Exercise 14.1

Q.1     Give five examples of data that you can collect from your day to day life.
Sol.

Five examples of data that we can gather from our day to day life are :
(i) Number of students in our class.
(ii) Number of fans in our school.
(iii) Electricity bills of our house for last two years.
(iv) Election results obtained from television or newspapers.
(v) Literacy rate figures obtained from educational survey.

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Q.2      Classify the data in Q. 1 above as primary or secondary data.
Sol.

Primary data : (i) , (ii) and (iii)
Secondary data : (iv) and (v)

Exercise 14.2

Q.1     The blood groups of 30 students of Class – VII 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?
Sol.        The frequency distribution table is as under :

Clearly, Most common Blood group – O,  Rarest Blood group – AB.

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Q.2      The distance (in km) of 40 engineers from their residence to their place of work were found as follows : 
             5         3           10         20           25           11           13         7          12         31 
             19       10          12         17           18           11           32         17        16         2 
             7         9            7           8             3             5            12         15        18         3 
             12       14          2           9             6             15          15         7          6           12

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?
Sol.

The minimum and maximum km in the given raw data are 2 and 32 respectively. It is given that 0 -5 is one of the class  intervals and the class size is same. So, the classes of equal size are
0– 5, 5 – 10, 10 – 15, 15 – 20, 20– 25, 25 – 30 and 30 – 35.
Thus, the frequency distribution table is as given under :

In this the upper limit of a class is not included in the class. Thus, in the class 0–5 of km, distance traveled by the engineer to their place of work, an engineer who is to travel 5 km is not included in this class. He is counted in the next class 5 – 10. It is known an exclusive method. It is observed that 27 engineers out of 40 lives at a distance not more than 15 km from their residence.

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Q.3    The relative humidity (in %) of a certain city a for month of 30 days was as follows :
           98.1     98.6     99.2    90.3    86.5     95.3     92.9    96.3     94.2      95.1
           89.2     92.3    97.1     93.5    92.7     95.1      97.2   93.3     95.2      97.3
           96.2     92.1     84.9     90.2   95.7      98.3     97.3   96.1     92.1      89
             (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 ?
Sol.

(i) The minimum and maximum relative humidity (in %) in the given raw data are 84.9 and 99.2 respectively. It is given that 84 – 86 is one of the class intervals and the class size is same. So, the classes of equal size are
84 – 86,      86 – 88,     88 – 90,    …, 98 -100.
Thus, the frequency distribution table is as under :

(ii) The data appears to be taken in the rainy season as  the relative humidity is high.
(iii) Range = 99.2 – 84.9 = 14.3

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Q.4     The heights of 50 students, measured to the nearest centimetre, have been found to be as follows :
              161        150        154         165        168         161      154       162        150        151
              162        164        171         165         158        154      156       172        160        170
              153        159        161         170         162        165      166       168        165        164
              154        152        153         156         158        162      160       161        173        166
              161        159        162         167         168        159      158       153        154        159
               (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 ?
Sol.

(i) The minimum and maximum heights in the given raw data are 150 cm and 173 cm respectively. It is given that 160 – 165 is one of the class intervals and the class size is same. So, the classes of equal size are 150 – 155, 155 – 160….., 170 – 175.
Thus, the frequency distribution table is as under

(ii) One conclusion that we can draw from the above table is that more than 50% of students are shorter than 165 cm.

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Q.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 :
               0.03        0.08       0.08           0.09        0.04       0.17
               0.16        0.05       0.02           0.06        0.18       0.20
               0.11        0.08       0.12            0.13        0.22      0.07
               0.08        0.01       0.10            0.06        0.09      0.18
               0.11         0.07      0.05            0.07        0.01      0.04
             (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 ?
Sol.

(i) The minimum and maximum  concentration of sulphur dioxide in the air  in parts per million is 0.01 and 0.22 respectively.
It is given that 0.00- 0.04 is one of the class intervals and the class size is the  same. So, the classes of equal size are
0.00 – 0.04, 0.04 – 0.08, …, 0.20 – 0.24
Thus, the frequency distribution table is as under :

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

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Q.6      Three coins were tossed 30 times. Each time the number of heads occurring was noted down as follows:
             0         1         2          2         1        2        3      1         3       0
             1          3        1          1         2        2        0      1         2       1
             3          0        0          1         1        2        3      2         2       0 
Prepare a frequency distribution table for the data given above.
Sol.        Frequency distribution table is as follows :

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Q.7    The value of π  upto 50 decimal places is given below :
     3.14159265358979323846264338327950288419716939937510.
          (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?
Sol.      (i) The frequency distribution table is as under :

(ii) The most frequently occurring digits are 3 and 9. The least occurring is 0.

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Q.8     Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
                   1        6          2        3          5        12        5        8        4        8
                   10      3          4        12        2        8         15       1        17      6
                   3        2          8        5          9        6          8        7        14      12
              (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?
Sol.

