Mean, Median, Mode and Range from Frequency Tables

Grade C

Score Frequency

0 42

1 422 543 994 645 286 21

Total: 350

Dr Payne is trialling a new painkiller. 350 patients take the tablets when they have a headacheThey record how long the tablet works for, to the nearest hour.

Score Frequency

0 42

1 422 543 994 645 286 21

Total: 350

The MODE is the score with the highest frequencyDon’t write down the value of the frequency.

Here the mode is 3

Score Frequency

0 42

1 422 543 994 645 286 21

Total: 350

The RANGE is the biggest score – the smallest score

Here the range is 6 – 0 = 6

Score Frequency Cumulative frequency

0 42 42

1 42 842 54 1383 99 2374 64 3015 28 3296 21 350

Total: 350

The MEDIAN is the middle value.For this we need an additional column for the cumulative frequency. This is the ongoing totals of the frequencies and should total the same as the frequency total.

To locate the position of the middle value work out (n+1) where n is the frequency total.2

Here, the position of the middle is:

350+1 = 175.5 2

Look down the cumulative frequency column for the smallest value bigger than 175.5 and read across to the score.

The median here is 3.

Score Frequency Score x frequency

0 42 0

1 42 422 54 1083 99 2974 64 2565 28 1406 21 126

Total: 350 969

To calculate the MEAN we need an additional column. For each row work out the value of the score x the frequencyThe mean is the Total of the score x frequency divided by the Total frequency

Here the mean is 969÷ 350 = 2.768571429Round your answer to a suitable number of decimal places (i.e 2.8)

Ages Frequency

16-18 56

19-21 6422-25 9226-29 12Total: 224

A survey is taken at a night club to ask people their age last birthday.The data was put into categories called class marks (grouped data)

Ages Frequency

16-18 56

19-21 6422-25 9226-29 12Total: 224

The MEDIAN will be a group and is found the same way as before. Look for the group with the highest frequency.

Here the median group is 22-25

Ages Frequency

16-18 56

19-21 6422-25 9226-29 12Total: 224

The RANGE will have two answers – an upper and lower bound.

The upper bound range is the biggest value – the smallest valueThe lower bound range is the smallest value of the biggest group – the biggest value of the smallest group.

Here the upper bound range is 29 – 16 = 13

The lower bound range is 26 – 18 = 8

Ages Frequency Cumulative frequency

16-18 56 56

19-21 64 12022-25 92 21226-29 12 224Total: 224

The MEDIAN is given as a group and is calculated in the same way as before. Work out the cumulative frequencyFind the position of the middle by (n+1)

2Read off the group that is in the middle position

Here the position of the middle is found by 224+1 = 112.5

2

Look down the cumulative frequency column for the smallest value bigger than 112.5 and read across to the group

The median is the 19-21 group

Ages Frequency Midpoint Frequency x Midpoint

16-18 56 17 952

19-21 64 20 128022-25 92 23.5 216226-29 12 27.5 330Total: 224 4724

The MEAN cannot be calculated exactly as the data is grouped. So we can only work out an estimate for the mean. This is done in a similar way as before however, we first must work out the midpoint of each group.To do this add together the smallest and largest value in each group and divide by 2Now multiply together the midpoint values with the corresponding frequency values.

The mean estimate is the Total of the frequency x midpoint divided by the Total of the frequency.

Here the mean is 4724 ÷ 224 =21.08928571 Which to a suitable degree of accuracy is 21.1

Exam Question

Exam Question

Money x Frequency0

32244030

Total = 126

Mean = 126 ÷ 30 = 4.2

Total = 30

4.20

Exam Question

Exam Question

Midpoint65758595

105

90 < t < 100

Midpoint x frequency780

1650195522801995

Total = 8660

8660 ÷ 100 = 86.6

Total = 100

86.6

Exam Question

Cumulative frequency20374981

106

(106+1) = 53.5 2

30 < t ≤ 40

Midpoint5

15253545

Midpoint x Frequency100255300

11201125

Total = 2900Total = 106

2900 ÷ 106 = 27.3584905727.4