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Chapter 25

Measures of Central Tendency (Mean) — Assertion-Reason Type Questions

Class - 10 RS Aggarwal Mathematics Solutions



Assertion–Reason Type Questions

Question 1

Assertion (A): The mean of first 9 natural numbers is 4.5.

Reason (R): Mean = Sum of all observationsTotal number of observations\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}

  1. Both A and R are true, and R is the correct explanation of A.

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

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

First 9 natural numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9

By formula,

Mean =Sum of all observationsTotal number of observations=1+2+3+4+5+6+7+8+99=459=5.\Rightarrow \text{Mean } = \dfrac{\text{Sum of all observations}}{\text{Total number of observations}}\\[1em] = \dfrac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{9}\\[1em] = \dfrac{45}{9}\\[1em] = 5.

∴ (A) is false, (R) is true.

Hence, option 4 is the correct option.

Question 2

Assertion (A) : For a grouped frequency distribution, we use Mean = A + (ΣftΣf)\Big(\dfrac{\Sigma f t}{\Sigma f}\Big) × h to find the mean using step deviation method.

Reason (R) : Here t = xAh\dfrac{x − A}{h}.

  1. Both A and R are true, and R is the correct explanation of A.

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

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

The standard formula to calculate the mean (xˉ\bar{x}) using the step-deviation method is:

xˉ=A+(ftf)×h\bar{x} = A + \Big( \dfrac{\sum ft}{\sum f} \Big) \times h

∴ Assertion (A) is true.

The step-deviation is defined as the difference between the class mark (x) and the assumed mean (A), divided by the class size (h):

t=xAht = \dfrac{x - A}{h}

∴ Reason (R) is true.

Reason (R) defines the step-deviation t that appears in the formula stated in Assertion (A), so R is the correct explanation of A.

Hence, option 1 is the correct option.

Question 3

Assertion (A) : If xi's are the mid-points of the class intervals of a grouped data, fi's are the corresponding frequencies and x̄ is the mean, then Σfi(xi − x̄) = 1.

Reason (R) : The sum of the deviations from the mean is 0.

  1. Both A and R are true, and R is the correct explanation of A.

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

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

The expression fi(xixˉ)\sum f_i(x_i - \bar{x}) represents the sum of the deviations of all observations from their mean, weighted by their frequencies.

fi(xixˉ)=fixifixˉ=fixixˉfi .....(1)\Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \sum f_i \bar x \\[1em] = \sum f_ix_i - \bar x \sum f_i \text{ .....(1)}

We know that,

xˉ=fixifixˉfi=fixi\bar x = \dfrac{\sum f_i x_i}{\sum f_i} \\[1em] \bar x \sum f_i = \sum f_i x_i

Substituting the values in equation (1),

fi(xixˉ)=xˉfixˉfi=0\therefore \sum f_i(x_i - \bar x) = \bar x \sum f_i - \bar x \sum f_i = 0

Since the value is 0 (not 1),

∴ Assertion (A) is false.

The sum of the deviations from the mean is 0. This is a fundamental and correct property of the arithmetic mean.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

Question 4

Assertion (A) : Out of 25 numbers, the mean of 15 of them is 18. If the mean of the remaining numbers is 13, then the mean of the 25 numbers is 14.

Reason (R) : Mean of the variates x1, x2, …, xn having corresponding frequencies f1, f2, …, fn is given by x̄ = (f1x1+f2x2++fnxnf1+f2++fn)\Big(\dfrac{f_1 x_1 + f_2 x_2 + \dots + f_n x_n}{f_1 + f_2 + \dots + f_n}\Big).

  1. Both A and R are true, and R is the correct explanation of A.

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

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Mean = Sum of termsNumber of terms\dfrac{\text{Sum of terms}}{\text{Number of terms}}

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of remaining 10 numbers is 13

∴ Sum of remaining terms = 13 × 10 = 130

Sum of 25 terms = 270 + 130 = 400.

Mean = 40025\dfrac{400}{25} = 16

Since the mean of the 25 numbers is 16 (not 14),

∴ Assertion (A) is false.

The mean of variates having corresponding frequencies is correctly given by xˉ=fixifi\bar x = \dfrac{\sum f_ix_i}{\sum f_i}.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

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