Statistics Questions and Answers Class 10 Maths Chapter 13 Kerala Syllabus Solutions

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 13 Statistics Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 13 Statistics Questions and Answers

Statistics Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 13 Statistics Solutions

Class 10 Maths Chapter 13 Kerala Syllabus – Not a Correct Average & Another Average

(Textbook Page No. 279)

Question 1.
The distances covered by Ahirath in long jump practice are 6.10, 6.20, 6.18, 6.20, 6.25, 6.21, 6.15, 6.10 in metres. Find the mean and median. Why is it that there is not much difference between these?
Answer:
Mean = \(\frac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}\) = 6.17
Arranging them in ascending order,
6.10, 6.10, 6.15, 6.18, 6.20, 6.20, 6.21, 6.25
Median = \(\frac{6.18+6.2}{2}\) = 6.19
The numbers on either side of the middle value are arranged roughly in equal amounts above and below it.
Therefore, the mean and the median are approximately equal to each other.
Also, there are no extremely large or extremely small numbers in the group.

Question 2.
The table below gives the rainfall during the first week of June 2025 in various districts of Kerala.

Calculate the mean and median rainfall in Kerala during this week. Why is the mean less than the median?
Answer:
Mean = \(\frac{108.7+89.4+74.8+72+42.6+35.7+66.4+73.5+69.1+50.5+43.6+93.1+39+37.5}{14}\) = 63.99
Arranging them in ascending order,
35.7, 37.5, 39, 42.6, 43.6, 50.5, 66.4, 69.1, 72.0, 73.5, 74.8, 89.4, 93.1, 108.7
Median = \(\frac{66.4+69.1}{2}\) = 67.75
The mean is the average calculated considering all the numbers.
The sum of the values less than the middle number is smaller, and the sum of the values greater than the middle number is larger.
Therefore, the mean is less than the median.

Question 3.
Prove that for a set of numbers in an arithmetic sequence, the mean and median are equal.
Answer:
When the number of terms is odd, there will be one middle term.
The mean obtained by dividing the sum of all the terms by their number is the same as this middle term.
This is a special property of an arithmetic sequence – that is, the mean and the median are the same.
When the number of terms is even, the terms can be paired from the ends so that each pair has the same sum.
The median is half the sum of the two middle terms, and this value is also equal to the mean.

SCERT Class 10 Maths Chapter 13 Solutions – Frequency and Median

(Textbook Page No. 282)

Question 1.
35 households in a neighbourhood are sorted according to their monthly income in the table below.

Calculate the median income.
Answer:

Total number of families = 35 (odd number)
Median = \(\frac{35+1}{2}=\frac{36}{2}\) = 18th term
Therefore the median = Rs. 2000

Question 2.
The table below shows the workers in a factory sorted according to their daily wages:

Calculate the median daily wage.
Answer:

Total number of workers = 30 (even number)
The terms in the middle are the 15th and 16th terms.
From the table, it is clear that the monthly income of both the 15th and 16th workers is 1000.
Therefore the median = Rs. 1000

Question 3.
The table below gives the number of babies born in a hospital during a week, sorted according to their birth weight.

Calculate the median birth-weight.
Answer:

Total number of babies = 70 (even number)
The terms in the middle are the 35th and 36th terms.
From the table, it is clear that the weight of the babies of both the 35th and 36th babies is 3.000 kg.
Therefore the median = 3.000 Kg

Class 10 Maths Kerala Syllabus Chapter 13 Solutions – Classes and Median

(Textbook Page No. 287-288)

Question 1.
The table shows some households sorted according to their usage of electricity:

Calculate the median usage of electricity.
Answer:

Total number of houses = 39
The 20th house lies in the middle, and the median class is 280 – 300.
If 20 units is equally distributed to 10 houses, then one part equals \(\frac {20}{10}\) = 2
It is assumed that the electricity consumption within the median class is in an arithmetic sequence.
The consumption of 17th house is 280 + 1 = 281 units
f = 281, d = 2
Then the 20th house represents the fourth term.
Hence the median = f + 3d = 281 + 6 = 287 units

Question 2.
The table below shows the children in a class sorted according to their marks in the math exam:

Calculate the median mark of the class.
Answer:

The total number of children = 36
The 18th and 19th marks lie in the middle.
Median class = 20 – 30
If the 10 marks are equally distributed among 10 students, then one part = 1 mark.
The 13th mark is = 20 + \(\frac {1}{2}\) = 20\(\frac {1}{2}\)
f = 20.5 and d = 1
x₆ = f + 5d = 20.5 + 5 = 25.5
x₇ = 26.5
∴ Median = 26

Question 3.
The table below gives the details of the income tax paid by the employees in an office in a year:

Calculate the median income tax paid.
Answer:

