Statistics – Exercise 14.2 – (MATHEMATICS) – 9th Class

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Exercise 14.2 

1. The blood groups of 30 students of Class VIII arc 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?

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Sol.

Blood GroupTally MarksFrequency
AIIII    IIII9
BIIII I6
ABIII3
OIIII  IIII II12
Total = 30

As the frequency of blood group 0 is highest, i.e., 12 and that of AB is shortest, hence 0 is the most common and AB is the rarest blood group.


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?

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Sol.

Distance (in km)Tally MarksFrequency
0-5IIII5
5-10IIII   IIII  I11
10-15IIII   IIII   I11
15-20IIII   IIII9
20-25I1
25-30I1
30-35II2
Total = 40

(i) 11 engineers each have distance 5-10 km and 10-15 km from residence to the place of work.
(ii) One engineer each has distance between 20-25 km and 25-30 km from residence to the place of work.


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, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?

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Sol.

Relative humidity (in%)Tally MarksFrequency
84-86I1
86-88I1
88-90II2
90-92II2
92-94IIII  II7
94-96IIII  I6
96-98IIII  II7
98-100IIII4
Total = 30

(i) The given data is about the month of September (rainy season).
(ii) Range = 99.2 – 84.9 = 14.3.


4. The heights of 50 students, measured to the nearest centimetres, 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 as 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?

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Sol. (i) Grouped Frequency Distribution Table:

Height (in cm) (C.I)Tally MarksNumber of students (Frequency)
150-155IIII  IIII II12
155-160IIII  IIII9
160-165IIII  IIII  IIII14
165-170IIII  IIII10
170-175IIII5
Total = 50

(ii) We conclude that the height of more than 50% of the students are shorter than 165 cm each.


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?

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Sol. (i) Groupe’l Frequency Distribution Table:

Concentration (in ppm) (C.I)Tally MarksNumber of days (frequency)
0.00 – 0.04IIII4
0.04 – 0.08IIII   IIII9
0.08 – 0.12IIII   IIII9
0.12 – 0.16II2
0.16 – 0.20IIII4
0.20 – 0.24II2
Total = 30

(ii) Concentration was more than 0.11 (ppm) for (.2 + 4 + 2) days, i.e., 8 days.


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.

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Sol. Frequency Distribution Table:

Total = 30

Number of headsTally MarksNumber of times (frequency)
0IIII   I6
1IIII    IIII10
2IIII    IIII9
3IIII5


7. The value of π up to 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510

(i) Make a frequency of distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?

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Sol. (i) Frequency Distribution Table:

DigitsTally MarksNumber of times
0II2
1IIII5
2IIII5
3IIII   III8
4IIII4
5IIII5
6IIII4
7IIII4
8IIII5
9IIII   III8
Total  =  50

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


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

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Sol. (i) Frequency Distribution Table :

Number of hours (C.I.)Tally MarksNumber of children (frequency)
0 – 5IIII   IIII10
5 – 10IIII   IIII  III13
10 – 15IIII5
15 – 20II2
Total = 30

(ii) 2 children watched TV for 15 hours or more.


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.

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Sol. Grouped Frequency Distribution:

Life (in years)Tally Marks Number of batteries
2.0 – 2.5II2
2.5 – 3.0IIII   I6
3.0 – 3.5IIII  IIII  IIII14
3.5 – 4.0IIII   IIII  I11
4.0 – 4.5IIII4
4.5 – 5.0III3
Total  = 40

 


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