Question:
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 \(n\) numbers 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 __________.
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 \(n\) numbers 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 __________.
Show Hint
Find the overall mean (5) and \(\sum x^2\) from the variance, then compute the new \(\sum x'^2\) and the new mean.
Updated On: Nov 4, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 68
Solution and Explanation
Original mean \(= \frac{2n\cdot6 + n\cdot3}{3n} = 5\). Variance 4 gives \(\frac{\sum x^2}{3n} = 4 + 25 = 29\), so \(\sum x^2 = 87n\).
Sum of first 2n numbers \(= 12n\); sum of last n numbers \(= 3n\).
New set: \(\sum x'^2 = \sum_{first}(x+1)^2 + \sum_{last}(x-1)^2 = \sum x^2 + 2(12n) + 2n - 2(3n) + n = 87n + 24n + 2n - 6n + n = 108n\).
New mean \(= \frac{(12n + 2n) + (3n - n)}{3n} = \frac{16}{3}\).
\(k = \frac{108n}{3n} - \left(\frac{16}{3}\right)^2 = 36 - \frac{256}{9} = \frac{68}{9}\).
\(9k = 68\).
Answer: 68
Sum of first 2n numbers \(= 12n\); sum of last n numbers \(= 3n\).
New set: \(\sum x'^2 = \sum_{first}(x+1)^2 + \sum_{last}(x-1)^2 = \sum x^2 + 2(12n) + 2n - 2(3n) + n = 87n + 24n + 2n - 6n + n = 108n\).
New mean \(= \frac{(12n + 2n) + (3n - n)}{3n} = \frac{16}{3}\).
\(k = \frac{108n}{3n} - \left(\frac{16}{3}\right)^2 = 36 - \frac{256}{9} = \frac{68}{9}\).
\(9k = 68\).
Answer: 68
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Statistics Questions
- For the following data table
\(x_i\)
\(f_i\)
0 - 4 2 4 - 8 4 8 - 12 7 12 - 16 8 16 - 20 6 Find the value of 20M (where M is median of the data)
- The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If \( \mu \) and \( \sigma^2 \) denote the mean and variance of the correct observations respectively, then \( 15(\mu + \mu^2 + \sigma^2) \) is equal to \(\ldots\)
- Let \( a_1, a_2, \ldots, a_{10} \) be 10 observations such that \(\sum_{k=1}^{10} a_k = 50 \quad \text{and} \quad \sum_{1 \leq k < j \leq 10} a_k \cdot a_j = 1100\). Then the standard deviation of \( a_1, a_2, \ldots, a_{10} \) is equal to:
- Let M denote the median of the following frequency distribution.Then 20M is equal to:
\(x_i\)
\(f_i\)
0 - 4 2 4 - 8 4 8 - 12 7 12 - 16 8 16 - 20 6 - Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and \(\frac{205}{7}\)respectively. Then the mean deviation about the mean of these 7 observations is :
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions