Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3

Get the most accurate TN Board Solutions for Class 11 Business Maths Chapter 08 Descriptive Statistics and Probability here. Updated for the 2026-27 academic session, these solutions are based on the latest TN Board textbooks for Class 11 Business Maths. Our expert-created answers for Class 11 Business Maths are available for free download in PDF format.

Detailed Chapter 08 Descriptive Statistics and Probability TN Board Solutions for Class 11 Business Maths

For Class 11 students, solving TN Board textbook questions is the most effective way to build a strong conceptual foundation. Our Class 11 Business Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 08 Descriptive Statistics and Probability solutions will improve your exam performance.

Class 11 Business Maths Chapter 08 Descriptive Statistics and Probability TN Board Solutions PDF

 

Question 1. Which of the following is a positional measure?
(a) Range
(b) Mode
(c) Mean deviation
(d) Percentiles
Answer: (d) Percentiles
In simple words: Percentiles tell us the relative standing of a data point within a dataset, showing its position compared to other values. They divide the data into 100 equal parts.

๐ŸŽฏ Exam Tip: Positional measures like percentiles, quartiles, and deciles help understand the distribution and spread of data by pinpointing specific locations within the sorted data set.

 

Question 2. When calculating the average growth of the economy, the correct mean to use is?
(a) Weighted mean
(b) Arithmetic mean
(c) Geometric mean
(d) Harmonic mean
Answer: (c) Geometric mean
In simple words: The geometric mean is best when you are looking at things that grow or change over time, like economic growth rates or interest rates. It gives a more accurate average for such situations.

๐ŸŽฏ Exam Tip: Remember to use the geometric mean for data involving rates of change or ratios, as it handles multiplicative relationships better than the arithmetic mean.

 

Question 3. When an observation in the data is zero, then its geometric mean is:
(a) Negative
(b) Positive
(c) Zero
(d) Cannot be calculated
Answer: (c) Zero
In simple words: If you have any zero in your set of numbers, and you calculate the geometric mean, the answer will always be zero. This is because the geometric mean involves multiplying all the numbers together.

๐ŸŽฏ Exam Tip: The geometric mean is only defined for non-negative numbers, and if any value is zero, the product of all values becomes zero, making the geometric mean zero.

 

Question 4. central tendency is:
(a) Arithmetic mean
(b) Harmonic mean
(c) Geometric mean
(d) Median
Answer: (a) Arithmetic mean
In simple words: The arithmetic mean is one of the main ways to find the central tendency of data, which is like finding the typical or average value in a group of numbers. It is simply the sum of all values divided by the count of values.

๐ŸŽฏ Exam Tip: The question is asking for one of the measures of central tendency, and the arithmetic mean is a very common and fundamental one, often simply referred to as "the mean."

 

Question 5. The harmonic mean of the numbers 2, 3, 4 is:
(a) \( \frac{12}{13} \)
(b) 12
(c) \( \frac{36}{13} \)
(d) \( \frac{13}{36} \)
Answer: (c) \( \frac{36}{13} \)
Hint:
The harmonic mean (HM) is calculated by dividing the number of values by the sum of their reciprocals.
\( \mathrm{HM} = \frac{n}{\frac{1}{X_1} + \frac{1}{X_2} + \ldots + \frac{1}{X_n}} \)
Here, \( n = 3 \) and the numbers are 2, 3, 4.
\( \mathrm{HM} = \frac{3}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \)
First, find a common denominator for the fractions in the bottom, which is 12.
\( = \frac{3}{\frac{6}{12} + \frac{4}{12} + \frac{3}{12}} \)
\( = \frac{3}{\frac{6+4+3}{12}} \)
\( = \frac{3}{\frac{13}{12}} \)
To divide by a fraction, multiply by its reciprocal.
\( = 3 \times \frac{12}{13} \)
\( = \frac{36}{13} \)
In simple words: To find the harmonic mean, you take the total count of numbers and divide it by the sum of 1 divided by each number. For 2, 3, and 4, this calculation gives 36 divided by 13.

๐ŸŽฏ Exam Tip: Remember the formula for harmonic mean is \( n \) divided by the sum of the reciprocals. Always simplify the sum of fractions in the denominator first before performing the final division.

