Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n number is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to ______.

<p>Let X1, ., X2n be the first 2n observations and Y1, ., Yn be the last n observations.<br>Given:<br>ΣXi / 2n = 6 =&gt; ΣXi = 12n (i)<br>ΣYi / n = 3 =&gt; ΣYi = 3n (ii)</p><p>Combined mean: (ΣXi + ΣYi) / 3n = 5 =&gt; ΣXi + ΣYi = 15n. This is consistent with (i) and (ii).</p><p>Combined variance: (ΣXi^2 + ΣYi^2) / 3n - (mean)^2 = 4<br> (ΣXi^2 + ΣYi^2) / 3n - 5^2 = 4<br>ΣXi^2 + ΣYi^2 = (4 + 25) * 3n = 87n.</p><p>New observations are Xi + 1 and Yi - 1.<br>New mean = (Σ (Xi + 1) + Σ (Yi - 1) / 3n = (ΣXi + 2n + ΣYi - n) / 3n = (15n + n) / 3n = 16n / 3n = 16/3.</p><p>New variance k:<br>k = (Σ (Xi + 1)^2 + Σ (Yi - 1)^2) / 3n - (new mean)^2<br>k = (Σ (Xi^2 + 2Xi + 1) + Σ (Yi^2 - 2Yi + 1) / 3n - (16/3)^2<br>k = (ΣXi^2 + ΣYi^2 + 2ΣXi - 2ΣYi + 2n + n) / 3n - 256/9<br>k = (87n + 2 (12n) - 2 (3n) + 3n) / 3n - 256/9<br>k = (87n + 24n - 6n + 3n) / 3n - 256/9<br>k = 108n / 3n - 256/9 = 36 - 256/9 = (324 - 256)/9 = 68/9.</p><p>So, k = 68/9, which means 9k = 68.</p>

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