Here, n = 92
The 46th and 47th incomes lie in the middle, and the median class is Rs. 4000 – 5000
If Rs. 1000 is distributed equally among 20 people, then one part = Rs. 50
The 34th income is 4000 + 25 = 4025
f = 4025, d = 50
x₁₃ = f + 12d
= 4025 + 12 × 50
= 4025 + 600
= 4625
x₁₀ = 4625 + 50 = 4675
Median = \(\frac{4625+4675}{2}\) = 4650

Statistics Class 10 Notes Pdf

Class 10 Maths Chapter 13 Statistics Notes Kerala Syllabus

Introduction
In this unit, we discuss certain measures that are based on numerical data collected from a group. These measures reflect the general characteristics of the group. In your previous classes, you have already learned about the measure called the mean. The mean of a group of numbers is obtained by dividing the sum of the numbers by the total number of values in the group.
The marks obtained by a student in seven examinations are given below.
10, 18, 14, 11, 17, 11, 15
Mean = \(\frac{10+18+14+11+17+11+15}{7}\) = 13.7
Therefore, the mean of the marks is 13.7.
When the marks are arranged in order, the number that comes in the middle is called the median. Since there are seven numbers here, there will definitely be one number exactly in the middle. The order is 10,11, 11,14, 15, 17, 18. The fourth number comes in the middle, so the median is 14. One important thing to note is that when the numbers are arranged in order and form an arithmetic sequence, the mean and the median will be equal.

Median is the middlemost number when the given set of values is arranged in ascending or descending order.

Steps to calculate the median of a given set of values.

• Step 1: Arrange the given set of values in ascending order
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position(s).
• Step 4: Calculate the middle value (That is, the median)

If n is odd, then there will be only one middle value, and it is the median.

If n is even, then there will be two middle values, and the average of these two values is the median.

In statistics, frequency simply means how many times something shows up in a set of data.

How to write a cumulative frequency column in a frequency table?

• In the first row, write the first frequency.
• In the second row, write the sum of the first and second frequencies.
• In the third row, write the sum of the first, second, and third frequencies, and so on.

Steps to calculate the median from a frequency table.

• Step 1: Write the cumulative frequency table.
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position (s).
• Step 4: Calculate the middle value (That is the median!)

Steps to calculate the median from a frequency table with classes.

• Step 1: Write the cumulative frequency table.
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position(s).
• Step 4: Identify the class where the middle value(s) belong. (This class is known as the median class.)
• Step 5: Find out the class width and number of values in this class.
• Step 6:Divide the class width by the number of values in this class and find its half.
• Step 7: Calculate the value corresponding to the first observation in the median class.
• Step 8: Calculate the median value.

Not A Correct Average & Another Average
The purpose of calculating the mean is to reduce a whole collection of numbers to a single number, which gives a general understanding of a situation. But numbers in the collection that are very much less or very much more than others affect the mean a lot.

Median: The median is the middlemost number when the given set of values is arranged in ascending or descending order.

Question 1.
The median of the first n odd numbers?
(a) 2n
(b) n²
(c) 3n
(d) n
Answer:
Sum of the first n odd numbers = n²
It is an arithmetic sequence.
Mean = \(\frac{n^2}{n}\) = n
Median = n

Question 2.
The median of the first n even numbers
(a) n + 1
(b) n
(c) n – 1
(d) 2n + 1
Answer:
Sum = n(n + 1)
Median = \(\frac{n(n+1)}{n}\) = n + 1

Question 3.
The algebraic form of an arithmetic sequence is 3n + 2.
(a) What is the 11th term?
(b) What is the median of the first 21 terms?
Answer:
(a) x₁₁ = 3 × 11 + 2 = 33 + 2 = 35
(b) The median of 21 terms is the 11th term.
Median = 35

Question 4.
A company has 10 workers. Among them, three earn a daily wage of Rs. 500 each, and the remaining workers earn Rs. 800 each.
(a) What is the median daily wage?
(b) How many workers earn less than the median wage?
Answer:
(a) When arranged in ascending order, the middle numbers are the 5th and 6th terms, both of which are 800.
Hence the median = Rs. 800
(b) 3 workers earn less than the median wage.

