Estimated Mean from Grouped Data
Estimated mean from grouped data is a method for approximating the mean of continuous or grouped data where individual values are unknown, only the class interval and frequency for each group. You use the midpoint of each class interval to represent every value in that group, then calculate (sum of frequency x midpoint) divided by total frequency. It is called an 'estimate' because the true individual values inside each group are not known exactly.
Before you start
Make sure you're comfortable with these topics first:
Method
- Find the midpoint of each class interval by adding the lower and upper bounds and dividing by 2.
- Multiply each midpoint by its frequency to get frequency x midpoint for every row.
- Add up the frequency x midpoint column to get the total.
- Add up the frequency column to get the total frequency.
- Divide the total of frequency x midpoint by the total frequency to get the estimated mean.
- Remember to call it an 'estimated' mean in your answer, since grouping loses the exact individual values.
Worked example
The table shows the time, in minutes, spent on homework by 40 students: 0-20 (freq 8), 20-40 (freq 14), 40-60 (freq 12), 60-80 (freq 6). Estimate the mean time spent on homework.
- Midpoints: 0-20 -> 10, 20-40 -> 30, 40-60 -> 50, 60-80 -> 70.
- Frequency x midpoint: 8x10=80, 14x30=420, 12x50=600, 6x70=420.
- Sum of frequency x midpoint = 80+420+600+420 = 1520.
- Total frequency = 8+14+12+6 = 40.
- Estimated mean = 1520/40 = 38 minutes.
- Final answer: estimated mean = 38 minutes.
Practice questions
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Q1Find the midpoint of the class interval 10-20.Show answer
Answer: 15 ((10+20)/2 = 15)
Q2Find the midpoint of the class interval 30-50.Show answer
Answer: 40 ((30+50)/2 = 40)
Q3A class interval 0-10 has frequency 6. Find frequency x midpoint.Show answer
Answer: 30 (midpoint = 5, 6 x 5 = 30)
Q4A table has class intervals with frequency x midpoint totals of 45, 120, 90 and frequencies 5, 8, 6. Find the estimated mean.Show answer
Answer: 13.42 (2 dp) (sum fx = 45+120+90 = 255, total freq = 5+8+6 = 19, 255/19 = 13.42)
Q5The table shows weights, in kg, of 20 parcels: 0-5 (freq 4), 5-10 (freq 10), 10-15 (freq 6). Estimate the mean weight.Show answer
Answer: 8 kg (midpoints 2.5, 7.5, 12.5; fx = 10+75+75 = 160; total freq = 20; 160/20 = 8)
Q6The table shows ages, in years, of 50 people at a gym: 10-20 (freq 6), 20-30 (freq 18), 30-40 (freq 16), 40-50 (freq 10). Estimate the mean age.Show answer
Answer: 31 years (midpoints 15, 25, 35, 45; fx = 90+450+560+450 = 1550; total freq = 50; 1550/50 = 31)
Exam-style questions
Written in the style of a GCSE Statistics exam paper, with a full mark scheme.
The table shows the distance, in km, travelled to work by 30 employees: 0-5 (freq 9), 5-10 (freq 12), 10-15 (freq 9). Estimate the mean distance travelled.
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The table shows the mass, in grams, of 60 apples: 100-150 (freq 10), 150-200 (freq 22), 200-250 (freq 20), 250-300 (freq 8). (a) Find the midpoint of each class interval. (b) Estimate the mean mass of an apple.
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The table shows the number of hours of sunshine recorded on 20 days: 0-2 (freq 4), 2-4 (freq x), 4-6 (freq 7), where x is unknown and the total frequency is 20. (a) Find x. (b) Estimate the mean number of hours of sunshine, to 1 decimal place.
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See real GCSE Statistics past-paper questions, with official mark schemes →
Free printable worksheet
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This topic is chapter 5 of GCSE Statistics Foundation Workbook 2 and chapter 9 of GCSE Statistics Higher Workbook 2, the whole course as one free printable PDF.
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