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Measures of Central Tendency: Mean, Median, Mode and Range with solved examples
Typology: Exercises
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Unit 1 Measures of Central
Tendency
Section
1.1 Mean, Median, Mode and Range
1.2 Finding the Mean from Tables and Tally Charts
1.3 Calculations with the Mean
1.4 Mean, Median and Mode for Grouped Data
In this unit, you will find out how to calculate statistical quantities which summarise the important characteristics of data.
The mean , median and mode are three different ways of describing the average.
Find (a) the mean (b) the median (c) the mode (d) the range of this set of data. 5, 6, 2, 4, 7, 8, 3, 5, 6, 6
(a) The mean is 5 6 2 4 7 8 3 5 6 6 10
=
52 10 = 5 2.
(b) To find the median, place all the numbers in order. 2, 3, 4, 5, 5, 6, 6, 6, 7, 8
As there are two middle numbers in this example, 5 and 6,
median =
5 + 6 2
=
11 2 = 5 5.
(d) Calculate the probability that a household chosen at random from those in the survey would have (i) exactly 4 children (ii) more than 4 children.
(a) Mode = 4 (as its frequency is highest)
=
34 10 = 3.
(ii) Median: first put the data in numerical order.
1, 1, 2, 3, 3, 4, 4, 4, 5, 7
median = =
3 + 4 2
3 5.
(c) The mode gives the value that occurs most frequently.
3 10
0 3. (4 occurs 3 times)
2 10
0 2. (5 and 7)
Exercises
8, 6, 7, 6, 5, 4 12 , 7 12 , 6 12 , 8 12 , 10
7, 5, 5 12 8, 9, 7, 5, 6, 8 12 6
For this data, find (a) the mean (b) the median (c) the mode (d) the range.
1 2 4 43
£ 12 , £ 15 , £ 15 , £ 14 , £ 13 , £ 14 , £ 13 , £ 13 (a) Find (i) the mean (ii) the median (iii) the mode.
(b) Which average would you use if you wanted to claim that the staff were: (i) well paid (ii) badly paid?
(c) What is the range?
Worker Mon Tue Wed Thu Fri Rachel 20 21 22 20 21 John 30 15 12 36 28
(a) Find the mean and range for Rachel and John. (b) Who is more consistent? (c) Who makes the most parts in a week?
Company A 20 5 20 20 20 6 20 20 20 8 Company B 17 18 15 16 18 18 17 15 17 18
(a) Find the mean, median and mode for each company's seeds. (b) Which company does the mode suggest is best? (c) Which company does the mean suggest is best? (d) Find the range for each company's seeds.
The weights, in grams, of seven sweet potatoes are 260, 225, 205, 240, 232, 205, 214 What is the median weight?
Here are the number of goals scored by a school football team in their matches this term. 3, 2, 0, 1, 2, 0, 3, 4, 3, 2 (a) Work out the mean number of goals. (b) Work out the range of the number of goals scored.
(a) The weights, in kilograms, of the 8 members of Hereward Hous e tug-of-war team at a school sports event are 75, 73, 77, 76, 84, 76, 77, 78. Calculate the mean weight of the team.
(b) The 8 members of Nelson House tug of war team have a mean weight of 64 kilograms. Which team do you think will win a tug-of-war between Hereward House and Nelson House? Give a reason for your answer.
The range of the class sizes for Year 9 is 3. (c) What does this tell you about the class sizes in Year 9 compared with those in Year 8?
Kelly 28 24 21 27 24 26 Rory 33 19 16 32 34 18
Kelly had a mean score of 25 with a range of 7. (a) Calculate Rory's mean score and range. (b) Which student would you choose to represent the school? Explain the reason for your choice, referring to the mean scores and ranges.
No. of Goals Frequency
0 8 1 10 2 12 3 3 4 5 5 2
No. of Goals Frequency No. of Goals ⋅ Frequency
0 8 0 ⋅ 8 = 0 1 10 1 ⋅ 10 = 10 2 12 2 ⋅ 12 = 24 3 3 3 ⋅ 3 = 9 4 5 4 ⋅ 5 = 20 5 2 5 ⋅ 2 = 10
TOTALS 40 73
(Total matches) (Total goals)
The rules say that the highest mark and the lowest mark are to be deleted. 5.3, 5.7, 5.9, 5.4, 4.5, 5.7, 5.8, 5.
