Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively. Then the mean of 4 numbers $x, y, x + y + 1, |x – y|$ is :
A) $11$ B) $9$ C) $10$ D) $12$
A) $11$ B) $9$ C) $10$ D) $12$
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Solution Steps
1
Sum of Observations from Mean
Mean $\bar{x} = \frac{\sum x_i}{n} = \frac{7}{2}$. With $n=8$:
$\sum x_i = 8 \times \frac{7}{2} = 28$.
Summing given numbers: $(-10) + (-7) + (-1) + 9 + 2 + 16 + x + y = 28$.
$9 + x + y = 28 \implies x + y = 19$.
2
Sum of Squares from Variance
Variance $\sigma^2 = \frac{\sum x_i^2}{n} – (\bar{x})^2$.
$\frac{293}{4} = \frac{100 + 49 + 1 + 81 + 4 + 256 + x^2 + y^2}{8} – (\frac{7}{2})^2$.
$\frac{293}{4} + \frac{49}{4} = \frac{491 + x^2 + y^2}{8} \implies \frac{342}{4} = \frac{491 + x^2 + y^2}{8}$.
$684 = 491 + x^2 + y^2 \implies x^2 + y^2 = 193$.
3
Solving for x and y
$(x+y)^2 = x^2 + y^2 + 2xy \implies 19^2 = 193 + 2xy$.
$361 – 193 = 2xy \implies 168 = 2xy \implies xy = 84$.
Numbers with sum 19 and product 84 are 7 and 12.
4
Final Calculation
The 4 numbers are: $7, 12, (19+1)=20, |7-12|=5$.
Mean $= \frac{7 + 12 + 20 + 5}{4} = \frac{44}{4} = 11$.
Mean = 11 (Option A)
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Theory
1. Arithmetic Mean ($\bar{x}$)
The arithmetic mean is the sum of all observations divided by the number of observations. It is the most common measure of central tendency. For a dataset where some values are unknown, the mean provides a linear equation involving the sum of those unknowns.
2. Statistical Variance ($\sigma^2$)
Variance measures the spread of data points around the mean. The computational formula $\sigma^2 = \frac{\sum x_i^2}{n} – (\bar{x})^2$ is derived from the average of squared deviations. It effectively provides a second-degree equation (sum of squares) for unknown variables in a dataset.
3. Algebraic Identities in Statistics
To find two unknown variables $x$ and $y$, statisticians often use the identity $(x+y)^2 = x^2 + y^2 + 2xy$. This allows the transition from knowing the sum and sum of squares (from mean and variance) to knowing the product $xy$, which simplifies solving the variables using a quadratic equation.
4. Absolute Difference $|x-y|$
The absolute difference represents the distance between two points on a number line. In statistics, it is often used as a measure of dispersion (Mean Absolute Deviation). In this problem, it serves as a transformation of the original variables $x$ and $y$ to form a new data set.
5. Properties of Sum of Squares
The sum of squares $\sum x_i^2$ is always non-negative. In variance problems, if $\sum x_i^2$ results in a value smaller than $n(\bar{x})^2$, it indicates a calculation error, as variance cannot be negative. This serves as a vital “sanity check” during JEE exams.
6. Transformation of Data Sets
Creating a new data set from old variables (like $x, y, x+y+1$) is a common JEE tactic. The mean of the new set is calculated independently by summing the newly derived values. This tests both the student’s ability to solve for unknowns and their understanding of basic definitions.
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FAQs
1
Why did we multiply 7/2 by 8?
⌄
Since Mean = Sum / Total Number, the Sum = Mean × Total Number. Here, $3.5 \times 8 = 28$.
2
What is the formula for variance used here?
⌄
We used $\sigma^2 = \frac{\sum x_i^2}{n} – (\bar{x})^2$, which is the simplified algebraic form.
3
Can variance be negative?
⌄
No, variance is the average of squared deviations, so it must always be $\ge 0$.
4
How did we get $x^2 + y^2 = 193$?
⌄
By substituting the sum of the other 6 squares (491) and the mean (3.5) into the variance formula.
5
What if I take $x=12$ and $y=7$?
⌄
The final set of 4 numbers will still be {12, 7, 20, 5}, and the mean will remain 11.
6
What is the value of $x+y+1$ in this case?
⌄
Since $x+y=19$, $x+y+1$ is simply $19+1 = 20$.
7
How is $|x-y|$ calculated?
⌄
It is $|12-7| = 5$ or $|7-12| = |-5| = 5$. Both give the same positive value.
8
Is there any other identity to find $|x-y|$?
⌄
Yes, $(x-y)^2 = (x+y)^2 – 4xy$. Here, $(x-y)^2 = 361 – 336 = 25$, so $|x-y|=5$.
9
What does $n=8$ represent?
⌄
It represents the total count of numbers in the initial data set provided in the question.
10
Is this question from a specific JEE paper?
⌄
Yes, this type of Statistics problem is a standard feature in recent JEE Main Mathematics papers.
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