Measures of central tendency | Grade 10

Your class has just had a test. The teacher says, "The average was 7." A friend says, "Half the class scored at least 7.5." A third says, "Most students got 8." Are they contradicting each other? No: each is describing the "centre" of the same data in a different way. This lesson shows the four standard ways, and when each one is the honest choice.

The mean

The mean uses every value, so it is sensitive to extreme ones.

The median and the quartiles

Sort the data from smallest to largest.

The median splits the data into two halves; the quartiles split it into four quarters.

The mode

The mode is the value that occurs most often. A data set can have no clear mode, or two modes. The mode is the only measure that also works for categories (the most popular colour, the most common shoe size).

Data in a frequency table

Example A frequency table

The scores of 20 students are summarised below. Find $\bar{x}$ , $M$ , $Q_1$ , $Q_3$ and the mode.

Score	5	6	7	8	9	10
Frequency	2	3	5	6	3	1

Solution.

$n=2+3+5+6+3+1=20$ . Sum $=10+18+35+48+27+10=148$ , so $\bar{x}=\frac{148}{20}=7.4$ .
Cumulative counts: $2, 5, 10, 16, 19, 20$ . The 10th value is $7$ and the 11th is $8$ , so $M=7.5$ .
Lower half = first 10 values: the 5th is $6$ and the 6th is $7$ , so $Q_1=6.5$ .
Upper half = last 10 values: the 15th and 16th are both $8$ , so $Q_3=8$ .
The highest frequency is $6$ , so the mode is $8$ .

Effect of an outlier

Key terms

English	Tiếng Việt
mean	số trung bình cộng
median	trung vị
quartile	tứ phân vị
mode	mốt
frequency	tần số
frequency table	bảng tần số
outlier	giá trị ngoại lệ
data set	mẫu số liệu

Summary

The mean is the sum divided by the count; with a frequency table, use $\bar{x}=\frac{\sum f_ix_i}{n}$ .
The median is the middle value of the sorted data; $Q_1$ and $Q_3$ are the medians of the lower and upper halves.
The mode is the most frequent value.
Extreme values pull the mean much more than the median.

Practice

Find the mean of $6, 8, 9, 10, 12$ .
Find the median of $3, 9, 4, 7, 1, 6$ .
Find the mode of $2, 3, 3, 5, 5, 5, 7$ .
For $2, 4, 5, 7, 8, 10, 11, 13, 15$ find $M$ , $Q_1$ and $Q_3$ .
The table gives the number of books read by 22 students. Find $\bar{x}$ (to $0.01$ ), $M$ and the mode.

Books 1 2 3 4 5

Students 3 5 8 4 2
Five numbers have mean $8$ . Four of them are $6, 7, 9, 10$ . Find the fifth.
A class has 20 girls with mean score $7.0$ and 10 boys with mean score $8.5$ . Find the mean of the whole class.
For $12, 13, 14, 15, 16, 17, 90$ find the mean and the median. Which describes the data better?
Challenge. Seven numbers have mean $10$ . After removing one number the mean of the other six is $9$ . Which number was removed?
Challenge. For $3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 20$ find $M$ , $Q_1$ and $Q_3$ .

Show solutions

$\frac{6+8+9+10+12}{5}=\frac{45}{5}=9$ .
Sorted: $1, 3, 4, 6, 7, 9$ . The middle values are $4$ and $6$ , so $M=5$ .
The value $5$ occurs three times: mode $=5$ .
$n=9$ : $M=8$ (5th value). Lower half $2, 4, 5, 7$ gives $Q_1=\frac{4+5}{2}=4.5$ . Upper half $10, 11, 13, 15$ gives $Q_3=\frac{11+13}{2}=12$ .
$n=22$ ; sum $=3+10+24+16+10=63$ , so $\bar{x}=\frac{63}{22}\approx 2.86$ . Cumulative counts $3, 8, 16, 20, 22$ : the 11th and 12th values are both $3$ , so $M=3$ . Highest frequency $8$ , so mode $=3$ .
Total $=5\cdot 8=40$ . Known sum $=32$ . The fifth number is $8$ .
Total $=20\cdot 7+10\cdot 8.5=140+85=225$ ; mean $=\frac{225}{30}=7.5$ .
Mean $=\frac{177}{7}\approx 25.29$ ; median $=15$ . The extreme value $90$ pulls the mean above all but one of the values, so the median describes the data better.
Total of seven $=70$ ; total of six $=54$ . The removed number is $70-54=16$ .
$n=12$ : the 6th and 7th values are $10$ and $12$ , so $M=11$ . Lower half $3, 4, 6, 7, 9, 10$ gives $Q_1=\frac{6+7}{2}=6.5$ . Upper half $12, 13, 15, 16, 18, 20$ gives $Q_3=\frac{15+16}{2}=15.5$ .