# Assertion Reason Maths Chapter 14: Statistics CBSE 10th

Assertion Reason Mathematics Chapter 14: Statistics CBSE 10th for Class 10th is very important as assertion reason and Case Study Based Passages have been introduced by CBSE in 2020. Assertion reason and Case Study Type of questions helps students to think Critically on every aspects of life.

Here is a collection of few questions for CBSE Class 10th Term 2 Exams. These Assertion Reason Questions are fully solved.

Directions: Mark the option which is most suitable :
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.

Question.1.
Assertion : Consider the following frequency distribution:

(b) : Clearly, Reason is correct.
The maximum frequency is 23 and the modal class is 12–15.
So, l = 12, f_{i} = 23, f_{0} = 21, f_{2} = 10 and h = 3
\therefore Mode = 12+\left(\frac{23-21}{2\times 23-21-10}\right) \times 3
= \left(12+3\times \frac{2}{15}\right)
= 12.4
\therefore Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

Question.2.
Assertion : If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.
Reason : Median = (mode + 2 mean)

(c) Assertion is true but reason is false.
Median =\frac{1}{3}(mode + 2 mean)
= \frac{1}{3}(60+2\times 66)
= 64

Question.3.
Assertion : The arithmetic mean of the following given frequency distribution table is 13.81.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Both assertion and reason are true, reason is the correct explanation of the assertion.

Question.4.
Assertion : If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27 then median is 30.
Reason : Median =(\frac{n+1}{2})^{th} value, if n is odd.

(d) Assertion is false but reason is true.
Arranging the terms in ascending order,
0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
Median value =(\frac{11+1}{2})^{th}
= 6^{th} value
= 27

Question.5.
Assertion : If the arithmetic mean of 5, 7, x, 10, 15 is x, then x = 9.25.
Reason : If x_{1}, x_{2}, x_{3}, …, x_{n} are n values of a variable X, then the arithmetic mean of these values is given by
\frac{x_{1}+x_{2}+x_{3}+…+x_{n}}{2n}.

(c) : If x_{1}, x_{2}, x_{3}, …, x_{n} are n values of a variable X, then the arithmetic mean of these values is given by
\frac{x_{1}+x_{2}+x_{3}+…+x_{n}}{n}
\therefore Reason is wrong.
Now, mean of given values = x [Given]
⇒ \frac{5+7+x+10+15}{5} = x
⇒ \frac{37+x}{5} = x
⇒ 37 + x = 5x
⇒ 37 = 4x
⇒ x = 9.25, which is correct
\therefore Assertion is correct but Reason is wrong.

Question.6.
Assertion : Consider the following frequency distribution:

(d) : Clearly, Reason is correct.
The maximum frequency is 12, which lies in the interval 20 – 25. So, the modal class is 20-25.
\therefore Assertion is wrong but Reason is correct.

Question.7.
Assertion : If for a certain frequency distribution, l = 24.5, h = 4, f_{0} = 14, f_{1} = 14, f_{2} = 15, then the value of mode is 25.
Reason : Mode of a frequency distribution is given by :
Mode = l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h.

(d) : Clearly, Reason is wrong.
Now, it is given that l = 24.5, h = 4, f_{0} = 14, f_{1} = 14, f_{2} = 15
\therefore Mode = 24.5+\left(\frac{14-14}{28-14-15}\right)\times 4
⇒ Mode = 24.5 + 0
⇒ Mode = 24.5
\therefore Assertion is wrong but Reason is correct.

Question.8.
Assertion : Consider the following frequency distribution:

(c) (c) : We know that, the class whose cumulative frequency is just greater than \left(\frac{n}{2}\right) is the median class.
So, Reason is wrong.
The cumulative frequency distribution table from the given data can be drawn as:

Question.9.
Assertion : If the median and mode of a frequency distribution are 8.9 and 9.2 respectively, then its mean is 9.
Reason : Mean, median and mode of a frequency distribution are related as:
mode = 3 median – 2 mean

(d) : Clearly, Reason is correct.
Using the relation given in Statement-II, we have
⇒ 2 Mean = 3 Median – Mode
⇒ 2 Mean = 3 \times 8.9 – 9.2
⇒ 2 Mean = 26.7 – 9.2
⇒ 2 Mean = 17.5
⇒ Mean = 8.75
\therefore Assertion is wrong but Reason is correct.

Question.10.
Assertion : The arithmetic mean of the following frequency distribution is 25.

(a) : Clearly, Reason is correct.
Now, the frequency distribution table from the given data can be drawn as :

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