© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Given Data
• We have a set of 3n numbers with variance 4.
• The first 2n numbers have mean 6.
• The remaining n numbers have mean 3.
• A new set is formed by adding 1 to each of the first 2n numbers and subtracting 1 from each of the remaining n numbers.
• We denote the variance of this new set by k and aim to find 9k.
Step 2: Represent the Data Mathematically
Let the first 2n observations be
x_{1}, x_{2}, \dots, x_{2n}
and the last n observations be
y_{1}, y_{2}, \dots, y_{n} .
Step 3: Use the Given Means to Find Summations
Given that the mean of the x_i is 6:
\frac{\sum x_i}{2n} = 6 \quad \Rightarrow \quad \sum x_i = 12n.
And the mean of the y_i is 3:
\frac{\sum y_i}{n} = 3 \quad \Rightarrow \quad \sum y_i = 3n.
Hence, the total sum of all 3n observations is:
\sum x_i + \sum y_i = 12n + 3n = 15n.
Thus, the overall mean of the 3n numbers is
\frac{15n}{3n} = 5.
Step 4: Use the Variance Formula for the Original Set
The variance is given by:
\text{Variance} = \frac{\sum x_i^2 + \sum y_i^2}{3n} - \Bigl(\text{mean}\Bigr)^2 = 4.
We know the mean is 5, so
\frac{\sum x_i^2 + \sum y_i^2}{3n} - 5^2 = 4.
That is,
\frac{\sum x_i^2 + \sum y_i^2}{3n} - 25 = 4.
\Rightarrow \frac{\sum x_i^2 + \sum y_i^2}{3n} = 29.
\Rightarrow \sum x_i^2 + \sum y_i^2 = 29 \cdot 3n = 87n.
Step 5: Compute the Mean of the New Set
In the new set, each of the first 2n numbers has been increased by 1, and each of the last n numbers has been decreased by 1. Thus, the sum of the new set’s observations is:
\sum (x_i + 1) + \sum (y_i - 1).
This equals:
\bigl(\sum x_i \bigr) + 2n \cdot 1 \;+\; \bigl(\sum y_i \bigr) - n \cdot 1.
We know \sum x_i = 12n and \sum y_i = 3n , so
12n + 2n + 3n - n = 12n + 3n + 2n - n = 16n.
Therefore, the mean of the new set (which still has 3n observations) is:
\frac{16n}{3n} = \frac{16}{3}.
Step 6: Express the Sum of Squares in the New Set
Each term of the new set is either (x_i + 1) or (y_i - 1) . Hence, the total sum of squares in the new set is:
\sum (x_i + 1)^2 + \sum (y_i - 1)^2.
Expanding,
\sum (x_i^2 + 2x_i + 1) + \sum (y_i^2 - 2y_i + 1).
= \sum x_i^2 + \sum y_i^2 + 2 \sum x_i - 2 \sum y_i + \bigl(2n + n\bigr).
We already have:
\sum x_i^2 + \sum y_i^2 = 87n,
\sum x_i = 12n,
\sum y_i = 3n.
So,
\sum x_i^2 + \sum y_i^2 + 2 \sum x_i - 2 \sum y_i + 3n = 87n + 2(12n) - 2(3n) + 3n.
= 87n + 24n - 6n + 3n = 108n.
Step 7: Calculate the Variance of the New Set (k)
The variance k is given by:
k = \frac{\sum (x_i + 1)^2 + \sum (y_i - 1)^2}{3n} - \Biggl(\frac{16}{3}\Biggr)^2.
Substitute the computed sum of squares and simplify:
k = \frac{108n}{3n} - \left(\frac{16}{3}\right)^2.
k = 36 - \frac{256}{9} = \frac{324}{9} - \frac{256}{9} = \frac{68}{9}.
Step 8: Find 9k
Since k = \frac{68}{9} , multiplying by 9 gives:
9k = 9 \times \frac{68}{9} = 68.
Final Answer
Therefore, 9k = 68.
