NCERT Solutions Class 10 Maths Chapter 14 Statistics Ex 14.3

NCERT Solutions Class 10 Maths 
Chapter – 14 (Statistics) 

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 14 Statistics Exercise 14.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 14 Statistics

• NCERT Class 10 Maths Solution Ex – 14.1
• NCERT Class 10 Maths Solution Ex – 14.2
• NCERT Class 10 Maths Solution Ex – 14.4

Exercise – 14.3

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.

Solution –  Find the cumulative frequency of the given data as follows:

From the table, it is observed that,
N = 68 and hence N/2 = 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 =
= 125 + [(34 − 22)/20] × 20
= 125 + 12
= 137
Therefore, median = 137
To calculate the mode:
Modal class = 125-145,
f_m or f₁ = 20,
f₀ = 13,
f₂ = 14
h = 20

Mode formula :
Mode = l+ [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Mode = 125 + [(20 – 13)/ (40 – 13 – 14)] × 20
= 125 + (140/13)
= 125 + 10.77
= 135.77
Therefore, mode = 135.77
Calculate the Mean :

x̄ = a + h (∑f_iu_i/∑f_i)
= 135 + 20 (7/68)
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.

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

Solution –  Given data,
n = 60
Median of the given data = 28.5

Where, N/2 = 30
Median class is 20 – 30 with a cumulative frequency = 25 + x
Lower limit of median class, l = 20,
cf = 5 + x,
f = 20 & h = 10
Median =
28.5 = 20 + [(30 − 5 − x)/20] × 10
8.5 = (25 – x)/2
17 = 25 – x
Therefore, x = 8.
Now, from cumulative frequency, we can identify the value of x + y as follows:
Since,
60 = 45 + x + y
Now, substitute the value of x, to find y
60 = 45 + 8 + y
y = 60 – 53
y = 7
Therefore, the value of x = 8 and y = 7.

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.

Solution –

Given data :
N = 100 and N/2 = 50
Median class = 35 – 40
l = 35,
cf = 45,
f = 33
h = 5
Median =
Median = 35 + [(50 – 45)/33] × 5
= 35 + (25/33)
= 35.76
Therefore, the median age = 35.76 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as 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.)
Solution –  Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

So, the data obtained are:
N = 40 and N/2 = 20
Median class = 144.5 – 153.5
l = 144.5,
cf = 17,
f = 12
h = 9
Median =
Median = 144.5 + [(20 – 17)/ 12] × 9
= 144.5 + (9/4)
= 146.75 mm
Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a lifetime of 400 neon lamps.

Find the median lifetime of a lamp.

Solution –

N = 400 & N/2 = 200
Median class = 3000 – 3500
l = 3000,
cf = 130,
f = 86
h = 500
Median =
Median = 3000 + [(200 – 130)/86] × 500
= 3000 + (35000/86)
= 3000 + 406.98
= 3406.98
Therefore, the median lifetime of the lamps = 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:

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:

N = 100 & N/2 = 50
Median class = 7 – 10
l = 7,
cf = 36,
f = 40
h = 3
Median =
Median = 7 + [(50 – 36)/40] × 3
Median = 7 + (42/40)
Median = 8.05
Calculate the Mode:
Modal class = 7-10,
l = 7,
f₁ = 40,
f₀ = 30,
f₂ = 16
h = 3

Mode = 7 + [(40 – 30)/(2 × 40 – 30 – 16)] × 3
= 7 + (30/34)
= 7.88
Therefore mode = 7.88
Calculate the Mean :

Mean = x̄ = ∑f_i x_i /∑f_i
Mean = 832/100 = 8.32
Therefore, mean = 8.32

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

Solution –

N = 30 and N/2= 15
Median class = 55 – 60
l = 55,
C_f = 13,
f = 6
h = 5
Median =
Median = 55 + [(15 – 13)/6] × 5
= 55 + (10/6)
= 55 + 1.666
= 56.67
Therefore, the median weight of the students = 56.67

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