NCERT Important Questions & Solutions for Class 9 Maths Chapter 14 Statistics

NCERT Important Questions & Solutions for Class 9 Maths Chapter 14 Statistics

You can find Chapter 2 Statistics Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.2

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:
The required frequency distribution table is

From the above table, we have The most common blood group is O. The rarest blood group is AB.

2.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 lowest value of observation = 84.9
The highest value of observation = 99.2
So, class intervals are 84 – 86, 86 – 88, 88 – 90, ……. , 98 – 100

(i) Thus, the required frequency distribution table is

(ii) Since, the relative humidity is high during the rainy season, so, the data appears to be taken in the rainy season.
(iii) Range = (Highest observation) – (Lowest observation) = 99.2 – 84.9 = 14.3

3.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 day’s was the the concentration of sulphur dioxide more than 0.11 parts per million ?
Solution:
(i) Here, the lowest value of the observation = 0.01
The highest value of the observation = 0.22
∴ Class intervals are 0.00 – 0.04, 0.04 – 0.08,……., 0.20 – 0.24
The required frequency distribution table is

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

4.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?
Solution:
(i) The required frequency distribution table

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

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

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.
Solution:
(i) The required graphical representation is shown as follows:

(ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) Two factors may be un education and poor background.

2.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?
Solution:
(i) The required bar graph is shown below:

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

3.The length of 40 leaves of a plant measured correct to one millimetre and the obtained data is represented in the following 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 the maximum number of leaves 153 mm long and Why?
Solution:
(i) The given frequency distribution table is not continuous. Therefore, first we have to modify it to be continuous distribution.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex. 126-127= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 118-0.5=117.5 & 126+0.5=126.5 so the new modified row is (117.5 – 126.5 ) so continue this in all the row of table.

Thus, the modified frequency distribution table is:

Now, the required histogram of the frequency distribution is shown below :

(ii) Yes, other suitable graphical representation is a ‘frequency polygon’.
(iii) No, it is not a correct statement. The maximum number of leaves lie in the class interval 145 – 153.

4.The runs scored by two teams A and B on the first 60 balls in a cricket match are given below

Represent the data of both the teams on the same graph by frequency polygons.
Solution:
The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex.6-7= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 1-0.5=0.5 & 6+0.5=6.5 so the new modified row is (0.5 – 6.5 ) so continue this in all the row of table.

Now also Find Class Mark for the given Number of balls. Class Mark= Upper limit+Lower limit/2 ( Ex. 1+6/2=7/2=3.5 )

Now, the required frequency polygons are as shown below:

5.100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet 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.
Solution:
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class size = 6 – 4 = 2.
Therefore, we have the following table for length of rectangles.

The required histogram is shown below:

(ii) The maximum frequency is 44, which is corresponding to the class interval 6 – 8.
∴ Maximum number of surnames lie in the class interval 6 – 8.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

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.
Solution:
To find the mean :
Here, n = 10

Thus, mean = 2.8

To find median:
Now arranging the given data in ascending order,
we have 0,1, 2, 3, 3, 3, 3, 4, 4, 5
∵ n = 10, an even number

Thus, median = 3

To find mode:
In the given data, the observation 3 occurs 4 times,
i.e., maximum number of times.
Thus, mode = 3

2.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
Solution:
Here, the given observations are in ascending order.
Since, n = 10 (an even number)

Since, median = 63 [Given]
∵ x + 1 = 63 ⇒ x = 63 – 1 = 62
Thus, the required value of x is 62.

3.Find the mean salary of 60 workers of a factory from the following table

Solution:

Thus, the required mean salary = Rs. 5083.33

NCERT Quick revision Notes For Chapter-4 Statistics

NCERT Solutions For Chapter-4 Statistics

NCERT MCQs For Chapter-4 Statistics