NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.3.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 14
Chapter NameStatistics
ExerciseEx 14.3
Number of Questions Solved7
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

NCERT Solutions for Class 10 Maths

Question 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.

Monthly consumption (in units)Number of consumers
65 – 854
85 – 1055
105 – 12513
125 – 14520
145 – 16514
165 – 1858
185 – 2054

Solution:
We may find class marks by using the relation

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 1s
Taking 135 as assumed mean (a) we may find di, ui, fiui, according to step deviation method as following

Monthly consumption 
(in units)
Number of consumers (i)xi class markdi xi – 135NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 1s.1 fiui 
65 – 85475– 60– 3– 12
85 – 105595– 40– 2– 10
105 – 12513115– 20– 1– 13
125 – 14520135000
145 – 1651415520114
165 – 185817540216
185 – 205419560312
Total687

From the table we may observe that

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 1s.2

Now from table it is clear that maximum class frequency is 20 belonging to class interval 125 – 145.
Modal class = 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 1s.3
We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.

Question 2.
If the median of the distribution is given below is 28.5, find the values of x and y.

Class intervalFrequency
0 – 105
10 – 20x
20 – 3020
30 – 4015
40 – 50y
50 – 605
Total60

Solution:
We may find cumulative frequency for the given data as following

Class intervalFrequencyCumulative frequency 
0 – 1055
10 – 20x5 + x
20 – 302025 + x
30 – 401540 + x
40 – 5y40 + x + y
50 – 60545 + x + y
Total (n)60

It is clear that n = 60

45 + x + y = 60

x + y = 15        (1)
Median of data is given as 28.5 which lies in interval 20 – 30.
So, median class = 20 – 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 2s
From equation (1)
8 + y = 15
y = 7
Hence values of x and y are 8 and 7 respectively.

Question 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 onwards but less than 60 year.

Age (in years)Number of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:
Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below

Age (in years)Number of policy holders (fi)Cumulative frequency (cf)
18 – 2022
20 – 256 – 2 = 46
25 – 3024 – 6 = 1824
30 – 3545 – 24 = 2145
35 – 4078 – 45 = 3378
40 – 4589 – 78 = 1189
45 – 5092 – 89 = 392
50 – 5598 – 92 = 698
55 – 60100 – 98 = 2100
Total (n)

Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 3s is 78 belonging to interval 35 – 40
So, median class = 35 – 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 3s1
So, median age is 35.76 years.

Question 4.

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)Number or leaves fi
118 – 1263
127 – 1355
136 – 1449
145 – 15312
154 – 1625
163 – 1714
172 – 1802

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)
Solution:

The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract 1/2=0.5  to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below

Length (in mm)Number or leaves fiCumulative frequency
117.5 – 126.533
126.5 – 135.553 + 5 = 8
135.5 – 144.598 + 9 = 17
144.5 – 153.51217 + 12 = 29
153.5 – 162.5529 + 5 = 34
162.5 – 171.5434 + 4 = 38
171.5 – 180.5238 + 2 = 40

From the table we may observe that cumulative frequency just greater then NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 4s  is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 4s1
So, median length of leaves is 146.75 mm.

Question 5.
Find the following tables gives the distribution of the life time of 400 neon lamps:

Life time (in hours)Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median life time of a lamp.
Solution:
We can find cumulative frequencies with their respective class intervals as below –

Life timeNumber of lamps (fi)Cumulative frequency
1500 – 20001414
2000 – 25005614 + 56 = 70
2500 – 30006070 + 60 = 130
3000 – 350086130 + 86 = 216
3500 – 400074216 + 74 = 290
4000 – 450062290 + 62 = 352
4500 – 500048352 + 48 = 400
Total (n)400

Now we may observe that cumulative frequency just greater than NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 5s is 216 belonging to class interval 3000 – 3500.
Median class = 3000 – 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 5s1
So, median life time of lamps is 3406.98 hours.

Question 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:

Number of letters1 – 44 – 77 – 1010 – 1313 – 1616 – 19
Number of surnames63040644

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.
Solution:
We can find cumulative frequencies with their respective class intervals as below

Number of lettersFrequency (fi)Cumulative frequency
1 – 466
4 – 73030 + 6 = 36
7 – 104036 + 40 = 76
10 – 131676 + 16 = 92
13 – 16492 + 4 = 96
16 – 19496 + 4 = 100
Total (n)100

Now we may observe that cumulative frequency just greater than  is 76 belonging to class NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 6s1 interval 7 – 10.
Median class = 7 – 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 6s2
Now we can find class marks of given class intervals by using relation
NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 6s3

Taking 11.5 as assumed mean (a) we can find di, ui and fiui according to step deviation method as below.

Number of
letters
Number of
surnames
xixi – a NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 6s4fiui
1 – 462.5-9-3-18
4 – 7305.5-6-2-60
7 – 10408.5-3-1-40
10 – 131611.5000
13 – 16414.5314
16 – 19417.5628
Total100-106

NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 6s5

We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 – 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.

Question 7

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight 
(in kg)
40 – 4545 – 5050 – 5555 – 6060 – 6565 – 7070 – 75
Number of students2386632

Solution:
We may find cumulative frequencies with their respective class intervals as below
Cumulative frequency just greater than NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 7s is 19, belonging to class interval 55 – 60.
Median class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
NCERT Solutions for Class 10 Maths Chapter 14 Statistics ex 14.3 7s1
So, median weight is 56.67 kg.

 

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