Statistics Practice Questions
Practice questions covering all aspects of statistics from the Cambridge IGCSE Mathematics 0580 syllabus.
Questions
The grouped frequency table shows test scores. Score: 0-20 (f=3), 20-40 (f=8), 40-60 (f=15), 60-80 (f=10), 80-100 (f=4). Estimate the mean score.
Draw a cumulative frequency curve for the data: 0<x≤10 (f=5), 10<x≤20 (f=12), 20<x≤30 (f=18), 30<x≤40 (f=10), 40<x≤50 (f=5). Find the median and interquartile range.
A histogram shows the following: class 0-10 has frequency density 1.2, class 10-25 has frequency density 2.4, class 25-30 has frequency density 3.6. Find the frequency for each class.
The table shows scores and their frequencies: 1(f=2), 2(f=5), 3(f=8), 4(f=3), 5(f=2). Find the variance.
The table shows the heights (cm) of 60 students. Heights 150–160: 8 students; 160–170: 20 students; 170–180: 24 students; 180–190: 8 students. Estimate the mean height.
Draw a cumulative frequency table for the following data: 0–10: 5, 10–20: 12, 20–30: 18, 30–40: 10, 40–50: 5. Find the median from the cumulative frequency curve.
A frequency table shows: length 10–14 cm (frequency 8), 14–16 cm (frequency 20), 16–20 cm (frequency 12). Draw a histogram for this data.
A data set has values 4, 7, 3, 9, 5, 6, 8, 2. Find the interquartile range.
The box-and-whisker plot for a data set shows: minimum 10, Q1 = 20, median = 30, Q3 = 45, maximum = 60. Find (a) the interquartile range, (b) the range.
The mean of 6 numbers is 14. A seventh number is added and the new mean is 15. Find the seventh number.
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Question 1 Solution
Step 1: Midpoints: 10, 30, 50, 70, 90
Step 2: Sum of fx = 10(3)+30(8)+50(15)+70(10)+90(4) = 30+240+750+700+360 = 2080
Step 3: Total frequency = 40
Step 4: Mean = 2080/40 = 52
Answer: 52
Question 2 Solution
Step 1: Cumulative frequencies: 5, 17, 35, 45, 50
Step 2: Plot at upper boundaries: (10,5), (20,17), (30,35), (40,45), (50,50)
Step 3: Median at n/2 = 25th value ≈ 25
Step 4: Q1 at n/4 = 12.5th value ≈ 16
Step 5: Q3 at 3n/4 = 37.5th value ≈ 33
Step 6: IQR = 33 − 16 = 17
Answer: Median ≈ 25, IQR ≈ 17
Question 3 Solution
Step 1: Frequency = frequency density × class width
Step 2: 0-10: 1.2 × 10 = 12
Step 3: 10-25: 2.4 × 15 = 36
Step 4: 25-30: 3.6 × 5 = 18
Answer: 12, 36, 18
Question 4 Solution
Step 1: Mean = (2+10+24+12+10)/20 = 58/20 = 2.9
Step 2: Sum of (x−mean)²×f = 2(1−2.9)² + 5(2−2.9)² + 8(3−2.9)² + 3(4−2.9)² + 2(5−2.9)²
Step 3: = 2(3.61) + 5(0.81) + 8(0.01) + 3(1.21) + 2(4.41)
Step 4: = 7.22 + 4.05 + 0.08 + 3.63 + 8.82 = 23.8
Step 5: Variance = 23.8/20 = 1.19
Answer: 1.19
Question 5 Solution
Step 1: Midpoints: 155, 165, 175, 185
Step 2: Σfx = 8×155 + 20×165 + 24×175 + 8×185 = 1240 + 3300 + 4200 + 1480 = 10220
Step 3: Σf = 60
Step 4: Mean = 10220 ÷ 60 = 170.3 cm (1 d.p.)
Answer: 170.3 cm
Question 6 Solution
Step 1: Cumulative frequencies: 5, 17, 35, 45, 50
Step 2: Total = 50, so median is at the 25th value
Step 3: 25th value falls in the 20–30 class
Step 4: Using interpolation or reading from the graph: median ≈ 23
Answer: Approximately 23 (accept 22–24)
Question 7 Solution
Step 1: Calculate frequency density = frequency ÷ class width
Step 2: 10–14: fd = 8 ÷ 4 = 2
Step 3: 14–16: fd = 20 ÷ 2 = 10
Step 4: 16–20: fd = 12 ÷ 4 = 3
Step 5: Draw bars with heights 2, 10, 3 on the frequency density axis
Answer: Frequency densities: 2, 10, 3
Question 8 Solution
Step 1: Ordered: 2, 3, 4, 5, 6, 7, 8, 9
Step 2: Lower quartile (Q1) = median of lower half {2,3,4,5} = 3.5
Step 3: Upper quartile (Q3) = median of upper half {6,7,8,9} = 7.5
Step 4: IQR = Q3 − Q1 = 7.5 − 3.5 = 4
Answer: 4
Question 9 Solution
Step 1: (a) IQR = Q3 − Q1 = 45 − 20 = 25
Step 2: (b) Range = maximum − minimum = 60 − 10 = 50
Answer: (a) 25 (b) 50
Question 10 Solution
Step 1: Sum of original 6 numbers = 6 × 14 = 84
Step 2: Sum of 7 numbers = 7 × 15 = 105
Step 3: Seventh number = 105 − 84 = 21
Answer: 21
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