 

Question 6. The geometric mean of two numbers 8 and 18 shall be:
(a) 12
(b) 13
(c) 15
Answer: (a) 12
Hint:
The geometric mean (GM) for two numbers is found by multiplying them and then taking the square root of the product.
\( \mathrm{GM} = \sqrt{X_1 \cdot X_2} \)
For the numbers 8 and 18:
\( \mathrm{GM} = \sqrt{8 \times 18} \)
\( = \sqrt{144} \)
\( = 12 \)
In simple words: To find the geometric mean of two numbers, you multiply them together and then find the square root of that result. For 8 and 18, their geometric mean is 12.

๐ŸŽฏ Exam Tip: The geometric mean is typically used for positive numbers and is defined as the \( n \)-th root of the product of \( n \) numbers. For two numbers, it's the square root of their product.

 

Question 7. The correct relationship among A.M., G.M.and H.M.is:
(a) A.M. < G.M. < H.M.
(b) G.M. > A.M. > H.M.
(c) H.M. > G.M. > A.M.
(d) A.M. > G.M. > H.M.
Answer: (d) A.M. > G.M. > H.M.
In simple words: For any set of positive numbers (that are not all the same), the arithmetic mean is always greater than the geometric mean, which is always greater than the harmonic mean. This order is a basic rule.

๐ŸŽฏ Exam Tip: This relationship (A.M. \( \ge \) G.M. \( \ge \) H.M.) is a fundamental inequality for positive real numbers and holds true, with equality only when all numbers are identical.

 

Question 8. Harmonic mean is the reciprocal of:
(a) Median of the values.
(b) Geometric mean of the values.
(c) Arithmetic mean of the reciprocal of the values.
(d) Quartiles of the values.
Answer: (c) Arithmetic mean of the reciprocal of the values.
In simple words: The harmonic mean is found by taking the average of the "flips" of your numbers, and then flipping that average back again. So it's the reciprocal of the arithmetic mean of the reciprocals.

๐ŸŽฏ Exam Tip: Understanding this definition helps clarify the harmonic mean's purpose, especially when dealing with rates or ratios where averaging reciprocals is more appropriate.

 

Question 9. Median is same as:
(b) Q2
(c) Q3
(d) D2
Answer: (b) Q2
In simple words: The median is the middle value in a sorted list of numbers. In statistics, this middle value is also known as the second quartile (Q2), which divides the data into two equal halves.

๐ŸŽฏ Exam Tip: Q2, the median, and the 50th percentile all represent the same point in a dataset: the value below which 50% of the data falls.

 

Question 10. The median of 10, 14, 11, 9, 8, 12, 6 is:
(a) 10
(b) 12
(c) 14
(d) 9
Answer: (a) 10
Hint:
First, arrange the numbers in ascending order (from smallest to largest):
6, 8, 9, 10, 11, 12, 14
Count the number of observations (n). Here, \( n = 7 \).
Since \( n \) is an odd number, the median is the \( \left(\frac{n+1}{2}\right)^{th} \) value.
\( = \left(\frac{7+1}{2}\right)^{th} \) value
\( = \left(\frac{8}{2}\right)^{th} \) value
\( = 4^{th} \) value
In the sorted list, the 4th value is 10.
In simple words: To find the median, first put all the numbers in order from smallest to largest. Then, find the number that is exactly in the middle. For this set of numbers, when put in order, 10 is the middle number.

๐ŸŽฏ Exam Tip: Always remember to arrange the data in ascending or descending order before finding the median. If there's an even number of values, the median is the average of the two middle values.

 

Question 11. The mean of the values 11, 12, 13, 14 and 15 is:
(a) 15
(b) 11
(c) 12.5
(d) 13
Answer: (d) 13
Hint:
The mean is calculated by summing all the values and dividing by the count of the values.
Sum of values = \( 11 + 12 + 13 + 14 + 15 = 65 \)
Number of values = 5
Mean = \( \frac{\text{Sum of values}}{\text{Number of values}} \)
Mean = \( \frac{65}{5} = 13 \)
In simple words: To find the mean (or average), you add up all the numbers and then divide by how many numbers you have. When you do this for 11, 12, 13, 14, and 15, the average is 13.

