JEE Main 2021 (31 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: 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 :

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

The definition requires knowing each individual $x_i$, which we don't have. The computational formula allows us to work with the aggregate properties (sum and sum of squares), which is exactly what we need when we only know the total mean and variance. It is the most efficient tool for this type of problem.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the dataset as a collection of seven points on a number line. The variance is a measure of how spread out these points are from their mean. To find the variance of a subset, you must first isolate the sum and the sum of squares of that subset by removing the contributions of the known observations from the total sum and total sum of squares.

Hint 2

The Arsenal: Utilize the computational formula for variance:

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

. You are given the total mean and total variance for

n = 7

. Calculate the total sum

\sum x_{i}

and the total sum of squares

\sum x_{i}^{2}

. Then, subtract the values corresponding to the two known observations (

6

and

8

) to find the sum and sum of squares for the remaining

5

observations.

Hint 3

The Execution: Start by setting up the equations:

\sum_{i = 1}^{7} x_{i} = 7 \times 8

and

\sum_{i = 1}^{7} x_{i}^{2} = 7 (σ^{2} + \overset{x}{ˉ}^{2})

. Once you have these, calculate

\sum_{i = 1}^{5} x_{i} = (\sum_{i = 1}^{7} x_{i}) - (6 + 8)

and

\sum_{i = 1}^{5} x_{i}^{2} = (\sum_{i = 1}^{7} x_{i}^{2}) - (6^{2} + 8^{2})

. Finally, apply the variance formula for the

5

observations:

σ_{n e w}^{2} = \frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} - (\frac{\sum _{i = 1}^{5} x _{i}}{5})^{2}

Visualized Solution

Given Data and Variables

Total observations ( $n$ ): $7$
Given Mean ( $\overset{x}{ˉ}$ ): $8$
Given Variance ( $σ^{2}$ ): $16$
Known observations: $6$ and $8$
Let remaining observations be: $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$

Mean Formula for $n = 7$

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
For 7 observations: $\frac{x _{1} + x _{2} + x _{3} + x _{4} + x _{5} + 6 + 8}{7} = 8$

Simplifying the Mean Equation

Summing known values: $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + 14 = 8 \times 7$
Result: $\sum_{i = 1}^{5} x_{i} + 14 = 56$

Finding $\sum_{i = 1}^{5} x_{i}$

Subtracting 14: $\sum_{i = 1}^{5} x_{i} = 56 - 14$
Equation 1: $\sum_{i = 1}^{5} x_{i} = 42$

Variance Formula for $n = 7$

Variance formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
For 7 observations: $16 = \frac{\sum _{i = 1}^{7} x _{i}^{2}}{7} - (8)^{2}$

Substituting Values for Variance

Expanding the sum of squares: $16 = \frac{\sum _{i = 1}^{5} x _{i}^{2} + 6 ^{2} + 8 ^{2}}{7} - 64$
Simplifying squares: $16 = \frac{\sum _{i = 1}^{5} x _{i}^{2} + 36 + 64}{7} - 64$

Simplifying the Variance Equation

Adding 64 to both sides: $16 + 64 = \frac{\sum _{i = 1}^{5} x _{i}^{2} + 100}{7}$
Multiplying by 7: $80 \times 7 = \sum_{i = 1}^{5} x_{i}^{2} + 100$
Result: $560 = \sum_{i = 1}^{5} x_{i}^{2} + 100$

Finding $\sum_{i = 1}^{5} x_{i}^{2}$

Subtracting 100: $\sum_{i = 1}^{5} x_{i}^{2} = 560 - 100$
Equation 2: $\sum_{i = 1}^{5} x_{i}^{2} = 460$

Variance of 5 Observations

New Variance formula: $σ_{n e w}^{2} = \frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} - (\frac{\sum _{i = 1}^{5} x _{i}}{5})^{2}$

Substituting Values for New Variance

Substituting from Eq 1 and Eq 2: $σ_{n e w}^{2} = \frac{460}{5} - (\frac{42}{5})^{2}$

Final Calculation

Simplifying terms: $92 - \frac{1764}{25}$
Taking LCM: $\frac{92 \times 25 - 1764}{25} = \frac{2300 - 1764}{25}$
Final Result: $\frac{536}{25}$

Conclusion

Key Takeaway: Variance of a subset requires finding both $\sum x_{i}$ and $\sum x_{i}^{2}$ from the original set.
Final Answer: $\frac{536}{25}$
Next Challenge: How does variance change if every observation is scaled by a constant $k$ ?

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We begin with a set of seven observations:

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 6, 8

. We are given the mean

\overset{x}{ˉ} = 8

and the variance

σ^{2} = 16

The mean provides the total sum of the observations. For

n = 7

, the sum is:

i = 1 \sum 7 x_{i} = 7 \times 8 = 56

Since two observations are

6

and

8

, the sum of the remaining five unknowns is:

i = 1 \sum 5 x_{i} = 56 - (6 + 8) = 56 - 14 = 42

The Power of Squares

To find the variance of the subset, we must first determine the sum of the squares of all seven observations. We use the standard variance formula:

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

Rearranging this to solve for the sum of squares, we get:

i = 1 \sum 7 x_{i}^{2} = n (σ^{2} + (\overset{x}{ˉ})^{2})

Substituting our known values:

i = 1 \sum 7 x_{i}^{2} = 7 (16 + 8^{2}) = 7 (16 + 64) = 7 (80) = 560

Now, we subtract the squares of the two removed observations (

6

and

8

) to isolate the sum of squares for the remaining five:

i = 1 \sum 5 x_{i}^{2} = 560 - (6^{2} + 8^{2}) = 560 - (36 + 64) = 560 - 100 = 460

Final Calculation

We now calculate the new variance

σ_{n e w}^{2}

for the five remaining observations using the sum and the sum of squares derived above:

σ_{n e w}^{2} = \frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} - (\frac{\sum _{i = 1}^{5} x _{i}}{5})^{2}

Substituting the values

460

and

42

σ_{n e w}^{2} = \frac{460}{5} - (\frac{42}{5})^{2}

Simplifying the expression:

σ_{n e w}^{2} = 92 - \frac{1764}{25}

Converting to a common denominator:

σ_{n e w}^{2} = \frac{2300 - 1764}{25} = \frac{536}{25}

The variance of the remaining five observations is $\frac{536}{25}$ (or $21.44$ ).

Common Traps & FAQs

Wait, why can't we just subtract the variance of the two known numbers from the total variance?

Variance is not an additive property like the sum of observations. It depends on the squared deviations from the mean. When you remove observations, the mean of the remaining set changes, and the squared deviations change entirely. You must work with the sum and sum of squares, which are additive properties that allow us to reconstruct the statistics of the subset.

Why do we use the formula

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

instead of the definition

\frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}

The definition requires knowing each individual

x_{i}

, which we don't have. The computational formula allows us to work with the aggregate properties (sum and sum of squares), which is exactly what we need when we only know the total mean and variance. It is the most efficient tool for this type of problem.

Wait, when calculating the new variance, why do we divide by

5

and not

7

The variance formula divides by the number of observations in the set. Since we are now calculating the variance for the remaining

5

observations, the new

n

5

. Using

7

would be a fundamental error in defining the spread of the new subset. Always ensure your

n

matches the number of observations in the specific set you are analyzing.

Animated Solution for Mathematics - Statistics: 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 :

Select Answer:

Visualized Solution

Given Data and Variables

Mean Formula for $n = 7$

Simplifying the Mean Equation

Finding $\sum_{i = 1}^{5} x_{i}$

Variance Formula for $n = 7$

Substituting Values for Variance

Simplifying the Variance Equation

Finding $\sum_{i = 1}^{5} x_{i}^{2}$

Variance of 5 Observations

Substituting Values for New Variance

Final Calculation

Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Squares

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 :

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-

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Animated Solution for Mathematics - Statistics: 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 :

Select Answer:

Visualized Solution

Given Data and Variables

Mean Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=7

Simplifying the Mean Equation

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑i=15​xi​

Variance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=7

Substituting Values for Variance

Simplifying the Variance Equation

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑i=15​xi2​

Variance of 5 Observations

Substituting Values for New Variance

Final Calculation

Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Squares

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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the data : 7, 8, 9, 7, 8, 7, λ, 8 is 8, then the variance of this data 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 :

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 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

.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

Mean Formula for $n = 7$

Finding $\sum_{i = 1}^{5} x_{i}$

Variance Formula for $n = 7$

Finding $\sum_{i = 1}^{5} x_{i}^{2}$

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-

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