NCERT Class 10 Maths – Statistics Exercise 13.2 Solutions

Q1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years):

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

Answer:

Mode

The maximum frequency is 23, so the modal class is 35–45.

Here $l=35$, $f_1=23$, $f_0=21$, $f_2=14$, $h=10$.

$\text{Mode}=35+\left(\frac{23-21}{2(23)-21-14}\right)\times 10$

$\text{Mode}=35+\frac{2}{11}\times 10=35+\frac{20}{11}\approx 36.82$

Mode ≈ 36.82 years

Mean

$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{2830}{80}=35.375$

Mean ≈ 35.38 years

Interpretation

The modal age (about 36.82 years) is the age of the maximum number of patients, while the mean age (about 35.38 years) is the average age of all patients. The mode is slightly greater than the mean here.

Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer:

Lifetimes of 225 Electrical Components

The maximum frequency is 61, so the modal class is 60–80.

Here $l=60$, $f_1=61$, $f_0=52$, $f_2=38$, $h=20$.

$\text{Mode}=60+\left(\frac{61-52}{2(61)-52-38}\right)\times 20$

$\text{Mode}=60+\frac{9}{32}\times 20=60+\frac{180}{32}=65.625$

Modal lifetime = 65.625 hours

This means the maximum number of electrical components have a lifetime of about 65.63 hours.

Q3: The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :

Answer:

Monthly Household Expenditure of 200 Families

Mode

The maximum frequency is 40, so the modal class is 1500–2000.

Here $l=1500$, $f_1=40$, $f_0=24$, $f_2=33$, $h=500$.

$\text{Mode}=1500+\left(\frac{40-24}{2(40)-24-33}\right)\times 500$

$\text{Mode}=1500+\frac{16}{23}\times 500=1500+\frac{8000}{23}\approx 1847.83$

Modal monthly expenditure ≈ ₹1847.83

Mean

$\bar{x}=\frac{532500}{200}=2662.5$

Mean monthly expenditure = ₹2662.50

Interpretation

The modal expenditure (about ₹1847.83) tells us the spending level of the largest group of families, while the mean expenditure (₹2662.50) is the average over all 200 families. The mean is higher than the mode because the higher expenditure classes pull the average up.

Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solutions:

Teacher–Student Ratio in Higher Secondary Schools

Mode

The maximum frequency is 10, so the modal class is 30–35.

Here $l=30$, $f_1=10$, $f_0=9$, $f_2=3$, $h=5$.

$\text{Mode}=30+\left(\frac{10-9}{2(10)-9-3}\right)\times 5$

$\text{Mode}=30+\frac{1}{8}\times 5=30+0.625=30.625$

Mode = 30.625 students per teacher

Mean

$\bar{x}=\frac{1022.5}{35}\approx 29.21$

Mean ≈ 29.21 students per teacher

Interpretation

The modal value (30.625) is the student–teacher ratio in the largest number of states/U.T.s, while the mean (about 29.21) is the average ratio across all states/U.T.s. The mode is slightly greater than the mean here.

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Answer: Given Data

Step 1: Identify the Modal Class

The maximum frequency is 18, so the modal class is 4000–5000.

Step 2: Write Down the Values

• Lower limit of modal class: $l=4000$
• Frequency of modal class: $f_1=18$
• Frequency of preceding class: $f_0=4$
• Frequency of succeeding class: $f_2=9$
• Class size: $h=1000$

Step 3: Apply the Mode Formula

$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

$\text{Mode}=4000+\left(\frac{18-4}{2(18)-4-9}\right)\times 1000$

$\text{Mode}=4000+\frac{14}{23}\times 1000$

$\text{Mode}=4000+\frac{14000}{23}\approx 4000+608.70$

$\boxed{{\displaystyle \text{Mode}\approx 6087\text{ runs}}}$

Interpretation

The mode of about 6087 runs means that the largest number of top batsmen (18 of the 49 listed) have scored runs in the range around 6087 — i.e., most batsmen fall in the 4000–5000 runs group, and the mode estimates the most typical score within that group.

Summary

Key tip: Always pick the modal class from the highest frequency first, then carefully read $f_0$f0​ and $f_2$f2​ from the classes immediately before and after it — a common exam mistake is using the wrong neighbouring frequencies.

Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Solution:

Given Data

A student observed cars passing through a spot for 100 periods, each of 3 minutes:

Step 1: Identify the Modal Class

The maximum frequency is 20, so the modal class is 40–50.

Step 2: Write Down the Values

• Lower limit of modal class: $l=40$
• Frequency of modal class: $f_1=20$
• Frequency of preceding class (30–40): $f_0=12$
• Frequency of succeeding class (50–60): $f_2=11$
• Class size: $h=10$

Step 3: Apply the Mode Formula

$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

$\text{Mode}=40+\left(\frac{20-12}{2(20)-12-11}\right)\times 10$

$\text{Mode}=40+\frac{8}{17}\times 10$

$\text{Mode}=40+\frac{80}{17}\approx 40+4.71$

$\boxed{{\displaystyle \text{Mode}\approx 44.71\text{ cars}}}$

Interpretation

The mode of about 44.71 cars means that, in most of the 3-minute observation periods, roughly 44–45 cars passed through the spot — the frequency of passing cars peaks in the 40–50 class.

Summary

Key tip: Even when the maximum frequency is not in the middle of the table (here it is 20 in the 40–50 class), always take $f_0$​ and $f_2$​ from the classes immediately before and after the modal class – 12 and 11 in this case.