๐ŸŽฏ Exam Tip: For an arithmetic progression (numbers with a constant difference), the mean is simply the middle term if the count of numbers is odd, as is the case here.

 

Question 12. If the mean of 1, 2, 3, ..., n is \( \frac{6 n}{11} \), then the value of n is:
(a) 10
(b) 12
(c) 11
(d) 13
Answer: (c) 11
Hint:
The mean of the first \( n \) natural numbers (1, 2, 3, ..., n) is given by the formula \( \frac{n+1}{2} \).
We are given that the mean is \( \frac{6n}{11} \).
So, we set the two expressions for the mean equal:
\( \frac{n+1}{2} = \frac{6n}{11} \)
Now, cross-multiply to solve for \( n \):
\( 11(n+1) = 2 \times 6n \)
\( 11n + 11 = 12n \)
Subtract \( 11n \) from both sides:
\( 11 = 12n - 11n \)
\( 11 = n \)
Therefore, \( n = 11 \).
In simple words: The average of the numbers from 1 up to \( n \) is usually \( \frac{n+1}{2} \). The problem tells us this average is \( \frac{6n}{11} \). By making these two equal and solving, we find that \( n \) must be 11.

๐ŸŽฏ Exam Tip: Remember the formula for the sum of the first \( n \) natural numbers \( \left(\frac{n(n+1)}{2}\right) \) and how it leads to the mean \( \left(\frac{n+1}{2}\right) \). This makes solving such problems much quicker.

 

Question 13. The harmonic mean is better than other means if the data are for:
(a) Speed or rates.
(b) Heights or lengths.
(c) Binary values like 0 and 1.
(d) Ratios or proportions.
Answer: (a) Speed or rates.
In simple words: The harmonic mean is especially useful when you are averaging rates, like speed, or other quantities that involve a reciprocal relationship. It gives a more meaningful average in these cases than other types of means.

๐ŸŽฏ Exam Tip: Always choose the correct type of mean based on the nature of the data. Harmonic mean is ideal for averaging rates (e.g., speed, efficiency) and time-based averages, while geometric mean is for growth rates, and arithmetic mean for simple sums.

 

Question 14. The first quartile is also known as:
(a) median
(b) lower quartile
(c) mode
(d) third decile
Answer: (b) lower quartile
In simple words: When you arrange a set of data in order, the first quartile (Q1) is the value that separates the lowest 25% of the data from the rest. It's often called the lower quartile.

๐ŸŽฏ Exam Tip: Quartiles divide a dataset into four equal parts. Q1 (lower quartile) marks the 25th percentile, Q2 (median) marks the 50th, and Q3 (upper quartile) marks the 75th percentile.

 

Question 15. If Q1 = 30 and Q3 = 50, the coefficient of quartile deviation is:
(a) 20
(b) 40
(c) 10
(d) 0.25
Answer: (d) 0.25
Hint:
The coefficient of quartile deviation is calculated using the formula:
Coefficient of Quartile Deviation \( = \frac{Q_3 - Q_1}{Q_3 + Q_1} \)
Given \( Q_1 = 30 \) and \( Q_3 = 50 \).
Substitute the values into the formula:
Coefficient of Quartile Deviation \( = \frac{50 - 30}{50 + 30} \)
\( = \frac{20}{80} \)
\( = \frac{1}{4} \)
\( = 0.25 \)
In simple words: To find the coefficient of quartile deviation, you subtract the first quartile from the third quartile, and then divide that result by the sum of the first and third quartiles. For the given values, this calculation gives 0.25.

๐ŸŽฏ Exam Tip: The coefficient of quartile deviation is a relative measure of dispersion, useful for comparing the variability of different datasets because it is a pure number, without units.

 

Question 16. If median = 45 and its coefficient is 0.25, then the mean deviation about median is:
(a) 11.25
(b) 180
(c) 0.0056
(d) 45
Answer: (a) 11.25
Hint:
The formula for the coefficient of mean deviation is:
Coefficient of M.D \( = \frac{\text{Mean Deviation}}{\text{Median}} \)
We are given:
Median \( = 45 \)
Coefficient of M.D \( = 0.25 \)
We need to find the Mean Deviation (MD).
Rearrange the formula to solve for MD:
MD \( = \text{Coefficient of M.D} \times \text{Median} \)
Substitute the given values:
MD \( = 0.25 \times 45 \)
MD \( = 11.25 \)
In simple words: The problem gives us the median and the coefficient of mean deviation. We know that the coefficient is found by dividing the mean deviation by the median. So, to find the mean deviation, we multiply the given coefficient by the median, which results in 11.25.

