NCERT Class 9 Maths Solutions Chapter - 14 Statistics, Ex - 14.2

Ex - 14.2

Question 1.  The blood groups of 30 students of Class VIII are recoded 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

Here, 9 students  have blood groups  A, 6 as B, 3 as AB and 12 as O.
So, the table representing the data is as follows:  

Blood group	Number of students
A	9
B	6
AB	3
O	12
Total	30

As 12 students have the blood group O and 3 have their blood group as AB, clearly that the most common blood group and the rarest blood group among these students is O and AB respectively.

Question 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 feature do you observe from this tabular representation?

Solution

Given that we have to construct a grouped frequency distribution table of class size 5. So, the  class intervals will be as 0 - 5, 5 - 10, 10 - 15, 15 - 20......
Required grouped frequency distribution table as following -

Distance (in km)	Tally marks	Number of engineers
0 - 5		5
5 - 10		11
10 -15		11
15 - 20		9
20 - 25		1
25 - 30		1
30 - 35		2
Total		40

Now  there are only 4 engineers whose homes are at more than or equal to 20 km distance, from their work place.
Most of the engineers are having their workplace up to 15 km distance, from their homes.

Question 3.  The relative humidity (in %) of a certain city for a 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

(ii)    Which month or season do you think this data is about?

(iii)   What is the range of this data?

Solution

(i)    To construct a grouped frequency distribution table of class size 2.
       Class intervals will be as follows 84 - 86, 86 - 88, and 88 - 90... ...

       

Relative humidity (in %)	Number of days (frequency )
84 - 86	1
86 - 88	1
88 - 90	2
90 - 92	2
92 - 94	7
94 - 96	6
96 - 98	7
98 - 100	4
Total	30

(ii)    Since relative humidity is high so the data must be of a month of rainy season.                     

(iii)    Range of data = maximum value - minimum value
                              = 99.2 - 84.9

                              = 14.3

Question 4.  The heights of 50 students, measured to the nearest centimeters, have been found  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?

Solution

(i)    We have to construct a grouped frequency distribution table taking class intervals as 160 - 165, 165 - 170, etc. Now by observing the data given as above we may construct the required table as below - 

       

Heights  (in cm)	Number of students (frequency )
150 - 155	12
155 - 160	9
160 - 165	14
165 - 170	10
170 - 175	5
Total	50

                                                                                  

(ii)    From the table we can see that  50% of 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:

        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?

Solution

To construct grouped frequency table class intervals to be taken as 0.00 - 0.04, 0.04 - 0.08, 

Concentration of SO2  (in ppm)	Number of days (frequency )
0.00 - 0.04	4
0.04 - 0.08	9
0.08 - 0.12	9
0.12 - 0.16	2
0.16 - 0.20	4
0.20 - 0.24	2
Total	30

                                                                                                            
Number of days for which concentration SO2 is more than 0.11 is number of days for which concentration is in between 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24.
So, required number of days = 2 + 4 + 2 = 8

Question 6.  Three coins were tossed 30 times simultaneously. 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.

Solution

By observing the data given above following frequency distribution table can be constructed

Number of heads	Number of times (frequency)
0	6
1	10
2	9
3	5
Total	30

Question 7.  The value of  up to 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?

Solution

(i)    By observation of digits after decimal point the following table is constructed 
            
        

Digit	Frequency
0	2
1	5
2	5
3	8
4	4
5	5
6	4
7	4
8	5
9	8
Total	50

                                                                                                  
(ii)    From the above table the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. So, the most frequently occurring digits are 3 and 9 and the least occurring digit is 0.

Question 8.  Thirty children were asked about the number of hours they watched TV programmers 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?

Solution

(i) Class intervals will be 0 - 5, 5 - 10, 10 -15.....
    The grouped frequency distribution table is as follows:
    

                      

Hours	Number of children
0 - 5	10
5 - 10	13
10 - 15	5
15 - 20	2
Total	30

                                                                                         
(ii) The number of children, who watched TV for 15 or more hours a week
        is 2 (i.e. number of children in class interval 15 - 20).

Question 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 intervals 2 - 2.5.

Solution

To construct a grouped frequency table of class size 0.5 and starting from class interval 2 - 2.5.  So, our class intervals will be as 2 - 2.5, 2.5 - 3, 3 - 3.5.......  Required grouped frequency distribution table is  as below -

Lives of batteries (in hours)	Number of batteries
2 - 2.5	2
2.5 - 3.0	6
3.0 - 3.5	14
3.5 - 4.0	11
4.0 - 4.5	4
4.5 - 5.0	3
Total	40

Ex-14.1

Ex-14.3