Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

The document Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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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.
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Sol. 
Median 
Let us prepare a cumulative frequency table:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Now, we have  Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
∵ 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,
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Mean Assumed mean (a) = 135
∵ Class interval (h) = 20
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Now, we have the following table:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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.
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Sol. Here, we have n = 60 [∵ ∑fi = 60]
Now, cumulative frequency table is:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Since, median = 28.5
∴ Median class is 20 − 30
We have: l = 20
h = 10
f = 20
cf = 5 + x
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
⇒ 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.
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Sol. The given table is cumulative frequency distribution. We write the frequency distribution as given below:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
∵ 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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
∑ fi =40 ⇒ n = 40
Now,
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

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

Sol. To compute the median, let us write the cumulative frequency distribution as given below:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
∑ fi = 400 ⇒ n = 400
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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:
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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  Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
l = 7
cf = 36
f = 40 and  h = 3
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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.
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev

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

Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
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
Ex 14.3 NCERT Solutions- Statistics Class 10 Notes | EduRev
Thus, the required median weight = 56.67 kg.

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