๐ŸŽฏ Exam Tip: Always ensure you use the correct formula and rearrange it properly when solving for a specific component. In this case, understanding the relationship between the coefficient, mean deviation, and median is key.

 

Question 17. The two events A and B are mutually exclusive if:
(a) P(A \( \cap \) B) = 0
(b) P(A \( \cap \) B) = 1
(c) P(A \( \cup \) B) = 0
(d) P(A \( \cup \) B) = 1
Answer: (a) P(A \( \cap \) B) = 0
In simple words: Mutually exclusive events are events that cannot happen at the same time. This means their intersection (A \( \cap \) B), which is the chance of both happening, is zero.

๐ŸŽฏ Exam Tip: For mutually exclusive events, there is no overlap in their outcomes, so the probability of both occurring simultaneously is 0. This is a fundamental concept in probability.

 

Question 18. The events A and B are independent if:
(a) P(A \( \cap \) B) = 0
(b) P(A \( \cap \) B) = P(A) \( \times \) P(B)
(c) P(A \( \cap \) B) = P(A) + P(B)
(d) P(A \( \cup \) B) = P(A) \( \times \) P(B)
Answer: (b) P(A \( \cap \) B) = P(A) \( \times \) P(B)
In simple words: Two events are independent if the occurrence of one does not affect the probability of the other happening. Mathematically, this means the probability of both events happening together is simply the product of their individual probabilities.

๐ŸŽฏ Exam Tip: Understand the key difference between mutually exclusive events (no overlap) and independent events (no influence). For independent events, the multiplication rule for intersection applies.

 

Question 19. If two events A and B are dependent then the conditional probability of P(B/A) is:
(a) P(A) P(B/A)
(b) \( \frac{P(A \cap B)}{P(B)} \)
(c) \( \frac{P(A \cap B)}{P(A)} \)
Answer: (c) \( \frac{P(A \cap B)}{P(A)} \)
In simple words: When events are dependent, the probability of event B happening given that event A has already occurred, written as P(B/A), is found by dividing the probability of both A and B happening by the probability of A happening.

๐ŸŽฏ Exam Tip: Conditional probability \( P(B|A) \) is critical for dependent events, as it tells us how the likelihood of B changes once we know A has happened. It's often read as "the probability of B given A."

 

Question 20. The probability of drawing a spade from a pack of card is:
(a) \( \frac{1}{52} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{1}{4} \)
Answer: (d) \( \frac{1}{4} \)
Hint:
A standard pack of cards has 52 cards in total.
There are 4 suits, and each suit has 13 cards.
The number of spade cards is 13.
The probability of drawing a spade card is the number of spade cards divided by the total number of cards.
Probability \( = \frac{\text{Number of spades}}{\text{Total number of cards}} \)
Probability \( = \frac{13}{52} \)
Simplify the fraction:
Probability \( = \frac{1}{4} \)
In simple words: A deck of cards has 52 cards, and 13 of them are spades. So, the chance of picking a spade is 13 out of 52, which simplifies to 1 out of 4.

๐ŸŽฏ Exam Tip: Always remember that a standard deck has 52 cards, with 13 cards in each of the four suits (hearts, diamonds, clubs, spades). This helps quickly calculate probabilities for card-related questions.

 

Question 21. If the outcome of one event does not influence another event then the two events are:
(a) Mutually exclusive
(b) Dependent
(c) Not disjoint
(d) Independent
Answer: (d) Independent
In simple words: When one event happening does not change the likelihood of another event happening, those two events are called independent. They do not affect each other at all.

๐ŸŽฏ Exam Tip: Independence is a crucial concept in probability. If events are independent, the conditional probability of one event given the other is simply the marginal probability of that event.

