Class 10 NCERT Solutions
Chapter 13: Statistics
Master the measures of central tendency, the calculation of mean, median, and mode for grouped data, and the logic of data distributions with our step-by-step logic.
Exercise 13.1
1. Find the mean number of plants per house from the following data. Which method did you use and why?
(Number of plants: 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14;
Number of houses: 1, 2, 1, 5, 6, 2, 3)
Using the direct method (since the numerical values of class intervals and frequencies are small):
$\sum f_i = 20$, $\sum f_i x_i = 1(1)+2(3)+1(5)+5(7)+6(9)+2(11)+3(13) = 1+6+5+35+54+22+39 = 162$
$$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = \mathbf{8.1} \text{ plants}$$
2. Find the mean daily wages of 50 workers of a factory using an appropriate method.
(Wages: ₹100–120, 120–140, 140–160, 160–180, 180–200; Workers: 12, 14, 8, 6, 10)
Using the step-deviation method with $a = 150$, $h = 20$:
$\sum f_i = 50$, $\sum f_i u_i = -24-14+0+6+20 = -12$
$$\text{Mean} = a+\frac{\sum f_i u_i}{\sum f_i}\times h = 150+\frac{-12}{50}\times20 = 150-4.8 = ₹\,\mathbf{145.20}$$
3. The mean pocket allowance is ₹18. Find the missing frequency $f$.
(Allowance: ₹11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25;
Children: 7, 6, 9, 13, $f$, 5, 4)
Using the step-deviation method with $a = 18$, $h = 2$:
$\sum f_i = 44+f$, $\sum f_i u_i = -21-12-9+0+f+10+12 = -20+f$
$$18 = 18+\frac{-20+f}{44+f}\times2 \Rightarrow -20+f = 0 \Rightarrow \mathbf{f = 20}$$
4. Find the mean heart beats per minute for 30 women using a suitable method.
(Heart beats: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86;
Women: 2, 4, 3, 8, 7, 4, 2)
Using the step-deviation method with $a = 75.5$, $h = 3$:
$\sum f_i = 30$, $\sum f_i u_i = -6-8-3+0+7+8+6 = 4$
$$\text{Mean} = 75.5+\frac{4}{30}\times3 = 75.5+0.4 = \mathbf{75.9} \text{ beats/min}$$
5. Find the mean number of mangoes kept in a packing box. Which method did you choose?
(Mangoes: 50–52, 53–55, 56–58, 59–61, 62–64; Boxes: 15, 110, 135, 115, 25)
Using the step-deviation method (chosen because $f_i$ values are large and the class-mark differences are uniform) with $a = 57$, $h = 3$:
$\sum f_i = 400$, $\sum f_i u_i = -30-110+0+115+50 = 25$
$$\text{Mean} = 57+\frac{25}{400}\times3 = 57+0.1875 \approx \mathbf{57.19} \text{ mangoes}$$
6. Find the mean daily expenditure on food by a suitable method.
(Expenditure in ₹: 100–150, 150–200, 200–250, 250–300, 300–350;
Households: 4, 5, 12, 2, 2)
Using the assumed mean method with $a = 225$:
$\sum f_i = 25$, $\sum f_i d_i = -400-250+0+100+200 = -350$
$$\text{Mean} = 225+\frac{-350}{25} = 225-14 = ₹\,\mathbf{211}$$
7. Find the mean concentration of $\text{SO}_2$ in the air from data collected for 30 localities.
(Concentration in ppm: 0.00–0.04, 0.04–0.08, 0.08–0.12, 0.12–0.16, 0.16–0.20, 0.20–0.24;
Frequency: 4, 9, 9, 2, 4, 2)
Using the step-deviation method with $a = 0.10$, $h = 0.04$:
$\sum f_i = 30$, $\sum f_i u_i = -8-9+0+2+8+6 = -1$
$$\text{Mean} = 0.10+\frac{-1}{30}\times0.04 = 0.10-0.001 \approx \mathbf{0.099} \text{ ppm}$$
8. Find the mean number of days a student was absent.
(Days: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40;
Students: 11, 10, 7, 4, 4, 3, 1)
Using the direct method:
$\sum f_i = 40$, $\sum f_i x_i = 33+80+84+68+96+99+39 = 499$
$$\text{Mean} = \frac{499}{40} = \mathbf{12.48} \text{ days}$$
9. Find the mean literacy rate of 35 cities.
(Literacy rate %: 45–55, 55–65, 65–75, 75–85, 85–95;
Cities: 3, 10, 11, 8, 3)
Using the assumed mean method with $a = 70$:
$\sum f_i = 35$, $\sum f_i d_i = -60-100+0+80+60 = -20$
$$\text{Mean} = 70+\frac{-20}{35} = 70-0.57 \approx \mathbf{69.43}\%$$Exercise 13.2
