Find the mode and mean of the ages of participants in the marathon.

Modal class 45-55: Mode = 45 + (2/11) × 10 ≈ 46.82 years. Mean = 3630 / 80 ≈ 45.38 years.

Determine the modal lifetimes of the 200 batteries.

Modal class 60-80: Mode = 60 + (9/32) × 20 = 65.625 hours.

Class 10 Maths Chapter 13 Exercise 13.2 NCERT Solutions

Question 1: Ages of Marathon Participants

The following table shows the ages of the participants in a local marathon:

Age (in years)	15–25	25–35	35–45	45–55	55–65	65–75
Number of Participants	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Step 1: Finding the Mode

The maximum class frequency is $23$, which lies in the class $45–55$.
Hence, the modal class is $45–55$.
Lower limit $l = 45$, class width $h = 10$, frequency of modal class $f_1 = 23$, frequency of preceding class $f_0 = 21$, frequency of succeeding class $f_2 = 14$.
$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ $$= 45 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10 = 45 + \left(\frac{2}{46 - 35}\right) \times 10 = 45 + \frac{20}{11} \approx 45 + 1.82 = 46.82\text{ years}$$

Step 2: Finding the Mean

Class marks $x_i$: $20, 30, 40, 50, 60, 70$.
$$\sum f_i = 6 + 11 + 21 + 23 + 14 + 5 = 80$$ $$\sum f_i x_i = 6(20) + 11(30) + 21(40) + 23(50) + 14(60) + 5(70)$$ $$= 120 + 330 + 840 + 1150 + 840 + 350 = 3630$$ $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3630}{80} = 45.375 \approx 45.38\text{ years}$$

Step 3: Interpretation

The mode of $46.82\text{ years}$ indicates that the highest number of participants are approximately $47\text{ years}$ old. The mean of $45.38\text{ years}$ represents the overall average age of all participants.

$$\text{Mode} \approx 46.82\text{ years}, \quad \text{Mean} \approx 45.38\text{ years}.$$

Question 2: Modal Lifetimes of Batteries

The following data gives the information on the observed lifetimes (in hours) of 200 batteries:

Lifetimes (in hours)	0–20	20–40	40–60	60–80	80–100	100–120
Frequency	10	35	52	61	38	4

Determine the modal lifetimes of the batteries.

Identifying Parameters

Maximum frequency is $61$, which belongs to the class $60–80$.
Modal class $= 60–80$.
Lower limit $l = 60$, class size $h = 20$.
$f_1 = 61$ (modal frequency), $f_0 = 52$ (preceding frequency), $f_2 = 38$ (succeeding frequency).

Calculation

$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ $$= 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$$ $$= 60 + \left(\frac{9}{122 - 90}\right) \times 20 = 60 + \frac{9 \times 20}{32} = 60 + \frac{180}{32} = 60 + 5.625 = 65.625\text{ hours}$$

$$\therefore \text{The modal lifetime of the batteries is } 65.625\text{ hours}.$$

Question 3: Weekly Household Savings

The following data gives the distribution of total weekly household savings of 200 families of a village. Find the modal weekly savings of the families. Also, find the mean weekly savings.

Savings (₹)	1000–1500	1500–2000	2000–2500	2500–3000	3000–3500	3500–4000	4000–4500	4500–5000
No. of Families	24	40	33	28	30	22	16	7

Step 1: Mode Calculation

Maximum frequency is $40$, corresponding to modal class $1500–2000$.
$l = 1500, h = 500, f_1 = 40, f_0 = 24, f_2 = 33$.
$$\text{Mode} = 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500$$ $$= 1500 + \frac{16}{80 - 57} \times 500 = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = ₹1847.83$$

Step 2: Mean Calculation (Step-Deviation)

Let assumed mean $a = 2750$ (mid-point of $2500–3000$), $h = 500$.
$$\sum f_i u_i = 24(-3) + 40(-2) + 33(-1) + 28(0) + 30(1) + 22(2) + 16(3) + 7(4)$$ $$= -72 - 80 - 33 + 0 + 30 + 44 + 48 + 28 = -185 + 150 = -35$$ $$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 - 87.5 = ₹2662.50$$

$$\text{Modal Weekly Savings} = ₹1847.83, \quad \text{Mean Weekly Savings} = ₹2662.50.$$

Question 4: Doctor-Patient Ratio

The following distribution gives the district-wise doctor-patient ratio in a state. Find the mode and mean of this data. Interpret the two measures.

Patients per Doctor	15–20	20–25	25–30	30–35	35–40	40–45	45–50	50–55
Number of Districts	3	8	9	10	3	0	0	2

Step 1: Mode Calculation

Maximum frequency is $10$, belonging to modal class $30–35$.
$l = 30, h = 5, f_1 = 10, f_0 = 9, f_2 = 3$.
$$\text{Mode} = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5 = 30 + \frac{1}{20 - 12} \times 5 = 30 + \frac{5}{8} = 30.625 \approx 30.6$$

Step 2: Mean Calculation

$$\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35$$ $$\sum f_i x_i = 3(17.5) + 8(22.5) + 9(27.5) + 10(32.5) + 3(37.5) + 0 + 0 + 2(52.5)$$ $$= 52.5 + 180 + 247.5 + 325 + 112.5 + 105 = 1022.5$$ $$\bar{x} = \frac{1022.5}{35} \approx 29.21 \approx 29.2$$

Step 3: Interpretation

Most districts have around $30.6$ patients per doctor (mode), while across all districts, the average ratio is $29.2$ patients per doctor.

$$\text{Mode} \approx 30.6, \quad \text{Mean} \approx 29.2.$$

Question 5: Football Goals Scored

The given distribution shows the number of goals scored by various teams in a league season. Find the mode of the data:

Goals Scored	30–40	40–50	50–60	60–70	70–80	80–90
Number of Teams	6	18	10	8	5	3

Mode Calculation

The maximum frequency is $18$, which belongs to the class $40–50$.
Hence, the modal class is $40–50$.
$l = 40, h = 10, f_1 = 18, f_0 = 6, f_2 = 10$.
$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ $$= 40 + \left(\frac{18 - 6}{2(18) - 6 - 10}\right) \times 10 = 40 + \left(\frac{12}{36 - 16}\right) \times 10$$ $$= 40 + \frac{12}{20} \times 10 = 40 + 6 = 46\text{ goals}$$

$$\therefore \text{The mode of the data is } 46\text{ goals}.$$

Question 6: Bike Traffic Passing Toll Booth

A student noted the number of bikes passing through a toll booth for 100 periods each of 5 minutes and summarised it in the table given below. Find the mode of the data:

Number of Bikes	0–10	10–20	20–30	30–40	40–50	50–60	60–70	70–80
Frequency	7	14	13	12	20	11	15	8

Calculation

The maximum frequency is $20$, which belongs to the class $40–50$.
Modal class $= 40–50$.
$l = 40, h = 10, f_1 = 20, f_0 = 12, f_2 = 11$.
$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ $$= 40 + \left(\frac{20 - 12}{2(20) - 12 - 11}\right) \times 10 = 40 + \left(\frac{8}{40 - 23}\right) \times 10$$ $$= 40 + \frac{80}{17} = 40 + 4.706 \approx 44.71\text{ bikes}$$

$$\therefore \text{The mode of the data is approximately } 44.7\text{ bikes}.$$

Statistics - Exercise 13.2

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