JEE Main 2019 (8 April Shift 1)

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Animated Solution for Mathematics - Statistics: 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 :

Q: Wait, do I need to solve for $a$ and $b$ individually? That looks like it might involve a quadratic equation.

Great observation! While you *could* solve for $a$ and $b$ individually by solving the quadratic equation $x^2 - (a+b)x + ab = 0$, it is unnecessary. The question only asks for the product $ab$. By using the identity $(a+b)^2 = a^2 + b^2 + 2ab$, we bypass the need to find the individual roots entirely. Work smarter, not harder!

Select Answer:

Visualized Solution

Identifying the Given Data

Total observations: $n = 7$
Mean: $\overset{x}{ˉ} = 8$
Variance: $σ^{2} = 16$
Known observations: $2, 4, 10, 12, 14$
Let the remaining two observations be $a$ and $b$ .

Applying the Mean Formula

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$

Substituting into Mean Formula

Substituting the values: $8 = \frac{2 + 4 + 10 + 12 + 14 + a + b}{7}$

Finding the Sum of Unknowns

$56 = 42 + a + b$
$a + b = 14$

Applying the Variance Formula

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

Substituting into Variance Formula

Substituting the values: $16 = \frac{\sum x _{i}^{2}}{7} - 8^{2}$

Simplifying the Variance Equation

$16 = \frac{\sum x _{i}^{2}}{7} - 64$
$80 = \frac{\sum x _{i}^{2}}{7}$

Total Sum of Squares

$\sum x_{i}^{2} = 80 \times 7 = 560$

Squares of Known Observations

$2^{2} + 4^{2} + 1 0^{2} + 1 2^{2} + 1 4^{2} = 4 + 16 + 100 + 144 + 196 = 460$

Finding $a^{2} + b^{2}$

$560 = 460 + a^{2} + b^{2}$
$a^{2} + b^{2} = 100$

Using the Algebraic Identity

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

Substituting the Known Values

$(14)^{2} = 100 + 2 ab$
$196 = 100 + 2 ab$

Solving for the Product

$2 ab = 196 - 100 = 96$
$ab = 48$

Final Conclusion

The product of the remaining two observations is $48$ .

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

The mean,

\overset{x}{ˉ} = 8

, represents the center of gravity of our data set. It provides the average value of all seven observations. The fundamental formula is:

\overset{x}{ˉ} = \frac{\sum x _{i}}{n}

With

n = 7

, we can write the equation as:

8 = \frac{2 + 4 + 10 + 12 + 14 + a + b}{7}

By cross-multiplying, we find that the total sum of all seven observations is

56

. Summing the known values (

2 + 4 + 10 + 12 + 14 = 42

), we determine that:

a + b = 56 - 42 = 14

This serves as our first anchor point: the sum of our unknowns is

14

The Power of Variance

Next, we utilize the variance,

σ^{2} = 16

. While the definition involves deviations, the computational formula is more efficient:

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

Substituting our known values, we get:

16 = \frac{\sum x _{i}^{2}}{7} - 8^{2}

Since

8^{2} = 64

, we add

64

to both sides to obtain

80 = \frac{\sum x _{i}^{2}}{7}

. Multiplying by

7

, we find the total sum of squares:

\sum x_{i}^{2} = 560

The Algebraic Bridge

We now possess two vital pieces of information: the sum of the unknowns (

a + b = 14

) and the sum of the squares of all observations (

\sum x_{i}^{2} = 560

). We must isolate the squares of our unknowns.

The sum of squares of all seven observations is the sum of the squares of the five knowns plus

a^{2} + b^{2}

. Calculating the squares of the knowns:

2^{2} + 4^{2} + 1 0^{2} + 1 2^{2} + 1 4^{2} = 4 + 16 + 100 + 144 + 196 = 460

Therefore,

560 = 460 + a^{2} + b^{2}

, which implies:

a^{2} + b^{2} = 100

Final Calculation

We now stand at the threshold of the solution with

a + b = 14

and

a^{2} + b^{2} = 100

. To find the product

ab

, we use the algebraic identity:

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

Substituting our known values into the identity:

(14)^{2} = 100 + 2 ab

196 = 100 + 2 ab

Subtracting

100

from both sides, we get

2 ab = 96

. Thus, the final result is:

$ab = 48$

Common Traps & FAQs

Wait, do I need to solve for

a

and

b

individually? That looks like it might involve a quadratic equation.

Great observation! While you *could* solve for

a

and

b

individually by solving the quadratic equation

x^{2} - (a + b) x + ab = 0

, it is unnecessary. The question only asks for the product

ab

. By using the identity

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

, we bypass the need to find the individual roots entirely. Work smarter, not harder!

Why is the variance formula

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

better than the definition

\frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}

Both are mathematically equivalent, but the computational formula

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

is a lifesaver in JEE exams. It allows you to work with the sum of squares directly without having to subtract the mean from every single observation, which saves time and reduces the chance of arithmetic errors.

What if I made a calculation error in the sum of squares? Is there a way to check?

Always double-check your arithmetic! Statistics problems are notorious for 'silly mistakes.' When you calculate

\sum x_{i}^{2}

, verify the squares of the known numbers (

2^{2} = 4, 4^{2} = 16, 1 0^{2} = 100, 1 2^{2} = 144, 1 4^{2} = 196

) before adding them. If your final

a^{2} + b^{2}

turns out to be negative, you know immediately that you've made a calculation error, as the sum of squares of real numbers cannot be negative.

Animated Solution for Mathematics - Statistics: 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 :

Select Answer:

Visualized Solution

Identifying the Given Data

Applying the Mean Formula

Substituting into Mean Formula

Finding the Sum of Unknowns

Applying the Variance Formula

Substituting into Variance Formula

Simplifying the Variance Equation

Total Sum of Squares

Squares of Known Observations

Finding $a^{2} + b^{2}$

Using the Algebraic Identity

Substituting the Known Values

Solving for the Product

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Variance

The Algebraic Bridge

Final Calculation

Similar Questions

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 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 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

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 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 standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is $70$ . If the correct mean is $16$ , then the correct variance is equal to :

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

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.

If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :

Animated Solution for Mathematics - Statistics: 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 :

Select Answer:

Visualized Solution

Identifying the Given Data

Applying the Mean Formula

Substituting into Mean Formula

Finding the Sum of Unknowns

Applying the Variance Formula

Substituting into Variance Formula

Simplifying the Variance Equation

Total Sum of Squares

Squares of Known Observations

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

Using the Algebraic Identity

Substituting the Known Values

Solving for the Product

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Variance

The Algebraic Bridge

Final Calculation

Similar Questions

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 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 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

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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of eight numbers 3,7,9,12,13,20,x and y be 10 and 25 respectively then xy is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If both the mean and the standard deviation of 50 observations x1​,x2​,...,x50​ are equal to 16, then the mean of (x1​−4)2,(x2​−4)2,...,(x50​−4)2 is :

Finding $a^{2} + b^{2}$

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.

If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :