Maths Class 10 Exercise 14.3
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 – 85 |
4 |
85 – 105 |
5 |
105 – 125 |
13 |
125 – 145 |
20 |
145 – 165 |
14 |
165 – 185 |
8 |
185 – 205 |
4 |
Solution: To calculate median:
Therefore, the median class = 125 – 145
So,
l = 125, n = 68, f = 20, cf = 22 and h = 20
=
125 + (34 – 22)/20 × 20
=
125 + 12
=
137
To calculate mean:
Width of the
class (h) =
20
= 7/68
= 0.102
= 135 + 20(0.102)
= 135 + 2.04
= 137.04
To calculate mode:
In the given
data, maximum frequency is 20 and it corresponds to the class interval 125 –
145.
Therefore,
the modal class
= 125 – 145
And l = 125, f_{1 }= 20, f_{0
}= 13, f_{2 }= 14 and h = 20
= 125 + 140/13
= 125 + 10.7692
= 125 + 10.77
= 135.77
Hence, median,
mean and mode of the given data is 137 units, 137.04 units and 135.77 units
respectively.
2. If the median of the
distribution given below is 28.5, then find the values of x and 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:
It is given that
the median of the distribution is 28.5, which lies in the interval 20 – 30.
Therefore, the median class = 20 – 30
So,
l =20, n = 60, f = 20, cf = 5 + x and h = 10
Since, 45 + x + y = 60
⇒ x + y
= 15 ………. (i)
⇒ 2(28.5) = 65 – x
⇒ 57 = 65 – x
⇒ x
= 65 – 57
⇒ x = 8
Putting
the value of x in equation
(i), we get
8
+ y = 15
⇒ y = 7
Hence,
the value of x and y is 8 and 7 respectively.
3. A life insurance agent found the following
data for distribution of ages of 100 policy holders. Calculate the median age,
if policies are only given to persons having age 18 years onwards but less than
60 years.
Age (in
years) |
Number of
policy holders |
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:
Here, Î£f_{i} = n = 100, therefore n/2
= 100/2 = 50, which lies in interval 35 – 40.
So,
Median class = 35 – 40
So,
l = 35, n = 100, f = 33, cf = 45 and h = 5
=
35 + 25/33
=
35 + 0.7575
=
35 + 0.76 (approx.)
=
35.76
Hence,
median age of given data is 35.76 years.
4. The lengths of 40 leaves of a plant are
measured correct to the nearest millimetre, and data obtained is represented in
the following table. Find the median length of the leaves.
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 |
Solution: Since the frequency distribution is
not continuous, so first of all, we shall change it into continuous
distribution.
Therefore,
Median class = 144.5 – 153.5
So,
l = 144.5, n = 40, f = 12, cf = 17 and h = 9
=
144.5 + 2.25
=
146.75
Hence,
median length of the leaves is 146.75 mm.
5. The following table gives the distribution
of the life time of 400 neon lamps. Find the median life time of a lamp.
Life time
(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 |
Solution:
Here, Î£f_{i} = n = 400, therefore n/2
= 400/2 = 200, which lies in interval 3000 – 3500.
Therefore, Median
class = 3000 – 3500
So,
l = 3000, n = 400, f = 86, cf = 130 and h = 500
=
3000 + 406.9767441
=
3000 + 406.98 (approx.)
=
3406.98
Hence,
median life time of a lamp is 3406.98 hours.
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 letters |
1 – 4 |
4 – 7 |
7 – 10 |
10 – 13 |
13 – 16 |
16 – 19 |
Number
of surnames |
6 |
30 |
40 |
16 |
4 |
4 |
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: To calculate median:
Therefore, Median
class = 7 – 10
So,
l = 7, n = 100, f = 40, cf = 36 and h = 3
=
7 + 1.05
=
8.05
To calculate mean:
From given data, assumed
mean (a) =
8.5
Width of the
class (h) =
3
= –6/100
= –0.06
= 8.5 + 3(– 0.06)
= 8.5 – 0.18
= 8.32
To calculate mode:
In the given
data, maximum frequency is 40 and it corresponds to the class interval 7 – 10.
Therefore,
the modal class
= 7 – 10
And l = 7, f_{1 }= 40, f_{0
}= 30, f_{2 }= 16 and h = 3
= 7 + 30/34
= 7 + 0.88
(approx.)
= 7.88
Hence, median,
mean and mode of the given data is 8.05 letters, 8.32 letters and 7.88 letters,
respectively.
7. The distribution below gives the weights of
30 students of a class. Find the median weight of the students.
Weight (in kg) |
Number of students |
40 – 45 |
2 |
45 – 50 |
3 |
50 – 55 |
8 |
55 – 60 |
6 |
60 – 65 |
6 |
65 – 70 |
3 |
70 – 75 |
2 |
Solution:
Here,
Î£f_{i}
= n = 30, therefore n/2 = 30/2 = 15, which lies in interval 55 – 60.
Therefore, the median class = 55 – 60
So,
l = 55, n = 30, f = 6, cf = 13 and h = 5
=
55 + 10/6
=
55 + 1.66666
=
55 + 1.67 (approx.)
=
56.67
Hence,
median weight of the students is 56.67 kg.