NCERT Class 10 Maths – Exercise 13.3 Solutions

NCERT Class 10 Maths

Chapter 13 – Statistics | Exercise 13.3

(Rationalized Syllabus 2025-26)

💡 Formula for Median

$$ \text{Median} = l + \left( \frac{\frac{n}{2} – cf}{f} \right) \times h $$
  • $l$: Lower limit of median class
  • $n$: Number of observations
  • $cf$: Cumulative frequency of class preceding the median class
  • $f$: Frequency of median class
  • $h$: Class size
Q1

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.

1. Finding Median
$n = 68 \Rightarrow n/2 = 34$. Cumulative frequency just greater than 34 is 42.
Median Class: 125 – 145
  • $l = 125, h = 20, f = 20, cf = 22$
$$ \text{Median} = 125 + \left( \frac{34 – 22}{20} \right) \times 20 $$ $$ = 125 + 12 = 137 \text{ units} $$
2. Summary of Other Measures
Mean: $\approx 137.05$ units.
Mode: $135.76$ units.
✔️ Median = 137 units
✔️ Mean = 137.05 units
✔️ Mode = 135.76 units
Q2

If the median of the distribution given below is 28.5, find the values of x and y. (Total frequency = 60)

Table Construction
Class IntervalFrequency ($f$)Cumulative Freq ($cf$)
0-1055
10-20x5 + x
20-302025 + x
30-401540 + x
40-50y40 + x + y
50-60545 + x + y
Calculation
Given $n=60$, so $45 + x + y = 60 \Rightarrow x + y = 15$.
Median = 28.5, so Median Class is 20 – 30.
  • $l = 20, f = 20, cf = 5 + x, h = 10$
$$ 28.5 = 20 + \left( \frac{30 – (5 + x)}{20} \right) \times 10 $$ $$ 8.5 = \frac{25 – x}{2} \Rightarrow 17 = 25 – x \Rightarrow x = 8 $$
Since $x + y = 15 \Rightarrow 8 + y = 15 \Rightarrow y = 7$.
✔️ $x = 8, y = 7$
Q3

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.

1. Convert to Class Intervals
AgecfFrequency ($f$)
15-2022
20-2566 – 2 = 4
25-302424 – 6 = 18
30-354545 – 24 = 21
35-407878 – 45 = 33
40-458989 – 78 = 11
45-509292 – 89 = 3
50-559898 – 92 = 6
55-60100100 – 98 = 2
2. Calculation
$n = 100 \Rightarrow n/2 = 50$. Median Class is 35 – 40 ($cf > 50$ is 78).
  • $l = 35, h = 5, f = 33, cf = 45$
$$ \text{Median} = 35 + \left( \frac{50 – 45}{33} \right) \times 5 $$ $$ = 35 + \frac{25}{33} = 35 + 0.76 $$
✔️ Median Age = 35.76 years
Q4

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.

1. Adjustment
Data is discontinuous (118-126, 127-135). Subtract 0.5 from lower limit and add 0.5 to upper limit.
$n = 40 \Rightarrow n/2 = 20$. Median Class is 144.5 – 153.5.
  • $l = 144.5, h = 9, f = 12, cf = 17$
2. Calculation
$$ \text{Median} = 144.5 + \left( \frac{20 – 17}{12} \right) \times 9 $$ $$ = 144.5 + \frac{3}{12} \times 9 = 144.5 + 2.25 $$
✔️ Median Length = 146.75 mm
Q5

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

Calculation
$n = 400 \Rightarrow n/2 = 200$.
From cumulative frequencies, the class exceeding 200 is 3000 – 3500.
  • $l = 3000, h = 500, f = 86, cf = 130$
$$ \text{Median} = 3000 + \left( \frac{200 – 130}{86} \right) \times 500 $$ $$ = 3000 + \frac{70}{86} \times 500 = 3000 + 406.98 $$
✔️ Median Life Time = 3406.98 hours
Q6

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English surnames was obtained. Determine the median number of letters in the surnames.

Calculation
$n = 100 \Rightarrow n/2 = 50$.
Cumulative freq just > 50 is 76. Median Class is 7 – 10.
  • $l = 7, h = 3, f = 40, cf = 36$
$$ \text{Median} = 7 + \left( \frac{50 – 36}{40} \right) \times 3 $$ $$ = 7 + \frac{14}{40} \times 3 = 7 + 1.05 $$
✔️ Median = 8.05 letters
Q7

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

Calculation
$n = 30 \Rightarrow n/2 = 15$.
Cumulative freq just > 15 is 19. Median Class is 55 – 60.
  • $l = 55, h = 5, f = 6, cf = 13$
$$ \text{Median} = 55 + \left( \frac{15 – 13}{6} \right) \times 5 $$ $$ = 55 + \frac{2}{6} \times 5 = 55 + 1.666… $$
✔️ Median Weight = 56.67 kg
🎉 Exercise 13.3 Completed | Chapter 13 Finished!
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