(i) The minimum and maximum number of hours children watched TV programmes in the previous week are 1 hour and 17 hours, respectively. It is given that 5 – 10 is one of the class intervals and the class size is same. So, the classes of equal size are 0 –5, 5–10, 10–15 and 15–20.
Thus the frequency distribution table is as under

(ii) 2 children watched television for 15 or more hours a week.

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Q.9     A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows :
                 2.6         3.0          3.7          3.2          2.2         4.1           3.5           4.5
                 3.5         2.3          3.2          3.4          3.8         3.2           4.6           3.7
                 2.5         4.4          3.4          3.3          2.9         3.0           4.3           2.8
                 3.5         3.2          3.9          3.2          3.2         3.1           3.7           3.4
                 4.6         3.8          3.2          2.6          3.5         4.2           2.9           3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Sol. The minimum and maximum life in number of years of car batteries  are 2.2 years and 4.6 years. It is given that 2 – 2.5 is one of the class interval with uniform size of 0.5. So , the classes of equal size are 2.0 – 2.5, 2.5 – 3.0, 3.0 – 3.5, …., 4.5 – 5.0.
Thus the frequency distribution table is as under :

Exercise 14.3

Q.1     A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Sol.

(i) In the graph drawn causes of illness and death among women between the ages 15 –44 (in years) worldwide is denoted on X-axis and female fatality rate (%) is denoted on the Y-axis.
(ii) The major cause of women’s ill health and death worldwide is reproductive health conditions.
(iii) Two other factors which play a major role in the cause in (ii) above are neuropsychaitric conditions and other causes.

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Q.2     The following data on the number of girls to the nearest ten per thousand boys in different sections of the society is given below :

               (i) Represent the information above by a bar graph.
               (ii) Write two conclusions you can arrive at from the graph, with justification.
Sol.         (i) The required graph is as under :

In the graph drawn different sections of the society is denoted on the horizontal and the number of girls to the nearest ten per thousand boys is denoted on the vertical. Their intersection represent 900. Scale : 1 cm = 10 girls.

(ii) From, the graph we find that the number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimum in urban.

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Q.3    Given below are the seats won by different political parties in the polling outcome of a state assembly elections :

            (i) Draw a bar graph to represent the polling results.
              (ii) Which political party won the maximum number of seats?
Sol.

(i) In the graph drawn political party is denoted on the X-axis and the number of seat won on the Y-axis.
Scale : 1 cm = 10 seats.
The required graph is as follows :
(ii) Party A won the maximum number of seats.

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Q.4     The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the table  :

                 (i) Draw a histogram to represent the given data.
                 (ii) Is there any other suitable graphical representation for the same data?
                 (iii) Is it correct to conclude that maximum number of leaves are 153 mm long ? Why ?
Sol.

(i) The given frequency distribution is not continuous. So,we shall first convert it into a continuous frequency distribution.
The difference between the lower limit of a class and the upper limit of the preceding class is 1. i.e., h = 1. To convert the given frequency   distribution into a continuous frequency distribution, we subtract h2=12=0.5 from each of lower limit and add 0.5 to each upper limit. The distribution so obtained is given as under :

The histogram of the above frequency distribution is shown as under :

(ii) Frequency polygon is another method of representing frequency distribution graphically.
(iii) Yes, as in the class interval 145 – 153, the lengths of leaves is maximum.

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Q.5      The following table gives the life times of 400 neon lamps :

            (i) Represent the given information with the help of a histogram.
            (ii) How many lamps have a life time of more than 700 hours ?
Sol.        (i) The histogram of the given frequency distribution is shown as under :

(ii) Number of lamps having life more than 700 hours = 74 + 62 + 48 = 184.

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Q.6      The following two table gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons.
Sol.           First we obtain the class marks as given in the following table :

We represent class marks on X-axis on a suitable scale and the frequencies on Y-axis on a suitable scale.
To obtain the frequency polygon of section A, we plot the points (5, 3), (15, 9), (25, 17), (35, 12) and (45, 9), and join these points by line segments.
To obtain the frequency polygon of section B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10) and (45, 1) on the  same scale and join these points by dotted line segments.
The two frequency polygons are as shown below :

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Q.7     The runs scored by two teams A and B on the first 60 balls in a cricket match are given on the next page :

Represent the data of both the teams on the same graph by frequency polygons.
Sol.        First we obtain the class marks as given in the following table :

We represent class marks on x-axis on a suitable scale and frequencies on y-axis on a suitable scale.
To obtain the frequency polygon of team A, we plot the points (3, 2) (9.5, 1), (15.5, 8), (21.5, 9), (27.5, 4), (33.5, 5), (39.5, 6) , (45.5, 10), (51.5, 6) and (57.5, 2) and join these points by the line segments.
To obtain the frequency polygon of team B, we plot the points (3, 3) (9.5, 6), (15.5, 2) (21.5, 10), (27.5, 5), (33.5, 6) (39.5, 3), (45.5, 4) (51.5,8) and (57.5, 10).
The two frequency polygons are as shown below :

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Q.8      A random survey of the number of children of various age groups playing in a park was found as follows :

Draw a histogram to represent the data above,
Sol.

