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NVAMediumJEE 2023Measures of Dispersion

JEE Mathematics 2023 Question with Solution

If the variance of the frequency distribution

xi2345678fi3616α956\begin{array}{|c|c|c|c|c|c|c|c|} \hline x_i & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline f_i & 3 & 6 & 16 & \alpha & 9 & 5 & 6 \\ \hline \end{array}

is given, find the value of α\alpha.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: The frequency distribution has values xi=2,3,4,5,6,7,8x_i = 2,3,4,5,6,7,8 and frequencies fi=3,6,16,α,9,5,6f_i = 3,6,16,\alpha,9,5,6. The variance is given as 33.

Find: The value of α\alpha.

Take deviations from 55:

di=xi−5d_i = x_i - 5

Then the table used in the solution is:

xifidi=xi−5fidi2fidi23−327−936−224−12416−116−165α0006919975220108635418\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & d_i=x_i-5 & f_i d_i^2 & f_i d_i \\ \hline 2 & 3 & -3 & 27 & -9 \\ 3 & 6 & -2 & 24 & -12 \\ 4 & 16 & -1 & 16 & -16 \\ 5 & \alpha & 0 & 0 & 0 \\ 6 & 9 & 1 & 9 & 9 \\ 7 & 5 & 2 & 20 & 10 \\ 8 & 6 & 3 & 54 & 18 \\ \hline \end{array}

Variance Calculation

Using the variance formula for grouped frequency data with deviations:

σ2=∑fidi2∑fi−(∑fidi∑fi)2\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2

From the table,

∑fidi=−9−12−16+0+9+10+18=0\sum f_i d_i = -9-12-16+0+9+10+18 = 0

and

∑fidi2=27+24+16+0+9+20+54=150\sum f_i d_i^2 = 27+24+16+0+9+20+54 = 150

Also, the total frequency is

∑fi=3+6+16+α+9+5+6=45+α\sum f_i = 3+6+16+\alpha+9+5+6 = 45+\alpha

Final Equation

Substitute in the variance expression:

3=15045+α−(045+α)23 = \frac{150}{45+\alpha} - \left(\frac{0}{45+\alpha}\right)^2

So,

3=15045+α3 = \frac{150}{45+\alpha}

which gives

150=135+3α150 = 135 + 3\alpha

Hence,

3α=153\alpha = 15

and therefore

α=5\alpha = 5

So the required value is 55.

Common mistakes

  • Using ∑fi=150\sum f_i = 150 is incorrect because 150150 is the value of ∑fidi2\sum f_i d_i^2, not the total frequency. First compute total frequency separately as 45+α45+\alpha.

  • Ignoring the term (∑fidi∑fi)2\left(\frac{\sum f_i d_i}{\sum f_i}\right)^2 without checking it is a mistake. Here it becomes zero only after verifying that ∑fidi=0\sum f_i d_i = 0.

  • Taking deviations from a wrong assumed mean leads to wrong entries in the table. The solution uses di=xi−5d_i = x_i - 5, so each deviation and its square must be computed with respect to 55 only.

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