Find the mean, median and mode number of pets per household.
Worked Solution & Answer
(Mean = total of values ÷ number of values) B1
Answer: Mean: 1.43, Median: 1, Mode: 1
Averages from tables — Notes & Worked Examples Show / Hide
Explanation:
For an ungrouped frequency table, find the mean by multiplying each value x by its frequency f, then divide the total of the fx column by the total frequency Σf (never divide by the number of rows!).
For a grouped frequency table (with intervals like 0 < t ≤ 10), you do not know the exact values, so find the midpoint of each interval first and calculate Estimated Mean = Σ(f × midpoint)Σf. The modal class is the interval with the highest frequency.
Example 1: Find the mean number of goals per match for 20 matches.
| Goals (x) | 0 | 1 | 2 | 3 | Total |
|---|---|---|---|---|---|
| Freq (f) | 4 | 7 | 6 | 3 | 20 |
| f × x | 0 | 7 | 12 | 9 | 28 |
Mean = ΣfxΣf = 2820 = 1.4 goals
Example 2: Estimate the mean time from the grouped table.
| Time t | 0<t≤10 | 10<t≤20 | 20<t≤30 | Total |
|---|---|---|---|---|
| Freq (f) | 6 | 10 | 4 | 20 |
| Midpoint | 5 | 15 | 25 | — |
| f × Mid | 30 | 150 | 100 | 280 |
Estimated Mean = 28020 = 14 mins (Modal class: 10 < t ≤ 20)
Find the mean, median and mode number of goals scored.
Worked Solution & Answer
(Mean = total of values ÷ number of values) B1
Answer: Mean: 2.87, Median: 3, Mode: 3
The table shows information about the heights of 80 teenagers.
Work out an estimate for the mean height of the teenagers.
cm
Worked Solution & Answer
(155×8 + 165×14 + 175×24 + 185×30 + 195×4) ÷ 80 = 14 080 ÷ 80 = 176 cm M2 A1
Answer: 176 cm
The table shows the amount of snow, in cm, that fell each day for 30 days.
Work out an estimate for the mean amount of snow per day.
cm
Worked Solution & Answer
Σfx = 5×8 + 15×10 + 25×7 + 35×2 + 45×3 = 40 + 150 + 175 + 70 + 135 = 570 M2
Mean = 570 ÷ 30 = 19 cm A1
Mean = 570 ÷ 30 = 19 cm A1
Answer: 19 cm
The table shows information about the lengths, in minutes, of 50 telephone calls.
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the total length, in minutes, of these telephone calls. [3 marks]
minutes
Worked Solution & Answer
(a)
Highest frequency is 18 for 25 < m ≤ 30 B1
Answer (a): 25 < m ≤ 30
(b)
Σ(f × midpoint) = 8×2.5 + 2×7.5 + 6×12.5 + 4×17.5 + 12×22.5 + 18×27.5 = 20 + 15 + 75 + 70 + 270 + 495 = 945 minutes M2 A1
Answer (b): 945 minutes
Seija works at a weather station.
The table gives information about the temperature, T °C, at midday for each of 50 cities in the UK on Tuesday.
The table gives information about the temperature, T °C, at midday for each of 50 cities in the UK on Tuesday.
(a) Calculate an estimate for the mean temperature. [3 marks]
°C
(b) Is Seija correct?
Give a reason for your answer. [1 mark]
Give a reason for your answer. [1 mark]
Worked Solution & Answer
(a)
Σfx = 2(12.5) + 8(17.5) + 13(22.5) + 21(27.5) + 6(32.5) = 25 + 140 + 292.5 + 577.5 + 195 = 1230 M2
Mean = 1230 ÷ 50 = 24.6 °C A1
Mean = 1230 ÷ 50 = 24.6 °C A1
Answer (a): 24.6 °C
(b)
Cumulative frequencies: 2, 10, 23, 44, 50 ⇒ 25.5th value is in 25 < T ≤ 30 C1
Answer (b): No, she did not take into account the frequencies (the 25th and 26th values lie in the 25 < T ≤ 30 group)
The table shows information about the weights of 120 oranges.
(a) Find the class interval that contains the median. [1 mark]
(b) Calculate an estimate for the mean weight of the 120 oranges.
Give your answer correct to 3 significant figures. [3 marks]
Give your answer correct to 3 significant figures. [3 marks]
grams
Worked Solution & Answer
(a)
60.5th value lies in 100 < w ⩽ 150 (cum freq 34, 63) B1
Answer (a): 100 < w ≤ 150
(b)
Σfx = 75×34 + 125×29 + 175×27 + 225×19 + 275×11 = 18200 M2
Mean = 18200 ÷ 120 = 151.66... = 152 g (3 sf) A1
Mean = 18200 ÷ 120 = 151.66... = 152 g (3 sf) A1
Answer (b): 152 grams
The table shows some information about the hourly rates of pay of 60 workers.
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean hourly rate of pay of the 60 workers. [4 marks]
dollars
Worked Solution & Answer
(a)
Highest frequency is 18 in 10 < p ≤ 15 B1
Answer (a): 10 < p ≤ 15
(b)
(12.5×18 + 17.5×16 + 22.5×14 + 27.5×8 + 32.5×4) ÷ 60 = 1170 ÷ 60 = 19.5 M3 A1
Answer (b): 19.5 dollars
The table gives information about the speeds, in kilometres per hour, of 80 motorbikes as each pass under a bridge.
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean speed of the motorbikes as they pass under the bridge.
Give your answer correct to 3 significant figures. [4 marks]
Give your answer correct to 3 significant figures. [4 marks]
kilometres per hour
Worked Solution & Answer
(a)
Highest frequency is 23 in 70 < s ≤ 80 B1
Answer (a): 70 < s ≤ 80
(b)
(10×45 + 16×55 + 19×65 + 23×75 + 12×85) ÷ 80 = (450 + 880 + 1235 + 1725 + 1020) ÷ 80 = 5310 ÷ 80 = 66.375 ≈ 66.4 M3 A1
Answer (b): 66.4 kilometres per hour
The table gives information about the distances 100 adults travel to work.
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean distance. [4 marks]
km
Worked Solution & Answer
(a)
Highest frequency is 40 in 5 < d ≤ 10 B1
Answer (a): 5 < d ≤ 10
(b)
(2.5 × 26 + 7.5 × 40 + 12.5 × 16 + 17.5 × 10 + 22.5 × 8) ÷ 100 = 920 ÷ 100 = 9.2 km M3 A1
Answer (b): 9.2 km