1. Find the mode and the mean of the ages of patients admitted in a hospital. Compare and interpret the two measures.
(Age in years: 5–15, 15–25, 25–35, 35–45, 45–55, 55–65;
Patients: 6, 11, 21, 23, 14, 5)
Mode: Maximum frequency $= 23$, so modal class is $35$–$45$. With $l=35$, $f_1=23$, $f_0=21$, $f_2=14$, $h=10$:
$$\text{Mode} = 35+\frac{23-21}{46-21-14}\times10 = 35+\frac{20}{11} \approx 36.8 \text{ years}$$Mean: Using step-deviation with $a=40$, $h=10$: $\sum f_i=80$, $\sum f_i u_i = -18-22-21+0+14+10=-37$
$$\text{Mean} = 40+\frac{-37}{80}\times10 = 40-4.63 \approx 35.37 \text{ years}$$Interpretation: The maximum number of patients are admitted at age 36.8 years, while the average age of patients is 35.37 years.
2. Determine the modal lifetimes of 225 electrical components.
(Lifetime in hours: 0–20, 20–40, 40–60, 60–80, 80–100, 100–120;
Frequency: 10, 35, 52, 61, 38, 29)
Maximum frequency $= 61$, so modal class is $60$–$80$. With $l=60$, $f_1=61$, $f_0=52$, $f_2=38$, $h=20$:
$$\text{Mode} = 60+\frac{61-52}{122-52-38}\times20 = 60+\frac{9\times20}{32} = 60+5.625 = \mathbf{65.625} \text{ hours}$$
3. Find the modal monthly expenditure of 200 families. Also find the mean monthly expenditure.
(Expenditure in ₹: 1000–1500, 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000;
Families: 24, 40, 33, 28, 30, 22, 16, 7)
Mode: Maximum frequency $= 40$, so modal class is $1500$–$2000$. With $l=1500$, $f_0=24$, $f_1=40$, $f_2=33$, $h=500$:
$$\text{Mode} = 1500+\frac{16\times500}{80-24-33} = 1500+\frac{8000}{23} \approx ₹\,\mathbf{1847.83}$$Mean: Using step-deviation with $a=2750$, $h=500$: $\sum f_i=200$, $\sum f_i u_i = -35$
$$\text{Mean} = 2750+\frac{-35}{200}\times500 = 2750-87.5 = ₹\,\mathbf{2662.50}$$
4. Find the mode and mean of the teacher-student ratio data. Interpret the two measures.
(Students per teacher: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55;
States/UTs: 3, 8, 9, 10, 3, 0, 0, 2)
Mean: With $a=32.5$, $h=5$: $\sum f_i=35$, $\sum f_i u_i=-23$
$$\text{Mean} = 32.5+\frac{-23}{35}\times5 = 32.5-3.28 \approx 29.22$$Mode: Maximum frequency $= 10$, so modal class is $30$–$35$. With $l=30$, $f_1=10$, $f_0=9$, $f_2=3$, $h=5$:
$$\text{Mode} = 30+\frac{1}{20-9-3}\times5 = 30+\frac{5}{8} \approx 30.62$$Interpretation: Most states/UTs have 30.62 students per teacher; the average is 29.22 students per teacher.
5. Find the mode of runs scored by top batsmen in one-day international cricket matches.
(Runs: 3000–4000, 4000–5000, 5000–6000, 6000–7000, 7000–8000, 8000–9000, 9000–10000, 10000–11000;
Batsmen: 4, 18, 9, 7, 6, 3, 1, 1)
Maximum frequency $= 18$, so modal class is $4000$–$5000$. With $l=4000$, $f_1=18$, $f_0=4$, $f_2=9$, $h=1000$:
$$\text{Mode} = 4000+\frac{18-4}{36-4-9}\times1000 = 4000+\frac{14000}{23} \approx \mathbf{4608.7} \text{ runs}$$
6. Find the mode of the number of cars passing through a spot.
(Cars: 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)
Maximum frequency $= 20$, so modal class is $40$–$50$. With $l=40$, $f_1=20$, $f_0=12$, $f_2=11$, $h=10$:
$$\text{Mode} = 40+\frac{20-12}{40-12-11}\times10 = 40+\frac{80}{17} \approx \mathbf{44.7} \text{ cars}$$Exercise 13.3
1. Find the median, mean and mode of monthly electricity consumption of 68 consumers. Compare them.
(Units: 65–85, 85–105, 105–125, 125–145, 145–165, 165–185, 185–205;
Consumers: 4, 5, 13, 20, 14, 8, 4)
Median: $\dfrac{N}{2} = 34$. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68. Median class is $125$–$145$.
