# 4.2 Statistics II, PT3 Focus Practice 1

4.2 Statistics II, PT3 Focus Practice 1

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

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

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

Question 4:
Table below shows the scores for a group of pupils in a sport game.

 Score 1 2 3 4 5 Number of pupils 4 11 5 3 2
Calculate the percentage number of pupils who obtain scores more than the mode score.

Solution:
Mode score = 2
Number of pupils who obtain scores more than 2
= 5 + 3 + 2
= 10
Total number of pupils
= 4 + 11 + 5 + 3 + 2
= 25
Percentage number of pupils who obtain scores more than the mode score
$\begin{array}{l}=\frac{10}{25}×100%\\ =40%\end{array}$

Question 5:
The mean age of Alex, Michelle and their three children is 34 years. The mean age of their three children is 20 years.
Calculate the mean age of Alex and Michelle.

Solution:
Total age = 34 × 5 = 170 years
Total age of their three children
= 20 × 3
= 60 years
Total age of Alex and Michelle
= 170 – 60
= 110 years

Mean age of Alex and Michelle
$\begin{array}{l}=\frac{110}{2}\\ =55\text{years}\end{array}$