The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y where x

Q: Wait, why do we use the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of the definition $\frac{\sum (x_i - \bar{x})^2}{n}$?

That is a great question! Both formulas are mathematically equivalent. However, the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is often much faster for calculations because it avoids calculating the deviation of each data point from the mean individually. It is a powerful tool in your JEE arsenal—use it to save time!

Q: Why do we need to form a quadratic equation at the end?

When you have the sum and the product of two variables, the most elegant and direct way to find the variables themselves is to treat them as roots of a quadratic equation. It turns a potentially messy system of equations into a simple factorization problem. It is the hallmark of a sophisticated approach.

Q: What if I get confused about which variable is $x$ and which is $y$?

Always look back at the problem statement for constraints! Here, we are given $x < y$. Once you find the two roots of your quadratic equation, simply assign the smaller value to $x$ and the larger one to $y$. It is a simple check that ensures you do not lose marks on a silly mistake.

Question

The mean and variance of the data $4, 5, 6, 6, 7, 8, x, y$ where $x < y$ are $6$ and $\frac{9}{4}$ respectively. Then $x^4 + y^2$ is equal to

Options:

(A) 162

(B) 320

(C) 674

(D) 420

Sigma Percentile · Accepted Answer

### The Heartbeat of Data: Understanding Mean and Variance

Imagine you are standing in a field of numbers. You have a set of data: $4, 5, 6, 6, 7, 8, x, y$.

In the world of JEE Advanced, this is a puzzle waiting to be solved. We are given the mean and the variance, and our mission is to uncover the hidden values of $x$ and $y$.

### Phase 1: Decoding the Mean

The mean is the balance point of your data, representing the center of gravity. We know the mean $\bar{x} = 6$ and the number of observations $n = 8$.

The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$. Summing our data, we get:

$$36 + x + y$$

Setting the mean equal to $6$, we have:

$$\frac{36 + x + y}{8} = 6$$

Multiplying by $8$ gives $36 + x + y = 48$, which simplifies to our first anchor point:

$$x + y = 12$$

### Phase 2: The Variance Challenge

Now, let us tackle the variance. Given $\sigma^2 = \frac{9}{4}$, we use the computational formula:

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

First, we calculate the sum of the squares of our observations:

$$\sum x_i^2 = 4^2 + 5^2 + 6^2 + 6^2 + 7^2 + 8^2 + x^2 + y^2 = 226 + x^2 + y^2$$

Substituting this into our variance formula:

$$\frac{226 + x^2 + y^2}{8} - 36 = \frac{9}{4}$$

Adding $36$ to both sides:

$$\frac{226 + x^2 + y^2}{8} = 36 + \frac{9}{4} = \frac{144 + 9}{4} = \frac{153}{4}$$

Multiplying by $8$ yields $226 + x^2 + y^2 = 306$. Subtracting $226$ leaves us with:

$$x^2 + y^2 = 80$$

### Phase 3: The Algebraic Bridge

We now have the system $x + y = 12$ and $x^2 + y^2 = 80$. We utilize the identity $(x + y)^2 = x^2 + y^2 + 2xy$ as our bridge.

Substituting our known values:

$$12^2 = 80 + 2xy$$
$$144 = 80 + 2xy$$

This simplifies to $2xy = 64$, or:

$$xy = 32$$

### Phase 4: The Final Reveal

We treat $x$ and $y$ as roots of the quadratic equation $t^2 - (x+y)t + xy = 0$. Substituting our values, we get:

$$t^2 - 12t + 32 = 0$$

Factoring this quadratic, we find:

$$(t - 4)(t - 8) = 0$$

The roots are $4$ and $8$. Assuming $x < y$, we identify $x = 4$ and $y = 8$.

Finally, we calculate the requested value:

$$x^4 + y^2 = 4^4 + 8^2 = 256 + 64 = \mathbf{320}$$

Animated Solution for Mathematics - Statistics: The mean and variance of the data $4, 5, 6, 6, 7, 8, x, y$ where $x < y$ are $6$ and $\frac{9}{4}$ respectively. Then $x^{4} + y^{2}$ is equal to

Select Answer:

Visualized Solution

Analyzing the Given Data and Mean

Establishing the Relation for $x + y$

Setting up the Variance Formula

Calculating the Sum of Squares

Solving for $x^{2} + y^{2}$

Finding the Product $x y$

Forming a Quadratic Equation

Solving for $x$ and $y$

Final Calculation: $x^{4} + y^{2}$

The Sigma Insight: Variance and Standard Deviation

The Heartbeat of Data

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The mean and variance of the data 4,5,6,6,7,8,x,y where x<y are 6 and 49​ respectively. Then x4+y2 is equal to