Class 10  >  Mathematics (Maths) Class 10  >  NCERT Solutions: Statistics (Exercise 14.3)

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

Document Description: NCERT Solutions: Statistics (Exercise 14.3) for Class 10 2022 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for NCERT Solutions: Statistics (Exercise 14.3) have been prepared according to the Class 10 exam syllabus. Information about NCERT Solutions: Statistics (Exercise 14.3) covers topics like and NCERT Solutions: Statistics (Exercise 14.3) Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Statistics (Exercise 14.3).

Introduction of NCERT Solutions: Statistics (Exercise 14.3) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & NCERT Solutions: Statistics (Exercise 14.3) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
1 Crore+ students have signed up on EduRev. Have you?

NCERT TEXTBOOK QUESTIONS SOLVED
Page No. 287 - 288
EXERCISE 14.3

Q 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Sol. 
Median 
Let us prepare a cumulative frequency table:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

Now, we have  NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
∵ This observation lies in the class 125−145.
∴ 125−145 is the median class.
∴ l = 125, cf = 22
f = 20 and h = 20
Using the formula,
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Mean Assumed mean (a) = 135
∵ Class interval (h) = 20
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Now, we have the following table:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Mode 
∵ Class 125−145 has the highest frequency.
∴ This is the modal class.
We have:
h = 20
l = 125
f1 = 20
f0 = 13
f2 = 14
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
We observe that the three measures are approximately equal in this case.

Q 2. If the median of the distribution given below is 28.5, find the values of x and y.
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Sol. Here, we have n = 60 [∵ ∑fi = 60]
Now, cumulative frequency table is:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Since, median = 28.5
∴ Median class is 20 − 30
We have: l = 20
h = 10
f = 20
cf = 5 + x
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
⇒ 57 = 40 + 25 − x
⇒ x = 40 + 25 − 57 = 8
Also 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 − 45 − 8 = 7
Thus, x = 8
y = 7

Q 3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years on wards but less than 60 years.
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Sol. The given table is cumulative frequency distribution. We write the frequency distribution as given below:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
∵ The cumulative frequency just greater than n/2 i.e., just greater than 50 is 78.
∴ The median class is 78.
Now n/2 = 50, l = 35, cf = 45, f = 33 and h = 5
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Thus, the median age = 35.76 years.

Q 4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5−126.5, 126.5−135.5, ..., 171.5−180.5.]

Sol. After changing the given table as continuous classes we prepare the cumulative frequency table:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
∑ fi =40 ⇒ n = 40
Now,
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
The cumulative frequency just above n/2 i.e., 20 is 29 and it corresponds to the class 144.5−153.5.
So, 144.5−153.5 is the median class.
We have:
n/2 = 20, l = 144.5, f = 12, cf = 17 and h = 9
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

Page No. 289
Q 5. The following table gives the distribution of the life time of 400 neon lamps:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Find the median life time of a lamp

Sol. To compute the median, let us write the cumulative frequency distribution as given below:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
∑ fi = 400 ⇒ n = 400
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Since, the cumulative frequency just greater than n/2 i.e., greater than 200 is 216.
∴ The median class is 3000−3500
∴ l = 3000, cf = 130, f = 86, h = 500 and n/2 = 200
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Thus, median life = 3406.98 hours.

Q 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Sol. Median
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Since, the cumulative frequency just greater than n/2 i.e., greater than 50 is 76.
∴ The class 7 − 10 is the median class,
We have  NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
l = 7
cf = 36
f = 40 and  h = 3
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Mode
Since the class 7−10 has the maximum frequency.
∴ The modal class is 7−10
So, we have
l = 7, h = 3
f1 = 40,
f0 = 30
f2 = 16
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Thus, the required Median = 8.05,
Mean = 8.32 and Mode = 7.88.

Q 7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

Sol. We have

Weight (in kg)

Frequency

(f)

Cumulative

frequency

40-45

2

2 + 0= 2

45-50

3

2 + 3= 5

50-55

8

5 + 8 = 13

55-60

6

13 + 6 = 19

60-65

6

19 + 6 = 25

65-70

3

25 + 3 = 28

70-75

2

28 + 2 = 30

Total

If  = 30

n = 30

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
The cumulative frequency just more than n/2 i.e., more than 15 is 19, which corresponds to the class 55−60.
n/2 = 15.
l = 5
5f = 6
cf = 13 and h = 5
NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10
Thus, the required median weight = 56.67 kg.

The document NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

Related Searches

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

practice quizzes

,

video lectures

,

MCQs

,

study material

,

past year papers

,

Exam

,

mock tests for examination

,

Important questions

,

shortcuts and tricks

,

Viva Questions

,

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

NCERT Solutions: Statistics (Exercise 14.3) Notes | Study Mathematics (Maths) Class 10 - Class 10

,

Semester Notes

,

Objective type Questions

,

pdf

,

Sample Paper

,

Previous Year Questions with Solutions

,

Free

,

ppt

,

Extra Questions

,

Summary

;