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:
Class Interval  36  69  912  1215  1518  1821 
Frequency  2  5  21  23  10  12 
The mode of the above data is 12.4.
Reason : The value of the variable which occurs most often is the mode.
(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{2321}{2\times 232110}\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.
x  4  7  10  13  16  19 
f  7  10  15  20  25  30 
Reason : `\overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}`
(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:
Class Interval  1015  1520  2025  2530  3035 
Frequency  5  9  12  6  8 
The modal class is 1015.
Reason : The class having maximum frequency is called the modal class.
(d) : Clearly, Reason is correct.
The maximum frequency is 12, which lies in the interval 20 – 25. So, the modal class is 2025.
`\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{1414}{281415}\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:
Class Interval  04  48  812  1216  1620 
Frequency  6  3  5  20  10 
The median class is 1216.
Reason : Let n = `\sum f_{i}`. Then, the class whose cumulative frequency is just lesser than `\left(\frac{n}{2}\right)` is the median class.
(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:
Class Interval  Frequency  Cumulative Frequency 

04  6  6 
48  3  9 
812  5  14 
1216  20  34 
1620  10  44 
Here, `n` = 44 ⇒ `\frac{n}{2}`= 22, which lies in the interval 12 – 16.
So, it is the median class.
`\therefore` Assertion is correct but Reason is wrong.
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 StatementII, 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.
Class Interval  010  1020  2030  3040  4050 
Frequency  5  18  15  16  6 
Reason : Mean`(\overline{x})=\frac{\sum f_{i}x_{i}}{\sum f_{i}}`, where `x_{i}` = `\frac{1}{2}`(Lower limit+Upper limit) of the `i^{th}` class interval and `f_{i}` is its frequency.
(a) : Clearly, Reason is correct.
Now, the frequency distribution table from the given data can be drawn as :
Class Interval  Frequency (`f_{i}`)  `x_{i}`  `f_{i}x_{i}` 

010  5  5  25 
1020  18  15  270 
2030  15  25  375 
3040  16  35  560 
4050  6  45  270 
`\sum f_{i}`=60  `\sum f_{i}x_{i}`=1500 
`\therefore` Mean = `\frac{\sum f_{i}x_{i}}{\sum f_{i}}`
⇒ Mean = `\frac{1500}{60}`
⇒ Mean = 25, which is true.
`\therefore` Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

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