JEE Main 2022 (29 June Shift 1)

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Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Progressive Hints

Hint 1

Hint 1: To solve this, visualize the total set of 5 observations as a whole. You know their mean and variance, which allows you to calculate the sum of the observations and the sum of their squares. The key is to isolate the 5th observation by comparing the total sum of 5 with the sum of the first 4.

Hint 2

Hint 2: Recall the fundamental definition of variance:

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

. This formula is your bridge. Use it to find the sum of squares for the 5 observations, then subtract the square of the 5th observation to find the sum of squares for the first 4.

Hint 3

Hint 3: Once you have the sum of squares for the first 4 observations, apply the variance formula again specifically for those 4. You will have all the necessary components to find 'a'. Finally, calculate

4 a + x_{5}

to reach your answer.

Visualized Solution

Defining the 5 Observations

Let the 5 observations be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ .
Mean of 5 observations: $\overset{x}{ˉ}_{5} = \frac{24}{5}$
Variance of 5 observations: $σ_{5}^{2} = \frac{194}{25}$

Sum of 5 Observations

Formula for Mean: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\frac{\sum _{i = 1}^{5} x _{i}}{5} = \frac{24}{5}$
$\sum_{i = 1}^{5} x_{i} = 24$

Variance Formula for 5 Observations

Formula for Variance: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Substitute values: $\frac{194}{25} = \frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} - (\frac{24}{5})^{2}$

Sum of Squares of 5 Observations

$\frac{\sum x _{i}^{2}}{5} = \frac{194}{25} + \frac{576}{25}$
$\frac{\sum x _{i}^{2}}{5} = \frac{770}{25} = \frac{154}{5}$
$\sum_{i = 1}^{5} x_{i}^{2} = 154$

The First 4 Observations

Mean of first 4 observations: $\overset{x}{ˉ}_{4} = \frac{7}{2}$
Variance of first 4 observations: $σ_{4}^{2} = a$

Sum of First 4 Observations

$\frac{\sum _{i = 1}^{4} x _{i}}{4} = \frac{7}{2}$
$\sum_{i = 1}^{4} x_{i} = 4 \times \frac{7}{2} = 14$

Finding the 5th Observation $x_{5}$

$x_{5} = \sum_{i = 1}^{5} x_{i} - \sum_{i = 1}^{4} x_{i}$
$x_{5} = 24 - 14 = 10$

Sum of Squares of First 4 Observations

$\sum_{i = 1}^{4} x_{i}^{2} = \sum_{i = 1}^{5} x_{i}^{2} - x_{5}^{2}$
$\sum_{i = 1}^{4} x_{i}^{2} = 154 - 1 0^{2}$
$\sum_{i = 1}^{4} x_{i}^{2} = 154 - 100 = 54$

Variance of First 4 Observations $a$

Variance $a = \frac{\sum _{i = 1}^{4} x _{i}^{2}}{4} - (\overset{x}{ˉ}_{4})^{2}$
$a = \frac{54}{4} - (\frac{7}{2})^{2}$

Calculating $4 a$

$a = \frac{54}{4} - \frac{49}{4}$
$a = \frac{5}{4}$
$4 a = 5$

Final Calculation $(4 a + x_{5})$

We need to find: $4 a + x_{5}$
Substitute $4 a = 5$ and $x_{5} = 10$
$4 a + x_{5} = 5 + 10 = 15$
Final Answer: 15

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given five observations

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

with a mean

\overset{x}{ˉ}_{5} = \frac{24}{5}

and a variance

σ_{5}^{2} = \frac{194}{25}

. These parameters serve as the foundation for determining the sum of the observations and the sum of their squares.

The Master Equation

First, we calculate the total sum of the five observations using the mean formula

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

i = 1 \sum 5 x_{i} = 5 \times \frac{24}{5} = 24

Next, we utilize the variance formula

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

to find the sum of the squares:

\frac{194}{25} = \frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} - (\frac{24}{5})^{2}

Solving for the sum of squares:

\frac{\sum _{i = 1}^{5} x _{i}^{2}}{5} = \frac{194}{25} + \frac{576}{25} = \frac{770}{25} = \frac{154}{5}

i = 1 \sum 5 x_{i}^{2} = 154

Surgical Extraction of the Fifth Observation

We are given that the mean of the first four observations is

\overset{x}{ˉ}_{4} = \frac{7}{2}

. The sum of these four observations is:

i = 1 \sum 4 x_{i} = 4 \times \frac{7}{2} = 14

Since the sum of all five observations is

24

, the fifth observation

x_{5}

is:

x_{5} = 24 - 14 = 10

We now determine the sum of the squares of the first four observations:

i = 1 \sum 4 x_{i}^{2} = i = 1 \sum 5 x_{i}^{2} - x_{5}^{2} = 154 - 1 0^{2} = 54

Final Calculation

We define

a

as the variance of the first four observations. Applying the variance formula:

a = \frac{\sum _{i = 1}^{4} x _{i}^{2}}{4} - (\overset{x}{ˉ}_{4})^{2}

a = \frac{54}{4} - (\frac{7}{2})^{2} = \frac{54}{4} - \frac{49}{4} = \frac{5}{4}

The problem requires the value of

(4 a + x_{5})

. Substituting our derived values:

4 a + x_{5} = 4 (\frac{5}{4}) + 10 = 5 + 10 = 15

The final result is 15.

Common Traps & FAQs

Wait, why can't I just subtract the variance of the 4 observations from the variance of the 5?

That is a classic trap! Variance is not additive in that simple way. Variance depends on the squared deviations from the mean, and when you remove an observation, the mean itself changes. You must always work with the sum of observations and the sum of their squares to correctly account for the change in the mean.

Why do we need the sum of squares? Can't we just work with the mean?

The mean only tells us about the central tendency. Variance, however, is a measure of dispersion, which requires the second moment of the data (the sum of squares). Without the sum of squares, you cannot calculate the variance of a subset.

Is there a shortcut using the definition of variance directly?

While you could potentially use the property of variance for combined sets, for a small number of observations like 5, the most robust and error-proof method is to calculate the sum of squares directly. It keeps your logic clean and prevents algebraic slips.

Select Answer:

Visualized Solution

Defining the 5 Observations

Sum of 5 Observations

Variance Formula for 5 Observations

Sum of Squares of 5 Observations

The First 4 Observations

Sum of First 4 Observations

Finding the 5th Observation $x_{5}$

Sum of Squares of First 4 Observations

Variance of First 4 Observations $a$

Calculating $4 a$

Final Calculation $(4 a + x_{5})$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Master Equation

Surgical Extraction of the Fifth Observation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Defining the 5 Observations

Sum of 5 Observations

Variance Formula for 5 Observations

Sum of Squares of 5 Observations

The First 4 Observations

Sum of First 4 Observations

Finding the 5th Observation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x5​

Sum of Squares of First 4 Observations

Variance of First 4 Observations .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4a

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (4a+x5​)

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Master Equation

Surgical Extraction of the Fifth Observation

Final Calculation

Similar Questions

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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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a+b+ab is equal to :

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Variance of First 4 Observations $a$

Calculating $4 a$

Final Calculation $(4 a + x_{5})$

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to :