$$\text{Median} = 125+\frac{34-22}{20}\times20 = 125+12 = \mathbf{137} \text{ units}$$Mean: With $a=135$, $h=20$: $\sum f_i u_i = -12-10-13+0+14+16+12=7$
$$\text{Mean} = 135+\frac{7}{68}\times20 \approx 137.05 \text{ units}$$Mode: Maximum frequency $= 20$, modal class $125$–$145$. $l=125$, $f_1=20$, $f_0=13$, $f_2=14$, $h=20$:
$$\text{Mode} = 125+\frac{7}{13}\times20 \approx 135.76 \text{ units}$$Comparison: All three measures are approximately equal, indicating a nearly symmetric distribution.
2. If the median of the following distribution is 28.5, find the values of $x$ and $y$.
(Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60; Frequency: 5, $x$, 20, 15, $y$, 5; Total: 60)
Since total $= 60$: $45+x+y = 60 \Rightarrow x+y = 15 \quad \cdots(i)$
Median $= 28.5$ falls in class $20$–$30$ (cf $= 5+x$, $f = 20$, $l = 20$, $h = 10$):
$$28.5 = 20+\frac{30-(5+x)}{20}\times10 \Rightarrow 8.5 = \frac{25-x}{2} \Rightarrow 17 = 25-x \Rightarrow x = 8$$From (i): $y = 15-8 = 7$. Therefore $\mathbf{x = 8,\ y = 7}$.
3. Calculate the median age of 100 policy holders aged between 18 and 60 years.
(Ages below: 20, 25, 30, 35, 40, 45, 50, 55, 60; Holders: 2, 6, 24, 45, 78, 89, 92, 98, 100)
Converting to class intervals and computing cumulative frequencies, we find $\dfrac{N}{2} = 50$ falls in class $35$–$40$ (cf $= 45$, $f = 33$).
$$\text{Median} = 35+\frac{50-45}{33}\times5 = 35+\frac{25}{33} \approx \mathbf{35.76} \text{ years}$$
4. Find the median length of 40 leaves.
(Length in mm: 118–126, 127–135, 136–144, 145–153, 154–162, 163–171, 172–180;
Leaves: 3, 5, 9, 12, 5, 4, 2)
Convert to continuous classes (e.g., 117.5–126.5, …). Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40. $\dfrac{N}{2} = 20$ falls in class $144.5$–$153.5$ (cf $= 17$, $f = 12$, $h = 9$).
$$\text{Median} = 144.5+\frac{20-17}{12}\times9 = 144.5+2.25 = \mathbf{146.75} \text{ mm}$$
5. Find the median lifetime of 400 neon lamps.
(Lifetime in hours: 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000;
Lamps: 14, 56, 60, 86, 74, 62, 48)
Cumulative frequencies: 14, 70, 130, 216, 290, 352, 400. $\dfrac{N}{2} = 200$ falls in class $3000$–$3500$ (cf $= 130$, $f = 86$, $h = 500$).
$$\text{Median} = 3000+\frac{200-130}{86}\times500 = 3000+406.98 \approx \mathbf{3406.98} \text{ hours}$$
6. From 100 surnames, find the median, mean and modal number of letters.
(Letters: 1–4, 4–7, 7–10, 10–13, 13–16, 16–19;
Surnames: 6, 30, 40, 16, 4, 4)
Mean: With $a=8.5$, $h=3$: $\sum f_i u_i = -12-30+0+16+8+12=-6$
$$\text{Mean} = 8.5+\frac{-6}{100}\times3 = 8.5-0.18 = \mathbf{8.32}$$Mode: Maximum frequency $= 40$, modal class $7$–$10$. With $l=7$, $f_0=30$, $f_1=40$, $f_2=16$, $h=3$:
$$\text{Mode} = 7+\frac{10}{80-30-16}\times3 = 7+0.88 \approx \mathbf{7.88}$$Median: $\dfrac{N}{2}=50$ falls in class $7$–$10$ (cf $= 36$, $f = 40$, $h = 3$):
$$\text{Median} = 7+\frac{50-36}{40}\times3 = 7+1.05 = \mathbf{8.05}$$
7. Find the median weight of 30 students.
(Weight in kg: 40–45, 45–50, 50–55, 55–60, 60–65, 65–70, 70–75;
Students: 2, 3, 8, 6, 6, 3, 2)
Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. $\dfrac{N}{2} = 15$ falls in class $55$–$60$ (cf $= 13$, $f = 6$, $h = 5$).
$$\text{Median} = 55+\frac{15-13}{6}\times5 = 55+\frac{10}{6} \approx \mathbf{56.67} \text{ kg}$$