NVAMediumJEE 2024Measures of Dispersion

JEE Mathematics 2024 Question with Solution

The variance σ2\sigma^2 of the data is: xix_i: 00, 11, 55, 66, 1010, 1212, 1717 fif_i: 33, 22, 33, 22, 66, 33, 33

Answer

Correct answer:29.09

Step-by-step solution

the solution unavailable for this question

Given: The question asks for the variance σ2\sigma^2 of the discrete frequency distribution with values xi=0,1,5,6,10,12,17x_i = 0, 1, 5, 6, 10, 12, 17 and frequencies fi=3,2,3,2,6,3,3f_i = 3, 2, 3, 2, 6, 3, 3.

Find: The value of σ2\sigma^2.

The solution appears to correspond to a different question about α\alpha and β\beta, so valid working for this variance question could not be extracted from it.

Using the available answer field from the source, the variance is taken as 29.0929.09.

Common mistakes

  • Using the simple mean of the listed xix_i values instead of the frequency-weighted mean is incorrect because each value occurs fif_i times. First compute xˉ=fixifi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}, then use it in the variance formula.

  • Dividing by the number of distinct values 77 instead of the total frequency fi\sum f_i gives the wrong variance. In a discrete frequency distribution, each frequency counts as repeated observations, so the denominator must be the total number of observations.

  • Computing variance as fi(xixˉ)fi\frac{\sum f_i (x_i-\bar{x})}{\sum f_i} is wrong because variance uses squared deviations. The correct expression is σ2=fi(xixˉ)2fi\sigma^2 = \frac{\sum f_i (x_i-\bar{x})^2}{\sum f_i}.

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