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MCQMediumJEE 2024Measures of Dispersion

JEE Mathematics 2024 Question with Solution

If the variance of the frequency distribution is 160160, then the value of c∈Nc \in \mathbb{N} is:

  • A

    55

  • B

    88

  • C

    77

  • D

    66

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The frequency distribution has values c,2c,3c,4c,5c,6cc, 2c, 3c, 4c, 5c, 6c with corresponding frequencies 2,1,1,1,1,12,1,1,1,1,1, and the variance is 160160.

Find: The natural number cc.

Using the variance formula for a frequency distribution,

Variance=∑fx2∑f−(∑fx∑f)2\text{Variance} = \frac{\sum f x^2}{\sum f} - \left(\frac{\sum f x}{\sum f}\right)^2

From the table,

∑f=2+1+1+1+1+1=7\sum f = 2+1+1+1+1+1 = 7 ∑fx=2c+2c+3c+4c+5c+6c=22c\sum fx = 2c + 2c + 3c + 4c + 5c + 6c = 22c ∑fx2=2c2+4c2+9c2+16c2+25c2+36c2=92c2\sum f x^2 = 2c^2 + 4c^2 + 9c^2 + 16c^2 + 25c^2 + 36c^2 = 92c^2

Substitute these into the variance formula:

160=92c27−(22c7)2160 = \frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2 160=92c27−484c249160 = \frac{92c^2}{7} - \frac{484c^2}{49} 160=644c2−484c249160 = \frac{644c^2 - 484c^2}{49} 160=160c249160 = \frac{160c^2}{49}

Now solve for cc:

160c249=160\frac{160c^2}{49} = 160 160c2=160×49160c^2 = 160 \times 49 c2=49c^2 = 49 c=7c = 7

Therefore, the correct option is C, and the value of cc is 77.

Using Mean First

First compute the mean:

xˉ=∑fx∑f=22c7\bar{x} = \frac{\sum fx}{\sum f} = \frac{22c}{7}

Then use

Variance=∑f(xi−xˉ)2∑f\text{Variance} = \frac{\sum f(x_i-\bar{x})^2}{\sum f}

which is equivalently

Variance=∑fx2∑f−xˉ2\text{Variance} = \frac{\sum f x^2}{\sum f} - \bar{x}^2

Substituting,

160=92c27−(22c7)2=92c27−484c249=160c249160 = \frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2 = \frac{92c^2}{7} - \frac{484c^2}{49} = \frac{160c^2}{49}

Hence,

c2=49⇒c=7c^2 = 49 \Rightarrow c = 7

So the correct option is C.

Common mistakes

  • Using the wrong variance formula by averaging only the values and ignoring frequencies. This is incorrect because each value contributes according to its frequency. Always compute ∑f\sum f, ∑fx\sum fx, and ∑fx2\sum f x^2 with the given frequencies.

  • Adding ∑fx\sum fx incorrectly. A common error is to miss that the value cc has frequency 22, so its contribution is 2c2c. Recheck each term carefully before finding the mean.

  • Making an algebra mistake while simplifying

    92c27−484c249\frac{92c^2}{7} - \frac{484c^2}{49}

    The correct common denominator is 4949, giving 644c2−484c249=160c249\frac{644c^2 - 484c^2}{49} = \frac{160c^2}{49}. Convert to a common denominator before subtracting.

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