JEE Main 2023 (01 February Shift 1)

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Animated Solution for Mathematics - Statistics: The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Q: Wait, why do we reject the negative root here?

In this specific problem, we are not solving for $a$ and $b$ individually, so there is no 'negative root' to reject! We are using algebraic identities to find the sum of cubes directly. This is a powerful technique that saves time and avoids the complexity of solving for individual variables.

Select Answer:

Visualized Solution

Define the Missing Observations

Let the five observations be $1, 3, 5, a, b$ .
The missing observations are represented by variables $a$ and $b$ .

Apply the Mean Formula

Mean ( $\overset{x}{ˉ}$ ) = $\frac{\sum x _{i}}{n}$
Given Mean = $5$ and $n = 5$ .

Set up the Mean Equation

$\frac{1 + 3 + 5 + a + b}{5} = 5$

Simplify for $a + b$

$9 + a + b = 25$
$a + b = 16 \dots (1)$

Apply the Variance Formula

Variance ( $σ^{2}$ ) = $\frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Given Variance = $8$ and Mean ( $\overset{x}{ˉ}$ ) = $5$ .

Set up the Variance Equation

$\frac{1 ^{2} + 3 ^{2} + 5 ^{2} + a ^{2} + b ^{2}}{5} - 5^{2} = 8$

Simplify for $a^{2} + b^{2}$

$\frac{35 + a ^{2} + b ^{2}}{5} - 25 = 8$
$\frac{35 + a ^{2} + b ^{2}}{5} = 33$
$35 + a^{2} + b^{2} = 165$
$a^{2} + b^{2} = 130 \dots (2)$

Find the Product $ab$

Using $(a + b)^{2} = a^{2} + b^{2} + 2 ab$
Substitute $a + b = 16$ and $a^{2} + b^{2} = 130$ .

Calculate $ab$

$1 6^{2} = 130 + 2 ab$
$256 = 130 + 2 ab$
$2 ab = 126$
$ab = 63$

Sum of Cubes Identity

Identity: $a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - ab)$

Final Substitution

$a^{3} + b^{3} = 16 (130 - 63)$

Final Calculation

$a^{3} + b^{3} = 16 \times 67$
$a^{3} + b^{3} = 1072$

Conclusion & Key Takeaway

Key Takeaway: Use Mean and Variance to form equations for $\sum x_{i}$ and $\sum x_{i}^{2}$ .
Algebraic Shortcut: Instead of finding individual values, use identities like $(a + b)^{2}$ and $a^{3} + b^{3}$ to find symmetric sums.

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given a set of five observations:

1, 3, 5, a,

and

b

. We are provided with the mean

\overset{x}{ˉ} = 5

and the variance

σ^{2} = 8

Our objective is to determine the value of

a^{3} + b^{3}

The Anchor of the Mean

The mean

\overset{x}{ˉ}

is defined as the sum of all observations divided by the total count

n

. Given

n = 5

, we set up the following equation:

\overset{x}{ˉ} = \frac{1 + 3 + 5 + a + b}{5} = 5

Multiplying both sides by

5

, we obtain:

9 + a + b = 25

This simplifies to our first vital clue:

a + b = 16

The Measure of Chaos

Next, we utilize the variance formula, which measures the spread of the data:

σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}

Substituting our known values into the formula, we get:

\frac{1 ^{2} + 3 ^{2} + 5 ^{2} + a ^{2} + b ^{2}}{5} - 5^{2} = 8

Simplifying the squares, we have

1 + 9 + 25 = 35

. Thus:

\frac{35 + a ^{2} + b ^{2}}{5} - 25 = 8

Adding

25

to both sides yields:

\frac{35 + a ^{2} + b ^{2}}{5} = 33

Multiplying by

5

and subtracting

35

, we find the sum of the squares:

a^{2} + b^{2} = 165 - 35 = 130

The Algebraic Bridge

We now possess

a + b = 16

and

a^{2} + b^{2} = 130

. To find

a^{3} + b^{3}

, we first determine the product

ab

using the identity:

(a + b)^{2} = a^{2} + b^{2} + 2 ab

Substituting our known values:

1 6^{2} = 130 + 2 ab

256 = 130 + 2 ab

2 ab = 126 \Rightarrow ab = 63

The Grand Finale

Finally, we apply the sum of cubes identity to reach our result:

a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - ab)

Plugging in the values we have derived:

a^{3} + b^{3} = 16 (130 - 63)

a^{3} + b^{3} = 16 (67)

Performing the final multiplication, we arrive at the solution:

$a^{3} + b^{3} = 1072$

Common Traps & FAQs

Wait, why do we reject the negative root here?

In this specific problem, we are not solving for

a

and

b

individually, so there is no 'negative root' to reject! We are using algebraic identities to find the sum of cubes directly. This is a powerful technique that saves time and avoids the complexity of solving for individual variables.

Why can't we just use energy conservation?

Energy conservation is a principle for physical systems. Here, we are dealing with statistical data. The 'conservation' equivalent in statistics is the relationship between the sum of observations and the sum of their squares, linked by the mean and variance. Think of the mean and variance as the 'constraints' of your data system.

Is there a specific formula for variance I should memorize?

Yes! The most efficient formula for these types of problems is

σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}

. It directly links the variance to the sum of the squares of the observations, which is exactly what we need to solve for the missing variables.

Animated Solution for Mathematics - Statistics: The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Select Answer:

Visualized Solution

Define the Missing Observations

Apply the Mean Formula

Set up the Mean Equation

Simplify for $a + b$

Apply the Variance Formula

Set up the Variance Equation

Simplify for $a^{2} + b^{2}$

Find the Product $ab$

Calculate $ab$

Sum of Cubes Identity

Final Substitution

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Anchor of the Mean

The Measure of Chaos

The Algebraic Bridge

The Grand Finale

Similar Questions

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ , be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, \dots, 5$ is

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $(4 a + x_{5})$ is equal to:

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observation, then $a + 3 b - 5$ is equal to ______.

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to

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

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $α$ in this data is replaced by $β$ , then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals

Animated Solution for Mathematics - Statistics: The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Select Answer:

Visualized Solution

Define the Missing Observations

Apply the Mean Formula

Set up the Mean Equation

Simplify for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

Apply the Variance Formula

Set up the Variance Equation

Simplify for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2

Find the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab

Sum of Cubes Identity

Final Substitution

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Anchor of the Mean

The Measure of Chaos

The Algebraic Bridge

The Grand Finale

Similar Questions

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

.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

Simplify for $a + b$

Simplify for $a^{2} + b^{2}$

Find the Product $ab$

Calculate $ab$

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