Ncert Solutions For Class 10 Maths Ex 14.3

Ncert Solutions For Class 10 Maths Chapter 14 Ex 14.3

Question 1:

The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption(in units) No. of customers
65-85 4
85-105 5
105-125 13
125-145 20
145-165 14
165-185 8
185-205 4

 Solution:

     Class Interval             Frequency Cumulative frequency
65-85 4 4
85-105 5 9
105-125 13 22
125-145 20 42
145-165 14 56
165-185 8 64
185-205 4 68
N=68

Where, n = 68 and hence n2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

Median=l+(n2cff)Xh

=125+(342220)X20

=125+12=137

 

Mode = Modal class = 125-145, f1=20,f0=13,f2=14 & h = 20

Mode=l+(f1f02Xf1f0f2)Xh

=125+(20132X201314)X20

=125+713X20

=125+10.77

=135.77

Calculate the Mean:

Class Interval            fi        xi   di=xi-a ui=di/h fiui
65-85 4 75 -60 -3 -12
85-105 5 95 -40 -2 -10
105-125 13 115 -20 -1 -13
125-145 20 135 0 0 0
145-165 14 155 20 1 14
165-185 8 175 40 2 16
185-205 4 195 60 3 12
Sum fi= 68 Sum fiui = 7

 

x¯=a+fiuifiXh =135+768X20

=137.05

Mean, median and mode are more/less equal in this distribution.

 

Question 2:

If the median of a distribution given below is 28.5 then, find the value of an x &y.

      Class Interval        Frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60

Solution: 

n = 60

Where,n2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class = 20, cf = 5 + x , f = 20 & h = 10

Median=l+(n2cff)Xh

Or,28.5=20+(305x20)X10

Or,25x2

Or,   25-x=17

Or,    x= 25 – 17   =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

60=5+20+15+5+x+y

Or, 45+x+y=60

Or, x + y=60-45=15

Hence, y = 15 – x= 15 – 8 = 7

Hence, x = 8  &  y = 7

 

Question 3:

The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the  persons whose age is 18 years onwards but less than the 60 years.

Age(in years) Number of policy holder
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

Solution:

Class interval Frequency Cumulative frequency
15-20 2 2
20-25 4 6
25-30 18 24
30-35 21 45
35-40 33 78
40-45 11 89
45-50 3 92
50-55 6 98
55-60 2 100

Where, n = 100 and n2 = 50

Median class = 35-45

Then, l = 35, cf = 45, f = 33 & h = 5

Median=l+(n2cff)Xh

=35+(504533)X5 =35+(2533)

=35.75

 

Question 4:

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

Length(in mm) Number of leaves
118-126 3
127-135 5
136-144 9
145-153 12
154-162 5
163-171 4
172-180 2

Find the median length of leaves.             

Solution:

Class Interval Frequency Cumulative frequency
117.5-126.5 3 3
126.5-135.5 5 8
135.5-144.5 9 17
144.5-153.5 12 29
153.5-162.5 5 34
162.5-171.5 4 38
171.5-180.5 2 40

Where, n = 40 and n2 = 20

Median class = 144.5-153.5

then, l = 144.5, cf = 17, f = 12 & h = 9

Median=l+(n2cff)Xh

=144.5+(201712)X9 =144.5+(94)

=146.75

 

Question 5:

The following table gives the distribution of a life time of 400 neon lamps.

Lifetime(in hours) Number of lamps
1500-2000 14
2000-2500 56
2500-3000 60
3000-3500 86
3500-4000 74
4000-4500 62
4500-5000 48

Find the median lifetime of a lamp.

Solution:

Class Interval Frequency Cumulative
1500-2000 14 14
2000-2500 56 70
2500-3000 60 130
3000-3500 86 216
3500-4000 74 290
4000-4500 62 352
4500-5000 48 400

Where, n = 400 &n2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130, f = 86 & h = 500

Median=l+(n2cff)Xh

=3000+(20013086)X500

=3000+406.97

=3406.97

 

Question 6:

In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Number of letters 1-4 4-7 7-10 10-13 13-16 16-19
Number of surnames 6 30 40 16 4 4

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution: To Calculate median:

Class Interval Frequency Cumulative Frequency
1-4 6 6
4-7 30 36
7-10 40 76
10-13 16 92
13-16 4 96
16-19 4 100

Where, n = 100 &n2 = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

Median=l+(n2cff)Xh

=7+(503640)X3 =7+(1440)X3

=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode=l+(f1f02Xf1f0f2)Xh

=7+(40302X403016)X3

=7+1034X3

=7.88

Calculate the Mean:

 

Class Interval fi xi fixi
1-4 6 2.5 15
4-7 30 5.5 165
7-10 40 8.5 340
10-13 16 11.5 184
13-16 4 14.5 51
16-19 4 17.5 70
Sum fi = 100 Sum fixi = 825

 

x¯=a+fiuifiXh =825100

=8.25

 

Q7. The distributions of below gives a weight of 30 students of a class. Find the median weight of a students.

Weight(in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Number of students 2 3 8 6 6 3 2

 

Solution:

Class Interval Frequency Cumulative frequency
40-45 2 2
45-50 3 5
50-55 8 13
55-60 6 19
60-65 6 25
65-70 3 28
70-75 2 30

Where, n = 30 and n2= 15

median class = 55-60

therefore, l = 55, cf = 13, f = 6 & h = 5

Median=l+(n2cff)Xh

=55+(15136)X5 =55+(53)

=56.67

 

Question 7:

The distribution shows the pocket money of students in an area. The mean pocket money is Rs 18. Find the missing frequency f.

 

Daily pocket allowance (in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Number of children 07 06 09 013 F 05 04

Solution:

Class interval fi xi fixi
11-13 07 012 84
13-15 06 014 84
15-17 09 016 144
17-19 013 018 234
19-21 f 020 20f
21-23 05 022 110
23-25 04 024 96
Sum fi = 44+f Sum fixi = 752+20f

 

We have,

Mean pocket money, xi = 18(given)

Hence substituting for fixi,

 

18(44+f) = 752 + 20f

Or, 792 + 18f = 752 + 20f

Or, 2f = 40

f=20
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