In the given frequency distribution, we see that the class-sizes are different. Hence, we calculate the adjusted frequency for each class by using the formula :
Adjusted frequency for a class   =Minimumclass−sizeClass−sizeofthisclass×itsfrequency
In this problem, the minimum class- size = 2–1 = 1.

We thus obtain the following table of the adjusted frequency.We now represent the class intervals along the X-axis on a suitable scale and the corresponding adjusted frequencies on the Y-axis on a different suitable scale. Now, we draw rectangles with the class intervals as bases and the corresponding adjusted frequencies  as the heights.
The required histogram is as shown :

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Q.9    100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabets in the surnames was found as follows :

             (i) Draw a histogram to depict the given information.
                (ii) Write the class interval in which the maximum number of surnames lie.
Sol.

(i) In the given frequency distribution, we see that the class-sizes are different. Hence, we calculate the adjusted frequency for each class by using the formula :
Adjusted  frequency for a class  =Minimumclass−sizeClass−sizeofthisclass×itsfrequency
In this problem, the minimum class-size = 6 – 4 = 2.
We thus obtain the following table of the adjusted frequency.

We now represent the class intervals along the X-axis on a suitable scale and the corresponding adjusted frequencies on the Y-axis on a different suitable scale. Now, we draw rectangles with the class intervals as bases and the corresponding adjusted frequencies as the heights. The required histogram is as shown.

(ii) The maximum of surnames lies in 6 – 8 class interval.

 

Exercise 14.4

Q.1    The following number of goals were scored by a team in a series of 10 matches :
           2,        3 ,         4,         5,         0,         1,        3,         3,        4,        3.
Find the mean median and mode of these scores.
Sol.

Using x¯=x1+x2+....+x1010, the mean is
x¯¯¯=2+3+4+5+0+1+3+3+4+310=2810=2.8
To find the median, arrange the given data in ascending order, as follows :
0,  1,   2,   3,   3,   3,   3,   4,   4,    5.
There are 10 terms. So, there are two middle terms, i.e., the (102)thand(102+1)th
i.e., the 5th and 6th terms.
So, the median is the mean of the values of the 5th and 6th terms.
i.e., the median =3+32=3
Again, in the data 3 occurs most frequently i.e, 4 times.
So, mode = 3

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Q.2      In a Mathematics test given to 15 students, the following marks (out of 100) are recorded:
           41,  39,   48,   52,    46,   62,   54,   40,  96,  52,  98,  40,  42,  52,  60
Find the mean, median and mode of this data.
Sol.

Using x¯=x1+x2+.....+x1515  , the mean is
x¯¯¯=41+39+48+52+46+62+54+40+96+52+98+40+42+52+6015
=82215=54.8
To find the median, arrange the given data in ascending order, as follows :
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98.
Since the number of terms is 15, an odd number, we find out the median by finding the marks
obtained by (15+12)the  student, which is 8th student.
Therefore , the median marks = 52.
Again,  the data 52 occurs most frequently i.e., 3 times.
Therefore Mode = 52

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Q.3     The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
           29,     32,     48,     50,   x,    x + 2,    72,     78,    84,     95.
Sol.

The given data  arranged in ascending order is as
29, 32, 48,50, x, x + 2, 72, 78, 84, 95
There are 10 terms. So, there are two middle terms, i.e. the (102)th and (102+1)th  i.e., the 5th and 6th terms.
So, the median is the mean of the values of the 5th and 6th terms.
i.e., themedian=x+(x+2)2=x+1
But median = 63                                                [Given]
Therefore    x + 1 = 63
⇒ x = 63 – 1 = 62

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Q.4      Find the mode 
                14,  25,  14,  28,  18,  17,  18,  14,  23,  22,  14,  18
Sol.

Arranging the given data in ascending order as follows, we get
14, 14, 14, 14, 17 18, 18, 18, 22, 23, 25, 28.
Here , 14 occurs most frequently i.e., 4 times.
Therefore Mode = 14.

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Q.5      Find the mean salary of 60 workers of a factory from the following table :

Sol.          Calculation of Mean

Therefore Mean =∑fixi∑fi=30500060=5083.33
Thus, mean salary of 60 workers is Rs 5083.33 (approx.)

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Q.6      Give one example of a situation in which
               (i) the mean is an appropriate measure of central tendency.
              (ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Sol.

(i) The mean is an appropriate measure because of its unique value and can be used to compare different groups of data.
(ii) For the Measurement of qualitative characteristics e.g., beauty, honesty, intelligence etc., mean cannot be used.