Free Statistics 02 Practice Test - 9th Grade 

Mean, median and mode are used as measures of central tendency. Mean is the average of all the observations, mode is the observation having maximum frequency and median is the value which divides the data into two equal halves. 

For finding the median, the data should be arranged in ascending or descending order first.

Arranging the data in ascending order ,we get :
1, 2, 3, 3, 4, 5, 5, 7, 7, 9.

There are even number of observations.
Hence the  median is the mean value of $(\frac{n}{2}{)}^{th}$ and $\left[\right(\frac{n}{2})+1{]}^{th}$ term. 
Here the ${\frac{10}{2}}^{th}term$= 5${}^{th}$ term = 4
$\left[\right(\frac{n}{2})+1{]}^{th}$ term.  = 6${}^{th}$ term = 5
 i.e. $\left(\frac{4+5}{2}\right)=4.5$
Hence, median of the given data $=4.5$

The difference between the highest and lowest values in the data is called the 'range' of the data. So, here the maximum value is 33 and the minimum is 14. So the difference between them i.e. 19 is the range of the given data.

Mode is that value of the observation which occurs most frequently i.e. an observation with maximum frequency is called mode. In the above set of observations, '9' occurs most frequently and so it is the mode of the given data.

We know that,
Mean = $\frac{Sumofallobservations}{Numberofobservations}$

Sum of values of all the observations
$=Mean\times Numberofobservation$
$=15\times 8$
$=120$

So, sum of the first eight observations
$=Mean\times Numberofobservation$
$=7\times 8$
$=56$
Sum of the last eight observations
$=Mean\times Numberofobservation$
$=10\times 8$
$=80$ 

Hence, value of the $8^{th}$observation 
$=(80+56)-120$
$=16$

The number of people in a class is proportional to area.
Therefore, the class of 0-20 is twice the size of other classes.
So, total frequency of 0-20 class is  $5\times 2=10$

Hence option A is correct and B is incorrect. 
The class of 21-30 age group has the highest frequency. 
Hence option C is correct.
On summing up all the frequencies, it can be seen that D is incorrect

Mean of ungrouped frequency distribution is given by

$\overline{x}=\frac{{\sum }_{i=1}^n{f}_i{x}_i}{{\sum }_{i=1}^n{f}_i}$
$=\frac{(10\times 2+13\times 5+16\times 7+19\times 6)}{(2+5+7+6)}$
$=\frac{311}{20}$

Therefore, the mean of the data is 15.55. 

In the given table, the upper limit of a class does not coincide with the lower limit of the succeeding class, so the class intervals are not continuous and option A is correct. 
Also the number of students in the range of 160 cm -174 cm = 2+1+2 = 5. Hence, option B is correct.
Also, any data having discontinuous class intervals can be converted into one with continuous class intervals. Hence, C is incorrect.
Range of data = Total width of data = 189-150 = 39.