Question 1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Solution: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
From the above table, we have:
The most common blood group is O.
The rarest blood group is AB.
Question 2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution: The given distance (in km) are:
Here, the observation with minimum and maximum value are 2 and 32 respectively.
∵ The first class interval is 0–5,
∴ The classes are:
0–5, 5–10, 10–15, 15–20, 20–25, 25–30, 30–35.
The required table is
From the above table we observe that:
(i) Frequencies of class interval 5–10 and 10–15 are equal, i.e. 11 each. It shows that maximum number of engineers have their residences at 5 to 15 km away from their work place.
(ii) Frequencies of class intervals 25–30 and 25–30 are also equal, i.e. 1 each. It shows that minimum number of engineers have their residences at 20 to 30 km away from their work place.
Question 3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
(i) Construct a grouped frequency distribution table with classes 84–86, 86–88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution: Here, the minimum value of the observation = 84.9
The maximum value of the observation = 99.2
∵ Some of the classes are 84–86 and 86–88, etc.
∴ Class size = 86–84 = 2 or 88–86 = 2
So the classes are: 84–86, 86–88, 88–90, 90–92, …, 98–100.
(i) Thus, the required frequency table is:
(ii) Since, the relative humidity is high during the rainy season.
∴ The data appears to be taken in the rainy season.
(iii) Since, range = [Highest observation] – [Lowest observation] = 99.2 – 84.9
Question 4. The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160–165, 165–170, etc.
(ii) What can you conclude about their heights from the table?
Solution: (i) Here, the lowest value of the observation = 150
The Highest value of the observation = 173
∴ Classes are: 150–155, 155–160, … 170–175.
(ii) Conclusions: More than 50% of the students are shorter than 165 cm.
Question 5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00–0.04, 0.04–0.08, and so on. (ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
(ii) The concentration of sulphur dioxide more than 0.11 parts was for 8 days.
Question 6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
Prepare a frequency distribution table for the data given above.
Solution: The required frequency distribution is as under.
Question 7. The value of p up to 50 decimal places is given below:
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
(ii) The most frequently occurring digits are 3 and 9.
The least frequently occurring digit is 0.
Question 8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5–10.
(ii) How many children watched television for 15 or more hours a week?
Solution: Here, the lowest observation = 1
The highest observation = 17
∴ The classes are: 0–5, 5–10, 10–15, …, 15–20.
(i) The required frequency distribution is as under.
(ii) Number of children who watched TV for 15 or more hours a week = 2.
Question 9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2–2.5.
Solution: Here, the lowest observation = 2.2 years
The highest observation = 4.6 years
The classes are: 2–2.5, 2.5–3.0, 3.0–3.5, …, 4.5–5.0
Thus, required grouped frequency distribution table is as under.
GRAPHICAL REPRESENTATION OF DATA
The graphical representations of numerical data has more soothing effect on mind. Here, we shall use three types of graphs, namely: (i) Bar graph (ii) Histogram (iii) Frequency polygon.
Bar Graph A bar graph is a pictorial representation of data in which bars of uniform width are drawn with various heights. The height of a bar represents the frequency of the corresponding observation.
Note: I. Keep equal spacing between the bars on one axis, depicting the variable. The equal spacing between the bars is taken as per our convenience.
II. The height of the bars on the other axis depend upon the values of the variables.
Histogram : A histogram is a graphical representation of a frequency distribution in the form of rectangles (having no gaps between the consecutive rectangles) with class intervals as bases and the corresponding frequencies as heights.
Note: I. A histogram is used for continuous class intervals (exclusive form). If the given frequency distribution is in an inclusive form then convert it in an exclusive form.
II. Since, there are no gaps between the consecutive rectangles, the histogram appears like a solid figure.
Frequency Polygon: A frequency polygon is obtained by joining the mid-points of the upper sides of the adjacent rectangles of the histogram by means of line segment we complete the polygon by taking two more classes (called imagined classes), one at the beginning and other at the end.
A frequency polygon only can be drawn independently by plotting the class marks along x-axis, frequencies along y-axis, and joining the plotted points by line segments.
Note: In case both the histogram and frequency polygon are to be drawn, it is advisable first to draw histogram and then join the mid-points of the tops of the rectangles of the histogram to get frequency-polygon.