(i), (ii) and (iii) are primary data
(iv) and (v) are secondatary data.

O is the most common and AB is the rarest blood group among these students.

(ii) The two conclusions we can arrive at from the graph are as follows:
(a) The numbers of girls to the nearest ten per thousand boys is maximum in Scheduled Tribe section of the society and minimum in Urban section of the society.
(b) The number of girls to the nearest ten per thousand boys is the same for 'Non Sc/ST' and 'Non-backward Districts' sections of the society.

(i)

(ii) Political party A won the maximum number of seats.

Modified Continues Distribution

(ii) Frequency Polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the original interval 145-153 (or modified interval 144.5-153.5).

The given data is 
14,25,14,28,18,17,18,14,23,22,14,18
Arranging the data in ascending order, we have
14,14,14,14,17,18,18,18,22,23,25,28
Here, 14 occurs most frequently (4 times)
\(\therefore\) Mode = 14.

\(\therefore\)   \(\overset { - }{ x } =\frac { \sum _{ i=1 }^{ 8 }{ { f }_{ i }{ x }_{ i } } }{ \sum _{ i=1 }^{ 8 }{ { f }_{ i } } } =\frac { 305000 }{ 60 } \) = Rs 5083.33
Hence, the mean salary is Rs 5083.33.

(i) 4 students were born in the month of November.
(ii) Maximum number of students were born in the month of August.

Since we want to draw a frequency polygon without a histogram, let us find the class-marks of the classes given above, that is of 140 - 150, 150 - 160,....
For 140 - 150, the upper limit = 150, and the lower limit = 140
So, the class-mark = \(\frac{150+140}{2}=\frac{290}{2}=145\)

We can now draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis, and then plotting and joining the points B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments. We should not forget to plot the point corresponding to the class-mark of the class 130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is, A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant frequency polygon will be ABCDEFGH.

Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is 0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end point, i.e., B is joined to this mid-point with zero frequency on the negative direction of the horizontal axis. The point where this line segment meets the vertical axis is marked as A. Let L be the mid-point of the class succeeding the last class of the given data. Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig.

First of all we arrange the data in ascending order, as follows:
144 145 147 148 149 150 152 155 160
Since the number of students is 9, an odd number, we find out the median by finding the height of the \(\left(\frac{n+1}{2}\right) \text { th }=\left(\frac{9+1}{2}\right) \text { th }\) the 5th student, which is 149 cm.
So, the median, i.e., the medial height is 149 cm.

(ii) 74 + 62 + 48 = 184 have a life time of more than 700 hours

To present such a large amount of data, so that a reader can make sense of it easily, we condense it into groups like 0-5, 5-10, ..., 30-35 (since, our data is from 5 to 32). Now, using tally marks, the data (given) can be condensed in tabular form as

From the above table, we observe that
(i) frequencies of class intervals 5-10 and 10-15 are equal, i.e. 11 each. It shows that the maximum number of engineers have their residences within the distance of 5 to 15 km from their workplace.
(ii) frequencies of class intervals 20-25 and 25-30 are also equal, i.e. 1 each. It shows that the minimum number of engineers have their residences within 20 to 30 km from their workplace.

(i) We condense the given data into groups, like 84-86, 86-88, ... , 98-100 (since, our data is from 84.9 to 99.2). So, the class width in this case is 2.
Now, the given data can be condensed in tabular form as follows

(ii) From the table, we observe that the data appears to be taken in the rainy season as the relative humidity is high.
(iii) We know that, Range = Upper limit of data - Lower limit of data
= 99.2- 84.9 = 14.3

(i) We condense the given data into groups like 150-155, 155-160, ... , 170-175 (since, our data is from 150 to 172). The class width in this case is 5. Now, the given data can be condensed in tabular form as follows

(ii) From the table, our conclusion is that more than 50% of students (i.e. 12 + 9 + 14 = 35) are shorter than 165 cm height.