(a) (i) Find the mean of the six remaining marks. (ii) Find the range of the six remaining marks.
(b) Do you think it is better to count all eight marks, or to count only the six remaining marks? Use the means and the ranges to explain your answer.
(c) The eight marks obtained by Diana in the same competition have a mean of 5.2 and a range of 0.6. Explain why none of her marks could be as high as 5.9.
1.2 Finding the Mean from Tables and
Tally Charts
Often data are collected into tables or tally charts. This section considers how to find the mean in such cases.
A football team keep records of the number of goals it scores per match during a season. The list is shown opposite. Find the mean number of goals per match.
The previous table can be used, with a third column added.
The mean can now be calculated.
Mean =
73 40
= 1 825.
The first step is to draw out and complete a tally chart. The final column shown below can then be added and completed.
Number of Accidents Tally Frequency No. of Accidents ⋅ Frequency
0 |||| 4 0 ⋅ 4 = 0 1 |||| |||| 10 1 ⋅ 10 = 10 2 |||| |||| |||| |||| || 22 2 ⋅ 22 = 44 3 |||| |||| |||| |||| ||| 23 3 ⋅ 23 = 69 4 |||| |||| |||| | 16 4 ⋅ 16 = 64 5 |||| |||| |||| || 17 5 ⋅ 17 = 85 6 |||| | 6 6 ⋅ 6 = 36 7 | 1 7 ⋅ 1 = 7 8 | 1 8 ⋅ 1 = 8 TOTALS 100 323
Mean number of accidents per day =
323 100
= 3 23.
The marks obtained by 25 pupils on a test are shown below.
3 4 5 6 5 5 1 2 3 3 4 7 5 1 5 2 5 6 5 4 6 4 5 4 3
(a) Copy and complete the frequency table below to present the information given above.
Marks Frequency
1 2 2 2 3 4 4 - 5 - 6 3 7 1
(b) Using the frequency distribution, state
(i) the modal mark
(ii) the median mark
(iii) the range.
(c) On graph paper, draw a histogram to illustrate the frequency distribution. Use axes as labelled below.
(d) A pupil is chosen at random from the group of pupils. What is the probability that the pupil's mark is greater than 5?
(a)
(Check: total frequency = 2 + 2 + 4 + 5 + 8 + 3 + 1 = 25 )
Frequency
Number of marks
1
3
2
4
6
5
7
8
(^01 2 3 4 5 6 7 )
Marks Frequency
1 2 2 2 3 4 4 5 5 8 6 3 7 1
No. of TV Sets Frequency 0 2 1 30 2 52 3 8 4 5 5 3
Calculate the mean number of TV sets per household.
Number of calls per day 0 1 2 3 4 5 6 7 8
Frequency 3 4 7 8 12 10 14 3 1
Calculate the mean number of calls per day.
No. of Runs Frequency 0 3 1 2 2 1 3 6 4 5 5 4 6 2 7 1 8 1
Calculate the mean number of runs per over.
6 3 2 1 3 2 1 3 0 1 0 3 2 1 1 4 0 1 2 0 1 1 2 2 2 4 3 1 1 1 2 3 3 1 2 2 2 1 7 1
(a) Calculate the mean number of plants. (b) How many times was the number of plants seen greater than the mean?
0 1 2 4 1 0 2 1 1 0 1 2 1 3 1 0 0 0 0 5 2 1 3 2 0 1 0 1 2 1 1 0 0 3 0 1 2 1 0 0
Construct a table and calculate the mean number of planes which were late each day.
Calculate the mean number of repeats per packet.
Goals per Match Frequency
0 4 1 6 2 3 8 4 2 5 1
(a) In how many matches did they score 2 goals? (b) Calculate the mean number of goals per match.
Number of repeats
2
4
6
8
10
12
0 1 2 3 4 5 6
q^
y
(a) How many children wear a size 7 shoe? (b) How many children wear a shoe size smaller than size 7? (c) Which shoe size is the modal size? (d) What is the median shoe size? (e) What is the probability that a child selected at random wears: (i) a shoe size of 5? (ii) a shoe size larger than 6? (f) Which of these two averages, the mode and the median, would be of greater interest to the owner of a shoe shop who wishes to stock up on children's shoes? Give a reason for your answer.
