Free Statistics 02 Practice Test - 10th Grade 

$\overline{x}=A+\frac{\sum f_i{d}_i}{\sum f_i}$

In the formula, the A stands for the assumed mean.

The given table can be written as:
 $$\begin{array}{cccccc}\text{Marks}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \text{No. of students}& 5& 5& 13& 8& 9\\ \text{Cumulative Frequency}& 5& 10& 23& 31& 40\end{array}$$
No. of students can be calculated as follows:
5
10 - 5 = 5
23 - 10 = 13
31 - 23 = 8
40 - 31 = 9

$\therefore \frac{n}{2}=\frac{40}{2}=20$

Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.

Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.

The class interval 20 - 30 is the required answer.

The given table can be written as:
$$\begin{array}{ccccccc}\text{Height (in cm)}& \text{Less than 140}& \text{Less than 145}& \text{Less than 150}& \text{Less than 155}& \text{Less than 160}& \text{Less than 165}\\ \text{No. of girls}& 4& 11& 29& 40& 46& 51\end{array}$$

$$\begin{array}{ccccccc}\text{Height (in cm)}& \text{Less than 140}& 140-145& 145-150& 150-155& 155-160& 160-165\\ \text{Cumulative Frequency}& 4& 11& 29& 40& 46& 51\end{array}$$

n = 51

$\therefore \frac{n}{2}=\frac{51}{2}$
                                     = 25.5

Since there are a total of 51 observations, the median class is the one whose cumulative frequency is closest to and greater than 25.5. The cumulative frequency for the class '145 - 150' is 29 which is greater than 25.5. The median class is '140 - 145'. Thus, the class interval '145 - 150' is the required answer.

 $$\begin{array}{ccc}ClassInterval& \text{Frequency}& CumulativeFrequency\\ 60-70& 2& 2\\ 70-80& 3& 5\\ 80-90& 5& 10\\ 90-100& 16& 26\\ 100-110& 14& 40\\ 110-120& 13& 53\\ 120-130& 7& 60\end{array}$$ 

$\frac{n}{2}=30$.
So, the median class is the one whose cumulative frequency is closest to and greater than 30.
$\therefore$ the median class is 100-110 

$Median=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$ 
Where $l$ is lower class limit of median class.
          $n$ is total number of observations.
          $cf$ is the cumulative frequency of the class preceding the median class.
          $f$ is the frequency of the median class and h is the class size.

$⟹Median=100+\frac{30-26}{14}\times 10$ 

$\therefore Median=102.86$

The method in question is the step-deviation method.

Mean = 45 and Mode = 60
3 Median = Mode +2 Mean
3 Median = 60 +2(45)
3 Median = 150
Median=50

For grouped data, using the direct method first, we find the sum of the values of all the observations after multiplying them with their respective frequencies. Then we divide this result by the total number of observations (sum of all frequencies). 

The empirical relationship between Median, Mode and Mean is Mode = 3 Median - 2 Mean

$l+(\frac{f_1-f_0}{2f_1-f_0-f_2}$) $\times h$ is used to determine the mode .  
A mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.

Using, Mode =   $l+\frac{f_1-f_0}{{2f}_1-f_0-f_2}\times h$  
where,
Maximum class frequency, $f_1=8$

The class corresponding to the frequency $=20-40$

Lower limit of the modal class, $l=20$

Frequency of class preceding the modal class, $f_0=7$

Frequency of class succeeding the modal class, $f_2=2$                                

$Now,Mode=20+\frac{8-7}{16-7-2}\times 20$

 $Mode=20+\frac{1}{7}\times 20$

$\therefore Mode=22.86$