Question

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

Answer 1

Hint: Considering a set of $3{{n}}$ numbers having variance 4. Now determine the value of the mean and the variance and then the obtained value of variance is equated with k to determine the value of 9k.

Formula Used:
The mean is $\dfrac{{\sum {\left( {{x_i} + 1} \right)} + \sum {\left( {{y_i} - 1} \right)} }}{{3n}}$
The variance is $\dfrac{{\sum {{{\left( {{x_i} + 1} \right)}^2}} + \sum {{{\left( {{y_i} - 1} \right)}^2}} }}{{3n}}$

Complete step by step solution:
Let first 2n observations are ${x_1},{x_2} \ldots \ldots \ldots \ldots ,{x_{2n}}$ and last $n$ observations are ${y_1},{y_2} \ldots \ldots \ldots \ldots ,{y_n}$
Now, $\dfrac{{\sum {{x_i}} }}{{2n}} = 6,\;\;\;\dfrac{{\sum {{y_i}} }}{n} = 3$
Evaluating the values of the summation from above terms, we get
$ \Rightarrow \sum {{x_i}} = 12n,\;\;\sum {{y_i}} = 3n\;\;\;\;\;\therefore \dfrac{{\sum {{x_i}} + \sum {{y_i}} }}{{3n}} = \dfrac{{15n}}{{3n}} = 5$
Now considering a set of $3{{n}}$ numbers having variance 4
${{Now, }}\dfrac{{\sum {x_i^2} + \sum {y_i^2} }}{{3n}} - {5^2} = 4$
Evaluate the above term
$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n$
Simplify the expression, we get
$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 87n$
Now, mean is $\dfrac{{\sum {\left( {{x_i} + 1} \right)} + \sum {\left( {{y_i} - 1} \right)} }}{{3n}} = \dfrac{{15n + 2n - n}}{{3n}} = \dfrac{{16}}{3}$
Now, variance is $\dfrac{{\sum {{{\left( {{x_i} + 1} \right)}^2}} + \sum {{{\left( {{y_i} - 1} \right)}^2}} }}{{3n}} - {\left( {\dfrac{{16}}{3}} \right)^2}$
Expanding the numerator to determine the variance.
Variance $ = \dfrac{{\sum {x_i^2} + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n}}{{3n}} - {\left( {\dfrac{{16}}{3}} \right)^2}$
Evaluating the above expression to determine the variance.
Variance $ = \dfrac{{87n + 2(9n) + 3n}}{{3n}} - {\left( {\dfrac{{16}}{3}} \right)^2}$
Evaluating the above expression to determine the variance, we get
Variance $ = 29 + 6 + 1 - {\left( {\dfrac{{16}}{3}} \right)^2}$
Evaluate the above terms,
Variance$ = [324 - 256]/9$
As the considered value of variance is k. So, equate the value obtained with k, we get
Variance $= \dfrac{68}{9} = k$
Evaluating the value of 9k
9k = 68

Therefore, the correct value of 9k is 68.

Note: The mean $\dfrac{{\sum {\left( {{x_i} + 1} \right)} + \sum {\left( {{y_i} - 1} \right)} }}{{3n}}$and the variance $\dfrac{{\sum {{{\left( {{x_i} + 1} \right)}^2}} + \sum {{{\left( {{y_i} - 1} \right)}^2}} }}{{3n}}$ should be evaluated by the proposed value and the considered set. Also, such problems can be evaluated by considering values in the set and then evaluating them for satisfying the mean and the variance.