Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Class 9 Mathematics by Full Circle

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Class 9 : Additional Questions Solutions- Statistics Class 9 Notes | EduRev

The document Additional Questions Solutions- Statistics Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Question 1. Following are the marks obtained by 40 students of class IX in an examination:

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

(i) Present the data in the form of a frequency distribution using the same class size such as 0–5, 5–10, etc.
 (ii) How many students obtained marks below 15?
 Solution: 
The lowest value of the observation = 1
The highest value of the observation = 24
∴ To cover all the given data, the required classes are: 0–5, 5–10, …, 20–25.
(i) The required frequency distribution is

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

(ii) From the table, it is found that number of students who have obtained marks below 15 = 5 + 10 + 9        
= 24


Question 2. The house-tax bills (in rupees) of 30 houses in a locality are given below:

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Construct a frequency distribution table with class size 10.
 Solution: 
The lowest observation = 814
The highest observation = 912
Class size = 10
To cover the given data we have the classes as: 814–824, 824–834, …, 904–914.
Thus the required frequency distribution is as under:

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Note: A frequency distribution in which each upper limit of each class is excluded and the lower limit is included, is called an exclusive form or the continuous form whereas on the other hand a frequency distribution in which each upper limit as well as lower limit is included, is called an inclusive form.
 

Question 3. Following data gives the marks (out of 50), obtained by 30 students of a class in a test:

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Arrange the data using classes as 0–10, 11–20, etc.
 Solution:
The lowest observation = 1
The highest observation = 48
∴ To cover the given data the required classes will be: 0–10, 11–20, …, 41–50.
Now, we have:

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

 

Question 4. If the mean of the following data is 18.75, then find the value of p.

xi1015p2530
fi510782

Solution: Writing the given data as in the following table.

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

∴          Additional Questions Solutions- Statistics Class 9 Notes | EduRev

But the mean of the given data is 18.75.

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Additional Questions Solutions- Statistics Class 9 Notes | EduRev

Thus, the required value of p is 20.


Question 5. If the marks of 41 students of a class are given in the following table, then find the median of marks obtained.

Marks obtainedFrequency
30 
25 
27 
40 
32 
35 
10
2
5
4
12
8

Solution: Arranging the observations in ascending order, we have:

Marks obtained252730323540
Frequency25101284

Now, we prepare the cumulative frequency table as given below:

Marks obtained FrequencyCumulative frequency 
25
27
30
32
35
40
2
5
10
12
8
4
2
7
17
29
37
41

∵ Total number of observations = 41, (an odd number)

∴ Median of marks = marks obtained byAdditional Questions Solutions- Statistics Class 9 Notes | EduRev student = marks obtained by the 21st student
∵ 18th to 29th students get 22 marks. i.e. 21st student gets 22 marks.
∴ The median of the marks = 22.


Question 6. Prove that   Additional Questions Solutions- Statistics Class 9 Notes | EduRev = 0, where Additional Questions Solutions- Statistics Class 9 Notes | EduRev is the mean of the ‘n’ observations x1, x2, x3, x4,
…, xn.
Additional Questions Solutions- Statistics Class 9 Notes | EduRev

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