 

Question 22. Let a sample space of an experiment be \( S = \{E_1, E_2, \ldots, E_n\} \) then \( \sum_{i=1}^{n} P(E_i) \) is equal to:
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{3} \)
Answer: (b) 1
Hint:
The sum of the probabilities of all possible outcomes in a sample space must always be equal to 1. This is a fundamental axiom of probability.
\( \text{i.e., } \sum_{i=1}^{n} P(E_i) = 1 \)
In simple words: If you add up the probabilities of all the possible results from an experiment, the total will always be 1. This means that one of those results is guaranteed to happen.

๐ŸŽฏ Exam Tip: This is a core property of probability distributions. The total probability of all elementary events in a sample space must sum to 1, representing certainty.

 

Question 23. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is:
(a) \( \frac{1}{36} \)
(b) 0
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{6} \)
Answer: (a) \( \frac{1}{36} \)
Hint:
When a pair of dice is rolled, the total number of possible outcomes in the sample space is \( 6 \times 6 = 36 \).
The only even prime number is 2.
We need to obtain an even prime number on *each* die. This means both dice must show a 2.
The only event where this happens is (2, 2).
So, there is only 1 favorable outcome.
Probability \( = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \)
Probability \( = \frac{1}{36} \)
In simple words: When you roll two dice, there are 36 possible results. The only even number that is also a prime number is 2. For both dice to show an even prime number, both must show 2. There's only one way for this to happen (2, 2), so the probability is 1 out of 36.

๐ŸŽฏ Exam Tip: Remember that 2 is the only even prime number. When calculating probabilities for multiple independent events (like rolling two dice), multiply the probabilities of each individual event, or count the specific favorable outcomes in the full sample space.

 

Question 24. Probability of an impossible event is:
(a) 1
(b) 0
(c) 0.2
(d) 0.5
Answer: (b) 0
In simple words: An impossible event is something that can never happen. Because it can never occur, the chance or probability of it happening is zero.

๐ŸŽฏ Exam Tip: The probability of an event always lies between 0 and 1, inclusive. An impossible event has a probability of 0, and a certain event has a probability of 1.

 

Question 25. The probability that at least one of the events A, B occur is:
(a) P(A \( \cup \) B)
(b) P(A \( \cap \) B)
(c) P(A/B)
(d) (A \( \cup \) B)
Answer: (a) P(A \( \cup \) B)
In simple words: The phrase "at least one of the events A, B occur" means either A happens, or B happens, or both happen. In probability, this is represented by the union of events A and B, written as P(A \( \cup \) B).

๐ŸŽฏ Exam Tip: Remember that "at least one" implies the union of events, and for any two events A and B, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

TN Board Solutions Class 11 Business Maths Chapter 08 Descriptive Statistics and Probability

Students can now access the TN Board Solutions for Chapter 08 Descriptive Statistics and Probability prepared by teachers on our website. These solutions cover all questions in exercise in your Class 11 Business Maths textbook. Each answer is updated based on the current academic session as per the latest TN Board syllabus.

Detailed Explanations for Chapter 08 Descriptive Statistics and Probability

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 11 Business Maths chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 11 students who want to understand both theoretical and practical questions. By studying these TN Board Questions and Answers your basic concepts will improve a lot.

Benefits of using Business Maths Class 11 Solved Papers

Using our Business Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 11 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 08 Descriptive Statistics and Probability to get a complete preparation experience.

FAQs

Where can I find the latest Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 for the 2026-27 session?

The complete and updated Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 is available for free on StudiesToday.com. These solutions for Class 11 Business Maths are as per latest TN Board curriculum.

Are the Business Maths TN Board solutions for Class 11 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Business Maths concepts are applied in case-study and assertion-reasoning questions.

How do these Class 11 TN Board solutions help in scoring 90% plus marks?

Toppers recommend using TN Board language because TN Board marking schemes are strictly based on textbook definitions. Our Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 will help students to get full marks in the theory paper.

Do you offer Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 11 Business Maths. You can access Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 in both English and Hindi medium.

Is it possible to download the Business Maths TN Board solutions for Class 11 as a PDF?

Yes, you can download the entire Samacheer Kalvi Class 11 Business Maths Solutions Chapter 8 Descriptive Statistics and Probability Ex 8.3 in printable PDF format for offline study on any device.