1.3 Calculations with the Mean
This section considers calculations concerned with the mean, which is usually taken to be the most important measure of the average of a set of data.
The mean of a sample of 6 numbers is 3.2. An extra value of 3.9 is included in the sample. What is the new mean?
Total of original numbers = 6 ⋅3 2. = 19 2.
New total = 19 2. +3 9. = 23 1.
New mean =
23 1 7
.
= 3 3.
The mean number of a set of 5 numbers is 12.7. What extra number must be added to bring the mean up to 13.1?
Total of the original numbers = 5 ⋅12 7. = 63 5.
Total of the new numbers = 6 ⋅13 1. = 78 6.
Difference = 78 6. −63 5. = 15 1. So the extra number is 15.1.
Rohan's mean score in three cricket matches was 55 runs. (i) How many runs did he score altogether?
After four matches his mean score was 61 runs.
(ii) How many runs did he score in the fourth match?
(i) Mean = 55 = 3
total scored
so total scored = 3 ⋅ 55 = 165
(ii) Total scored = 4 ⋅ 61 = 244
Fourth match score = 244 − 165 = 79
Exercises
Also note that when we speak of someone by age, say 8, then the person could be any age from 8 years 0 days up to 8 years 364 days (365 in a leap year!). You will see how this is tackled in the following example.
The ages of students in a small primary school were recorded in the table below.
Age 5 – 6 7 – 8 9 – 10
Frequency 29 40 38
(a) Estimate the mean. (b) Estimate the median. (c) Find the modal class.
(a) To estimate the mean, we must use the mid-point of each interval; so, for example for '5 – 6', which really means 5 ≤ age< 7 the mid-point is taken as 6.
Class Interval Mid-point Frequency Mid-point ⋅ Frequency
5 – 6 6 29 6 ⋅ 29 = 174 7 – 8 8 40 8 ⋅ 40 = 320 9 – 10 10 38 10 ⋅ 38 = 380
Totals 107 874
Mean =
874 107 = 8 2. (to 1 decimal place)
(b) The median is given by the 54th value, which we have to estimate. There are 29 values in the first interval, so we need to estimate the 25th value in the second interval. As there are 40 values in the second interval, the median is estimated as being 25 40
of the way along the second interval. This has width 9 − 7 = 2 years, so the median is estimated by 25 40
⋅ 2 = 1 25.
from the start of the interval. Therefore the median is estimated as
7 + 1 25. =8 25. years
(c) The modal class is the 7 – 8 age group.
Worked Example 1 uses what are called continuous data , since height can be of any value. (Other examples of continuous data are weight, temperature, area, volume and time.)
The next example uses discrete data , that is, data which can take only a particular value, such as the integers 1, 2, 3, 4,... in this case.
The calculations for mean and mode are not affected but estimation of the median requires replacing the discrete grouped data with an approximate continuous interval.
The number of days that students were missing from school due to sickness in one year was recorded.
Number of days off sick 1 – 5 6 – 10 11 – 15 16 – 20 21 – 25
Frequency 12 11 10 4 3
(a) Estimate the mean. (b) Find the median class. (c) Find the modal class.
(a) The estimate is made by assuming that all the values in a class interval are equal to the midpoint of the class interval.
Class Interval Mid-point Frequency Mid-point ⋅ Frequency
1–5 3 12 3 ⋅ 12 = 36 6–10 8 11 8 ⋅ 11 = 88 11–15 13 10 13 ⋅ 10 = 130 16–20 18 4 18 ⋅ 4 = 72 21–25 23 3 23 ⋅ 3 = 69
Totals 40 395
Mean =
395 40 = 9 875. days
(b) As there are 40 students, we need to consider the mean of the 20th and 21st values. These both lie in the 6–10 class interval, which is really the 5.5–10.5 class interval, so this interval contains the median. [ You could also estimate the median as follows. As there are 12 values in the first class interval, the median is found by considering the 8th and 9th values of the second interval.
As there are 11 values in the second interval, the median is estimated as being 8 5 11. of the way along the second interval. But the length of the second interval is 10 5. − 5 5. = 5 , so the median is estimated by 8 5 11
5 3 86
. ⋅ ≈.
from the start of this interval. Therefore the median is estimated as 5 5. + 3 86. = 9 36. ] (c) The modal class is 1–5, as this class contains the most entries.
