Answer: Option A. -> 15.55
:
A

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. 

Answer: Option A. -> 350 students prefer to travel by bus.
:
A and B
We can observe from the graph:
Number of students using car = 250
Number of students using bike = 425
Number of students using bus = 350
Number of students using metro = 450
Number of students using cycle = 225
So, option A is correct. Also, cycle has the least frequency. So option B is also correct.
The frequency of metro is higher than the frequency of bike. So, option C is incorrect.
Also, D is incorrect as car doesn't have the highest frequency.

Answer: Option A. -> The class distribution is discontinuous
:
A, B, and D

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.

Answer: Option B. -> 15
:
B

Median is the middle observation of the given data, when arranged in either ascending or descending order.
Arranging the observations in ascending order, we get 
6, 9, 13 , 14, 15, 15, 18, 21, 22, 42

Since, there are even number of observations.
Now median, 
=  Average of ( $\frac{n}{2}$)^th observation and  $(\frac{n}{2}+1)$^th observation
=  $\frac{5^{th}term+6^{th}term}{2}$   
= $\frac{15+15}{2}$
= 15
Thus, median of the given data is 15.

Answer: Option A. -> True
:
A

Bar graphs are usually used to display "categorical data" i.e. data that fits into categories. Histograms, on the other hand, are usually used to present "continuous data", 

Answer: Option D. -> 15
:
D

Let, the observations be a, b, c, d, e.
We know that,
Mean = $\frac{Sumofallobservations}{Numberofobservations}$
Now,
mean of 5 observations = 12
$\Rightarrow \frac{a+b+c+d+e}{5}=12$ 
$\Rightarrow a+b+c+d+e=60$.....(i) 
It is also given that the mean of the first three observations is 10
$\Rightarrow \frac{a+b+c}{3}=10$
 $\Rightarrow a+b+c=30$...........(ii)
From (i) and (ii) we get
d + e = 30
Hence, the mean of the other two observations i.e. 'd' and 'e' is
$=\frac{d+e}{2}=15$

Answer: Option A. -> Bar Graph
:
A
The given description is the one about a bar graph. In a bar graph, rectangular bars of uniform width are drawn with equal spacing between them on one of the coordinate axes, usually the x-axis, to represent the type of variables. The frequency of variable is projected on the other axis i.e. the y-axis.

Answer: Option B. -> False
:
B
A bar graph is used to represent data that have different categories while histogram is used to represent classes that are continuous by bars which are side-by-side. Thus they both  are different.

Answer: Option A. -> The area under the frequency polygon is equal to the sum of areas of all bars of the histogram.
:
A and C
From the definition, the area under frequency polygon is same as the sum of the areas of all the bars of the histogram provided the class size of all classes is the same. Hence option A is correct.
Also, we can see that the bar of February is not the smallest. Hence, option B is incorrect.
The bar of June is highest, so option C is correct.
Also, the frequency polygon is correctly drawn joining all the class marks and starting and ending at 0 so option D is not correct.

Answer: Option A. -> The area under the frequency polygon is equal to the sum of areas of all bars of the histogram.
:

To obtain the median of a given set of data, we have to arrange the data in ascending or descending order. After arranging in ascending order, we get
45, 76, 87, 89, 101, 121, 134, 202, 212, 323.
Here, we have even number of observations i.e. n = 10. So, we will obtain the median by taking the mean of the value of $5^{th}$ and $6^{th}$ observation i.e. $\frac{101+121}{2}=111$