Question

Consider a data consisting of 10 observations \[ x_1, x_2, \dots, x_{10}, \] whose mean is \(5\) and variance is \(7\). If the mean and the variance of the first 8 observations \[ x_1, x_2, \dots, x_8 \] are \(4\) and \(3.5\), respectively, and \[ x_9<x_{10}, \] then the value of \[ 3x_9 + 2x_{10} \] is:

Hint:

Try shifting all the observations by the overall mean first. This makes the total sum zero and removes the mean-squared term when converting variance into sum of squares, keeping the numbers smaller and less error-prone. Once you know the sum and sum of squares of the two missing observations, write them as the average of the pair plus or minus a single unknown difference rather than setting up a full quadratic equation.

Answer 1

Correct Answer: 44

Step 1: Understanding the Question:
We are given statistical parameters for a set of 10 observations and a subset of 8 observations. We need to find the values of the remaining two observations, $x_9$ and $x_{10}$, and calculate a linear combination of them.

Step 2: Key Formula or Approach:

• Sum of observations: $\sum x_i = n \bar{x}$.

• Sum of squares of observations: $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$.

Step 3: Detailed Explanation:

• For $n=10$, $\bar{x}=5, \sigma^2=7 \implies \sum_{i=1}^{10} x_i = 10 \times 5 = 50$.
$\sum_{i=1}^{10} x_i^2 = 10(7 + 5^2) = 10(32) = 320$.

• For the first 8 observations, $\bar{x}_1=4, \sigma_1^2=3.5 \implies \sum_{i=1}^8 x_i = 8 \times 4 = 32$.
$\sum_{i=1}^8 x_i^2 = 8(3.5 + 4^2) = 8(19.5) = 156$.

• Now, find equations for $x_9$ and $x_{10}$:
$x_9 + x_{10} = \sum_{1}^{10} x_i - \sum_{1}^8 x_i = 50 - 32 = 18$.
$x_9^2 + x_{10}^2 = \sum_{1}^{10} x_i^2 - \sum_{1}^8 x_i^2 = 320 - 156 = 164$.

• Use the identity $(x_9 + x_{10})^2 - 2x_9x_{10} = x_9^2 + x_{10}^2$:
$18^2 - 2x_9x_{10} = 164 \implies 324 - 164 = 2x_9x_{10} \implies x_9x_{10} = 80$.

• $x_9$ and $x_{10}$ are roots of $t^2 - 18t + 80 = 0 \implies (t-10)(t-8) = 0$.

• Since $x_9 < x_{10}$, we have $x_9 = 8$ and $x_{10} = 10$.

• Value $= 3(8) + 2(10) = 24 + 20 = 44$.

Step 4: Final Answer:
The value of $3x_9 + 2x_{10}$ is 44.

Answer 2

Concept:

• Variance does not change when every observation in a data set is shifted by the same constant, since it depends only on how spread out the values are, not on their position. Shifting all values by the overall mean makes the mean of the shifted values zero, which removes the mean-squared term from the sum-of-squares formula for that set.
• Once the sum and sum of squares of exactly two unknown numbers are known, writing them as their average plus or minus a single deviation variable solves for them directly, without forming and factoring a quadratic equation.

Step 1: Shift every observation by the overall mean.
Let $y_i=x_i-5$ for $i=1,2,\ldots,10$.

Step 2: Use the shift to simplify the full data set.
Mean of $y$ (all $10$ values) $=5-5=0$. Variance is unaffected by a shift, so variance of $y=7$.
Since the mean of $y$ is $0$: $\sum_{i=1}^{10}y_i=10(0)=0$, and $\sum_{i=1}^{10}y_i^2=10(7)=70$ (the mean-squared term vanishes because the mean is $0$).

Step 3: Apply the same shift to the first $8$ observations.
Mean of $y$ for the first $8=4-5=-1$. Variance of $y$ for the first $8=3.5$ (shift-invariant).
$\sum_{i=1}^{8}y_i=8(-1)=-8$
$\sum_{i=1}^{8}y_i^2=8\left(3.5+(-1)^2\right)=8(4.5)=36$

Step 4: Subtract to isolate the last two shifted observations.
$y_9+y_{10}=0-(-8)=8$
$y_9^2+y_{10}^2=70-36=34$

Step 5: Write $y_9,y_{10}$ as their average plus or minus a deviation, and solve.
Average of the pair: $m=\dfrac{y_9+y_{10}}{2}=4$. Let $y_9=m-d$ and $y_{10}=m+d$, with $d>0$ since $x_9<x_{10}$ (a shift does not change the order).
$y_9^2+y_{10}^2=(m-d)^2+(m+d)^2=2m^2+2d^2=34$
$2(16)+2d^2=34$
$2d^2=2 \Rightarrow d^2=1 \Rightarrow d=1$
So $y_9=4-1=3$ and $y_{10}=4+1=5$.

Step 6: Shift back to the original observations and compute the required value.
$x_9=y_9+5=8$ and $x_{10}=y_{10}+5=10$
$3x_9+2x_{10}=3(8)+2(10)=24+20=44$

Final Answer: $44$