Find the median, mode, and mean of the monthly water consumption of 80 consumers.

Median = 110 + ((40 - 24)/20) × 20 = 126 units. Mode ≈ 123.33 units. Mean = 126 units.

If the median of the distribution is 30 and total frequency is 100, find x and y.

x + y = 25. Median lies in 30-40 => 30 = 30 + ((50 - (35+x))/30) × 10 => x = 15, y = 10.

Class 10 Maths Chapter 13 Exercise 13.3 NCERT Solutions

Question 1: Monthly Water Consumption

The following frequency distribution gives the monthly consumption of water (in units) of 80 consumers of a locality. Find the median, mean, and mode of the data and compare them:

Consumption (units)	50–70	70–90	90–110	110–130	130–150	150–170	170–190
Consumers	4	8	12	20	16	12	8

Cumulative Frequency & Median

Cumulative frequencies: $4, 12, 24, 44, 60, 72, 80$.
$N = 80 \implies \frac{N}{2} = 40$.
The cumulative frequency just greater than $40$ is $44$, which belongs to the class $110–130$.
Hence, the median class is $110–130$.
$l = 110, cf = 24, f = 20, h = 20$.
$$\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 110 + \left(\frac{40 - 24}{20}\right) \times 20 = 110 + 16 = 126\text{ units}$$

Mode & Mean Calculation

Mode: Modal class is $110–130$ ($f_1 = 20, f_0 = 12, f_2 = 16$).
$$\text{Mode} = 110 + \left(\frac{20 - 12}{2(20) - 12 - 16}\right) \times 20 = 110 + \frac{8}{12} \times 20 = 110 + 13.33 \approx 123.33\text{ units}$$ Mean: Let $a = 120, h = 20$. $\sum f_i u_i = 24$.
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{N}\right) \times h = 120 + \left(\frac{24}{80}\right) \times 20 = 120 + 6 = 126\text{ units}$$

$$\text{Median} = 126\text{ units}, \quad \text{Mode} \approx 123.33\text{ units}, \quad \text{Mean} = 126\text{ units}.$$

Question 2: Missing Frequencies ($x$ and $y$)

If the median of the distribution given below is 30, find the values of $x$ and $y$. Total frequency is 100.

Class Interval	0–10	10–20	20–30	30–40	40–50	50–60
Frequency	10	$x$	25	30	$y$	10

Step 1: Relation from Total Frequency

$$\sum f_i = 10 + x + 25 + 30 + y + 10 = 75 + x + y$$ $$75 + x + y = 100 \implies x + y = 25 \quad \text{--- (Equation 1)}$$

Step 2: Median Formula

The median is given as $30$, which lies in the class $30–40$.
$l = 30, h = 10, f = 30, cf = 10 + x + 25 = 35 + x$.
$$\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$$ $$30 = 30 + \left(\frac{50 - (35 + x)}{30}\right) \times 10$$ Subtract 30 from both sides: $$0 = \frac{15 - x}{3} \implies 15 - x = 0 \implies x = 15$$

Step 3: Finding $y$

Substitute $x = 15$ into Equation 1: $$15 + y = 25 \implies y = 10$$

$$\therefore x = 15 \quad \text{and} \quad y = 10.$$

Question 3: Insurance Policy Holders (Cumulative to Frequency)

An insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age:

Age (in years)	Below 20	Below 25	Below 30	Below 35	Below 40	Below 45
No. of Policy Holders	6	15	29	55	85	100

Converting to Class Intervals

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($cf$)
15–20	6	6
20–25	$15 - 6 = 9$	15
25–30	$29 - 15 = 14$	29
30–35	$55 - 29 = 26$	55
35–40	$85 - 55 = 30$	85
40–45	$100 - 85 = 15$	100

Median Calculation

$N = 100 \implies \frac{N}{2} = 50$.
Median class is $30–35$ (since $cf$ reaches 55 here).
$l = 30, cf = 29, f = 26, h = 5$.
$$\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 30 + \left(\frac{50 - 29}{26}\right) \times 5$$ $$= 30 + \frac{21 \times 5}{26} = 30 + \frac{105}{26} \approx 30 + 4.04 = 34.04\text{ years}$$

$$\therefore \text{The median age is approximately } 34.04\text{ years}.$$

Question 4: Length of Leaves (Continuous Boundaries)

The lengths of 50 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (mm)	110–119	120–129	130–139	140–149	150–159
No. of Leaves	5	10	15	12	8

Find the median length of the leaves.