Question 5.
25 numbers are written in a sequence. They form an arithmetic progression, and the median of the numbers is 36.
(a) What is the 13th number?
(b) What is the sum of the smallest and the largest numbers?
(c) When written in order, what is the sum of the numbers on either side of the middle number?
Answer:
(a) The 25 terms are in an arithmetic progression.
Its middle term is its median.
The middle term = 13th term
Therefore median = 36
(b) The sum of the smallest and the largest number = 2 × 13th term
= 2 × 36
= 72
(c) The sum of the numbers on either side of the middle number is 12th term + 14th term = 2 × 13th term
= 2 x 36
= 72

Question 6.
The temperatures recorded in a town over seven consecutive days are as follows:
26°C, 28°C, 25°C, 30°C, 27°C, 26°C, 25°C
(a) What is the median temperature?
(b) How many days have temperatures higher than the median temperature, and how many have lower temperatures?
(c) How many temperatures are lower than the median temperature?
Answer:
(a) Arrange the temperatures in ascending order.
25, 25, 26, 26, 27, 28, 30
There are 7 terms, and the middle term is the 4th term.
Therefore the median = 4th term = 26° C
(b) Higher temperature = 3 days
Lower temperature = 2 days
(c) Temperature lower than the median = 2 temperatures

Frequency and Median
In statistics, frequency simply means how many times something shows up in a set of data.
The marks obtained by 40 students of a class in a test are shown in the table below.
9 students scored 7 marks.
Similarly, 10 students scored 11 marks, 4 students scored 13 marks, 13 students scored 15 marks, and 4 students scored 19 marks.
This table is already arranged in ascending order of scores.
up to 7 marks, there are 9 students
up to 11 marks, 9 + 10 = 19 students
up to 13 marks, 19 + 4 = 23 students
up to 15 marks, 23 + 13 = 36 students
and up to 19 marks, all 40 students.
Let’s now represent this in a table.

Total score = 40
Here, the middle terms are 20 and 21, and the scores of both are 13.
Therefore the median = 13

Question 1.
The scores obtained by 40 students of a class in a quiz competition are given below.

(a) Using a suitable table, calculate the total marks obtained by the class.
(b) What is the mean of the marks?
(c) Find the median mark.
(d) Calculate the number of students who scored higher than the median mark.
Answer:
(a) Total Score = 4 × 5 + 6 × 10 + 9 × 10 + 10 × 7 + 15 × 8
= 20 + 60 + 90 + 70 + 120
= 360
(b) Mean = \(\frac {360}{40}\) = 9
(c) Consider the table

The total number of students is 40.
Therefore, the 20th and 21st terms lie in the middle.
From the 16th student up to the 25th student, the score is 9.
Hence, the median = 9.
(d) The number of students who scored above the median is 15.

Question 2.
The weights of 12 members of a team are given below:

(a) Prepare a table for calculating the median.
(b) What is the median of the weights?
(c) How many members have having median weight and below?
(d) How many members are there above the median weight?
Answer:
(a)

n = 12 (even)
Therefore 6th and 7th members come in the middle.
From the table, it is clear that a member weighs 70.
Median weight is 70.
(c) 7 members are weighing the median weight.
(d) There are 5 members with a median.

Question 3.
The daily wages of 200 workers in a factory are given below.

(a) Prepare the table for calculating the median.
(b) Find the median wage.
(c) How many workers are getting a median wage and below?
(d) How many workers are getting above the median wage?
Answer:
(a)

(b) n = 200 (even)
So the 100th and 101st wage comes in the middle.
From the table, it is clear that the wages of both workers are 500.
Median is 500.
(c) 134 workers have having daily wage below 500.
(d) There are 66 workers having wages above 500.

Classes and Median
The given data can be divided into different groups, and the number of items in each group can be counted and arranged in a table. Such a table is called a frequency table. A frequency table has two columns – the first column shows the groups, and the second column shows the frequency.

Let us calculate the median from the table given below:
The table shows the ages of employees and the number of people working in an organization.

First, we prepare a table by arranging the given data in ascending order.

There are a total of 46 workers.
The 23rd and 24th workers’ ages fall in the middle, and they belong to the 40 – 45 group.
The 40 – 45 group is called the median class.
It is assumed that the ages in the median class are in arithmetic sequence.
If the 5-year interval of the median class is divided equally for 10 workers, that is, 29 – 19 = 10
Then the share for one person is 5 ÷ 10 = \(\frac {1}{2}\) year.
The age of the 20th worker is 40 + (\(\frac {1}{2}\)) ÷ 2 = 40\(\frac {1}{4}\) years.
If we take the 20th worker’s age as the first term and the common difference as \(\frac {1}{2}\), then the age of the 23rd worker will be the fourth term of that arithmetic sequence.
f = 40\(\frac {1}{4}\), d = \(\frac {1}{2}\),
x₄ = f + 3d
= 40\(\frac {1}{4}\) + 3 × \(\frac {1}{2}\)
= 41\(\frac {3}{4}\)
The age of the 24th worker = \(41 \frac{3}{4}+\frac{1}{2}=42 \frac{1}{4}\) years.
The median is the average of the ages of the 23rd and 24th workers.
Median = \(\left(41 \frac{3}{4}+42 \frac{1}{4}\right) \div 2\) = 42