NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.3#
Exercise 13.3 covers finding the median of grouped data using the median formula.
Median of Grouped Data:
$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$where $l$ = lower class limit of median class, $n$ = total frequency, $cf$ = cumulative frequency of class preceding median class, $f$ = frequency of median class, $h$ = class size. The median class is the class where cumulative frequency first exceeds $\dfrac{n}{2}$.
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
| Monthly consumption (units) | $f_i$ | $cf$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|---|
| 65-85 | 4 | 4 | 75 | 300 |
| 85-105 | 5 | 9 | 95 | 475 |
| 105-125 | 13 | 22 | 115 | 1495 |
| 125-145 | 20 | 42 | 135 | 2700 |
| 145-165 | 14 | 56 | 155 | 2170 |
| 165-185 | 8 | 64 | 175 | 1400 |
| 185-205 | 4 | 68 | 195 | 780 |
| Total | 68 | 9320 |
Mean: $\bar{x} = \dfrac{9320}{68} = 137.06$ units
Median: $n = 68$, $\dfrac{n}{2} = 34$
Median class: 125-145 (cf before = 22, cf after = 42 ≥ 34)
$l = 125$, $cf = 22$, $f = 20$, $h = 20$
$$\text{Median} = 125 + \frac{34-22}{20} \times 20 = 125 + 12 = 137$$Mode: Modal class = 125-145
$l = 125$, $f_1 = 20$, $f_0 = 13$, $f_2 = 14$, $h = 20$
$$\text{Mode} = 125 + \frac{7}{13} \times 20 = 125 + 10.77 = 135.77$$Comparison: Mean ≈ 137.06, Median = 137, Mode ≈ 135.77. All three are approximately equal (~137), indicating a nearly symmetric distribution.
2. If the median of the distribution given below is 28.5, find the values of x and y. #
Answer
| Class interval | $f_i$ | $cf$ |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | x | 5+x |
| 20-30 | 20 | 25+x |
| 30-40 | 15 | 40+x |
| 40-50 | y | 40+x+y |
| 50-60 | 5 | 45+x+y |
| Total | 60 |
$\sum f_i = 5 + x + 20 + 15 + y + 5 = 45 + x + y = 60$
$$x + y = 15 \quad \cdots (1)$$Median = 28.5, so $\dfrac{n}{2} = 30$.
Median class = 20-30 (cf = 25+x, need cf ≥ 30 → x should be such that 5+x < 30 and 25+x ≥ 30 → x < 25 and x ≥ 5, so median class is indeed 20-30).
$l = 20$, $cf = 5+x$, $f = 20$, $h = 10$
$$28.5 = 20 + \frac{30-(5+x)}{20} \times 10 = 20 + \frac{25-x}{2}$$$$8.5 = \frac{25-x}{2}$$$$x = 25 - 17 = 8$$From (1): $y = 15 - 8 = 7$
∴ $x = 8$, $y = 7$.
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 years. #
Answer
| Age (years) | Number of policy holders | $cf$ |
|---|---|---|
| 18-20 | 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 |
$n = 100$, $\dfrac{n}{2} = 50$
Median class: 35-40 (cf = 45 < 50, next cf = 78 ≥ 50)
$l = 35$, $cf = 45$, $f = 33$, $h = 5$
$$\text{Median} = 35 + \frac{50-45}{33} \times 5 = 35 + \frac{25}{33} = 35 + 0.76 = 35.76 \text{ years}$$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. #
Answer
(Note: Convert to actual class intervals — “118-126” means 117.5-126.5 in continuous form)
| Length (mm) | $f_i$ | $cf$ |
|---|---|---|
| 118-126 | 3 | 3 |
| 127-135 | 5 | 8 |
| 136-144 | 9 | 17 |
| 145-153 | 12 | 29 |
| 154-162 | 5 | 34 |
| 163-171 | 4 | 38 |
| 172-180 | 2 | 40 |
Using actual boundaries (117.5-126.5, etc.), $h = 9$.
$n = 40$, $\dfrac{n}{2} = 20$
Median class: 145-153 (actual: 144.5-153.5), cf = 17 < 20, next = 29 ≥ 20.
$l = 144.5$, $cf = 17$, $f = 12$, $h = 9$
$$\text{Median} = 144.5 + \frac{20-17}{12} \times 9 = 144.5 + \frac{27}{12} = 144.5 + 2.25 = 146.75 \text{ mm}$$5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp. #
Answer
| Life time (hours) | $f_i$ | $cf$ |
|---|---|---|
| 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 |
$n = 400$, $\dfrac{n}{2} = 200$
Median class: 3000-3500 (cf = 130 < 200, next = 216 ≥ 200)
$l = 3000$, $cf = 130$, $f = 86$, $h = 500$
$$\text{Median} = 3000 + \frac{200-130}{86} \times 500 = 3000 + \frac{70 \times 500}{86} = 3000 + \frac{35000}{86} = 3000 + 406.98 = 3406.98 \text{ hours}$$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. #
Answer
| Number of letters | $f_i$ | $cf$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|---|
| 1-4 | 6 | 6 | 2.5 | 15 |
| 4-7 | 30 | 36 | 5.5 | 165 |
| 7-10 | 40 | 76 | 8.5 | 340 |
| 10-13 | 16 | 92 | 11.5 | 184 |
| 13-16 | 4 | 96 | 14.5 | 58 |
| 16-19 | 4 | 100 | 17.5 | 70 |
| Total | 100 | 832 |
Mean: $\bar{x} = \dfrac{832}{100} = 8.32$ letters
Median: $\dfrac{n}{2} = 50$; Median class: 7-10 (cf = 36 < 50, next = 76 ≥ 50)
$l = 7$, $cf = 36$, $f = 40$, $h = 3$
$$\text{Median} = 7 + \frac{50-36}{40} \times 3 = 7 + \frac{42}{40} = 7 + 1.05 = 8.05$$Mode: Modal class = 7-10
$l = 7$, $f_1 = 40$, $f_0 = 30$, $f_2 = 16$, $h = 3$
$$\text{Mode} = 7 + \frac{10}{34} \times 3 = 7 + 0.88 = 7.88$$7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. #
Answer
| Weight (kg) | $f_i$ | $cf$ |
|---|---|---|
| 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 |
$n = 30$, $\dfrac{n}{2} = 15$
Median class: 55-60 (cf = 13 < 15, next = 19 ≥ 15)
$l = 55$, $cf = 13$, $f = 6$, $h = 5$
$$\text{Median} = 55 + \frac{15-13}{6} \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67 \text{ kg}$$