–10: 19 ÷ 5 = 3.8; 10–20: 12 ÷ 10 = 1.2 [1]. The modal class is 5 ≤ d < 10 [1]. [2] (b) n/2 = 25; cumulative frequencies are 14, 33, 45, 50 so the median is in 5 ≤ d < 10 [1]. Median ≈ 5 + (25 − 14) ÷ 19 × 5 [1] = 7.89 km [1]. [3] (c) The last class has no upper limit, so it has no midpoint and the total distance cannot be estimated [1]. [1] Examiner insight: For an open-ended class, say the midpoint is unknown, not just that the mean is “not possible”.

11. (a) Σf = 30 + p [1]. Midpoints 5, 15, 25, 35, 45 give Σfx = 30 + 15p + 350 + 280 + 90 = 750 + 15p [1]. (750 + 15p) ÷ (30 + p) = 22.5, so 750 + 15p = 675 + 22.5p [1]. 75 = 7.5p, so p = 10 [1]. [4] (b) n = 40, so n/2 = 20; cumulative frequencies are 6, 16, 30, 38, 40, so the median is in 20 ≤ t < 30 [1]. Median ≈ 20 + (20 − 16) ÷ 14 × 10 [1] = 22.9 minutes [1]. [3] (c) The highest frequency is 14 and all widths are equal, so the modal class is 20 ≤ t < 30 [1]. [1] (d) New mean = (900 + 95) ÷ 41 = 24.3 minutes (up 1.8) [1]. With n = 41, n/2 = 20.5, so new median ≈ 20 + (20.5 − 16) ÷ 14 × 10 = 23.2 minutes (up 0.4) [1]. The mean is affected more, because it uses the size of every value and the median only depends on position [1]. [3] Examiner insight: In a comparison, give both new values, the size of each change, then a reason.

Where marks are usually lost

• Quoting the frequency (such as 22) as the mode instead of the value or class.
• Averaging two group means directly instead of using the totals.
• Using class limits (such as 6 and 10) instead of the midpoint 8, or dividing by the number of classes.
• Taking the median of a grouped table as the middle class instead of interpolating within it.
• Using n + 1 over 2 as the grouped position, or the wrong cumulative frequency before the class.
• Choosing the modal class by frequency when widths are unequal.
• Giving a grouped answer with no word “estimate”.
• Justifying an average with a general statement that does not use the data in the question.

Next steps

• Recap with the revision notes or return to the study guide.
• Open the Cambridge IGCSE Statistics hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

These questions follow Topic 4, Measures of central tendency (outcomes 4.1 to 4.4), of the Cambridge IGCSE Statistics 0479 syllabus for examination in 2027, published by Cambridge University Press & Assessment (Cambridge International Education).