12.2.1 Measures of Central Tendencies, PT3 Focus Practice - Mathematics

12.2.1 Measures of Central Tendencies, PT3 Focus Practice

Question 1:

Given that the mode of the set of data 7, 5, 3, 2, 3, y, 7, 6 and 5 is 7, find the mean.

Solution:

If the mode is 7, then y = 7

\begin{matrix} ∴ Mean = \frac{7 + 5 + 3 + 2 + 3 + 7 + 7 + 6 + 5}{9} = \frac{45}{9} = 5 \end{matrix}

$\begin{array}{l} ∴ Mean = \frac{7 + 5 + 3 + 2 + 3 + 7 + 7 + 6 + 5}{9} \\ = \frac{45}{9} \\ = 5 \end{array}$

Question 2:

The mean of the set of numbers 7, 2, x, 4, 5, 3, y is 5. The value of x + y is

Solution:

\begin{matrix} Mean = \frac{sum of all values of data}{total number of data} \frac{7 + 2 + x + 4 + 5 + 3 + y}{7} = 5 ∴ x + y + 21 = 7 \times 5 x + y = 35 - 21 x + y = 14 \end{matrix}

$\begin{array}{l} Mean = \frac{sum of all values of data}{total number of data} \\ \frac{7 + 2 + x + 4 + 5 + 3 + y}{7} = 5 \\ ∴ x + y + 21 = 7 \times 5 \\ x + y = 35 - 21 \\ x + y = 14 \end{array}$

Question 3:

Table below shows the number of story books read by a group of pupils in a week.

Number of books	1	2	3	4	5
Number of pupils	3	0	1	5	6

The median of the data is

Solution:

Number of books	1	2	3	4	5
Number of pupils	3	0	1	5	6
Position	1 – 3	3	4	5 – 9	10 – 15

Number of pupils = 3 + 0 + 1 + 5 + 6 = 15

Middle position situated at 8^th.

From the table, the position 8 has a value of 4, therefore the median of the data is 4.

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