NCERT Solutions class 10 maths chapter 14
Statistics
NCERT Solutions class 10 maths chapter 14 Exercise 14.3
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.
Answer 1
For Mean
Mean x̄ =
=135 + 20 x (
7
/
68
)
= 135 + 2.06
Mean=137.06
Here for median
n = 68 and
n
/
2
= 34So the median class is 125-145 with cumulative frequency = 42
l = 125, n = 68, Cf = 22, f = 20, h = 20
=125+(
(34−22)
/
20
) × 20=125+12 = 137
here median = 137
For mode
Modal class = 125-145,
f1=20, f2=14 , f0=13 and h = 20
Mode =
Mode = 125 + (
(20-13)
/
(40-13-14)
)×20=125+(
140
/
13
)=125+10.77
=135.77
mode = 135.77
Question 2
If the median of the distribution given below is 28.5, find the values of x and y.
Answer 2
Here We Have
45 + x + y = 60 [ n = 60 ]
x + y =60 -45
x + y =15 -------------(i)
Median of the given data = 28.5
n
/
2
= 30 Median lie between in class interval (20 - 30) Median class is 20 – 30 with a cf = 25+x
l = 20, Cf = 5+x, f = 20 & h = 10
On putting value
28.5=20+(
(30−5−x)
/
20
) × 108.5 =
(25 – x)
/
2
17 = 25-x
x =8
From eq(i)
x + y = 15
y = 15 - 8
y = 7
the value of x = 8 and y = 7
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.
Answer 3
Here n = 100 and
n
/
2
= 50 Median class = 35-45
l = 35, cf = 45, f = 33 and h = 5
Median = 35+(
(50-45)
/
33
) × 5 = 35 + (
5
/
33
) x 5 = 35.75
the median age = 35.75
Question 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 :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.)
Answer 4
n = 40 and
n
/
2
=
40
/
2
=20 Median class = 144.5-153.5 then, l = 144.5, cf = 17, f = 12 and h = 9
Median = 144.5+(
(20-17)
/
12
) × 9 = 144.5 + (
9
/
4
) = 146.75 <
, the median length of the leaves = 146.75 mm.
Question 5
The following table gives the distribution of the life time of 400 neon lamps :Find the median life time of a lamp
Answer 5
Given
n = 400 and n/2
9
/
4
=
400
/
2
=200 Median class = 3000 – 3500 , l = 3000, Cf = 130,
f = 86 and h = 500
Median = 3000 + (
(200-130)
/
86
) × 500 = 3000 + (
35000
/
86
) = 3000 + 406.97 = 3406.97
the median life time of the lamps = 3406.97 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:
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.
Answer 6
Givenn = 100 and
n
/
2
=
100
/
2
= 50
Median class = 7-10 ,l = 7, Cf = 36, f = 40 & h = 3 Median = 7+(
(50-36)
/
40
) × 3 Median = 7+
42
/
40
Median=8.05
For Mode:
Modal class = 7-10, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3
Mode =
Mode = 7+(
(40-30)
/
(2×40-30-16)
) × 3 = 7+(
30
/
34
)
= 7.88 mode = 7.88
Mean =
Mean = 825/100 = 8.25
mean = 8.25
Question 7
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Answer 7
n = 30 and
n
/
2
= 15 Median class = 55-60 , l = 55, Cf = 13, f = 6 and h = 5
Median = 55+(
(15-13)
/
6
)×5 Median=55 + (
10
/
6
) = 55+1.666 Median =56.67
The median weight of the students = 56.67