Conversion to Continuous Classes

Subtract $0.5$ from lower limits and add $0.5$ to upper limits:

Length (mm)	Frequency ($f$)	$cf$
109.5–119.5	5	5
119.5–129.5	10	15
129.5–139.5	15	30
139.5–149.5	12	42
149.5–159.5	8	50

Median Calculation

$N = 50 \implies \frac{N}{2} = 25$.
Median class is $129.5–139.5$.
$l = 129.5, cf = 15, f = 15, h = 10$.
$$\text{Median} = 129.5 + \left(\frac{25 - 15}{15}\right) \times 10 = 129.5 + \frac{100}{15} \approx 129.5 + 6.67 = 136.17\text{ mm}$$

$$\therefore \text{The median length of the leaves is } 136.17\text{ mm}.$$

Question 5: Lifetime of Neon Lamps

The following table gives the distribution of the lifetime of 500 neon lamps:

Life time (hours)	1500–2000	2000–2500	2500–3000	3000–3500	3500–4000	4000–4500	4500–5000
Number of Lamps	20	60	70	100	90	80	80

Find the median lifetime of a lamp.

Cumulative Frequency & Class Selection

Cumulative frequencies: $20, 80, 150, 250, 340, 420, 500$.
$N = 500 \implies \frac{N}{2} = 250$.
Median class is $3000–3500$.
$l = 3000, cf = 150, f = 100, h = 500$.

Calculation

$$\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 3000 + \left(\frac{250 - 150}{100}\right) \times 500$$ $$= 3000 + \left(\frac{100}{100}\right) \times 500 = 3000 + 500 = 3500\text{ hours}$$

$$\therefore \text{The median lifetime of a lamp is } 3500\text{ hours}.$$

Question 6: Surnames Letter Frequency

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	8	35	45	8	2	2

Determine the median number of letters. Find the mean number of letters. Also, find the modal size of the surnames.

(i) Median Calculation

$cf: 8, 43, 88, 96, 98, 100$.
$N = 100 \implies \frac{N}{2} = 50$. Median class is $7–10$.
$l = 7, cf = 43, f = 45, h = 3$.
$$\text{Median} = 7 + \left(\frac{50 - 43}{45}\right) \times 3 = 7 + \frac{21}{45} = 7 + 0.47 = 7.47\text{ letters}$$

(ii) Mode Calculation

Modal class is $7–10$ ($f_1 = 45, f_0 = 35, f_2 = 8, h = 3$).
$$\text{Mode} = 7 + \left(\frac{45 - 35}{2(45) - 35 - 8}\right) \times 3 = 7 + \frac{10}{90 - 43} \times 3 = 7 + \frac{30}{47} \approx 7.64\text{ letters}$$

(iii) Mean Calculation

Class marks: $2.5, 5.5, 8.5, 11.5, 14.5, 17.5$.
$$\sum f_i x_i = 8(2.5) + 35(5.5) + 45(8.5) + 8(11.5) + 2(14.5) + 2(17.5) = 751$$ $$\bar{x} = \frac{751}{100} = 7.51\text{ letters}$$

$$\text{Median} = 7.47, \quad \text{Mode} \approx 7.64, \quad \text{Mean} = 7.51.$$

Question 7: Weights of 40 Students

The distribution below gives the weights of 40 students of a class. Find the median weight of the students:

Weight (in kg)	40–45	45–50	50–55	55–60	60–65	65–70	70–75
Number of Students	3	4	10	8	8	4	3

Calculation

Cumulative frequencies: $3, 7, 17, 25, 33, 37, 40$.
$N = 40 \implies \frac{N}{2} = 20$.
The $cf$ just greater than 20 is 25, so the median class is $55–60$.
$l = 55, cf = 17, f = 8, h = 5$.
$$\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 55 + \left(\frac{20 - 17}{8}\right) \times 5 = 55 + \frac{15}{8} = 55 + 1.875 = 56.875\text{ kg}$$

$$\therefore \text{The median weight of the students is } 56.88\text{ kg}.$$

Statistics - Exercise 13.3

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