Assertion- We find midpoints of a class interval with the following formula- upper class limit-lower class limitReason- To draw frequency polygons independently without drawing histogram it is required to have the class mark-Which of the following is correct-

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Standard IX

Mathematics

Class Mark

Assertion: We...

Question

Assertion: We find midpoints of a class interval with the following formula.
$\frac{u p p e r c l a s s l i m i t + l o w e r c l a s s l i m i t}{2}$

Reason: To draw frequency polygons independently without drawing histogram it is required to have the class mark.

Which of the following is correct?

Both A and R are true and R is the correct explanation of A

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Both A and R are true, but R is not the correct explanation of A

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A is true and R is false

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A is false and R is true

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Solution

The correct option is B
Both A and R are true, but R is not the correct explanation of A

A: By definition, this is true
R: By definition, this is true
To draw the frequency polygon one has to find the class mark.
But, the first statemnt is independent of second statement. So, Reason is not correct explanation of Assertion.

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Similar questions

Assertion: We find midpoints of a class interval with the following formula.
$\frac{u p p e r c l a s s l i m i t + l o w e r c l a s s l i m i t}{2}$

Reason: To draw frequency polygons independently without drawing histogram it is required to have the class mark.

Which of the following is correct?

Q. Assertion: We find mid points of a class interval as upper as

\frac{u p p e r c l a s s l i m i t + l o w e r c l a s s l i m i t}{2}

Reason: To draw frequency polygons independently without drawing histogram it is required to have class mark.
Which of the following is correct

Q. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively
(a) upper limits of the classes
(b) lower limits of the classes
(c) class marks of the classes
(d) upper limits of preceding classes

Q. Question 22
For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abcissae are, respectively:

A) Upper limits of the classes
B) Lower limits of the classes
C) Class marks of the classes
D) Upper limits of preceding classes

A frequency polygon is constructed by plotting frequency of the class interval and the

(a) Upper limit of the class

(b) Lower limit of the class

(d) any values of the class

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Assertion: We find midpoints of a class interval with the following formula. upper class limit+lower class limit2 Reason: To draw frequency polygons independently without drawing histogram it is required to have the class mark. Which of the following is correct?