JEE Main 2023 (01 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is 4 and the mean is $\overset{x}{ˉ}$ , then $\overset{x}{ˉ} + x_{6}$ is equal to :

Q: Why is $d=2$ and not $-2$? Both satisfy $d^2=4$.

Great catch! The problem explicitly states that $x_1 < x_2 < \dots < x_7$. This means the sequence is strictly increasing. For an A.P. to be increasing, the common difference $d$ must be positive. Thus, we reject $d=-2$.

Select Answer:

Visualized Solution

$x_{1}, x_{2}, \dots, x_{7} in A.P.$

Given A.P. terms: $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}$
First term $x_{1} = 9$
Common difference = $d$
General term: $x_{k} = 9 + (k - 1) d$

$Common Difference d$

Since terms are in A.P., the difference between consecutive terms is constant.
$x_{2} - x_{1} = d$
$x_{3} - x_{2} = d$

$Identify the Mean \overset{x}{ˉ}$

For an A.P. with $n$ terms (where $n$ is odd), the mean $\overset{x}{ˉ}$ is the middle term.
Middle term is $x_{\frac{7 + 1}{2}} = x_{4}$
$\overset{x}{ˉ} = x_{4} = 9 + (4 - 1) d = 9 + 3 d$

$Set up Variance σ^{2}$

Standard Deviation $σ = 4 \Rightarrow$ Variance $σ^{2} = 16$
Variance formula: $σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})^{2}$
Substitute $n = 7$ : $σ^{2} = \frac{1}{7} \sum_{k = 1}^{7} (x_{k} - x_{4})^{2}$

$Calculate Deviations$

Deviation $x_{k} - \overset{x}{ˉ} = (9 + (k - 1) d) - (9 + 3 d)$
Simplifying: $x_{k} - \overset{x}{ˉ} = (k - 4) d$
Deviations are: $- 3 d, - 2 d, - d, 0, d, 2 d, 3 d$

$Simplify the Variance$

$σ^{2} = \frac{1}{7} [(- 3 d)^{2} + (- 2 d)^{2} + (- d)^{2} + 0^{2} + d^{2} + (2 d)^{2} + (3 d)^{2}]$
$σ^{2} = \frac{d ^{2}}{7} [9 + 4 + 1 + 0 + 1 + 4 + 9]$
$σ^{2} = \frac{28 d ^{2}}{7} = 4 d^{2}$

$Solve for d$

Given $σ = 4 \Rightarrow σ^{2} = 16$
Equating: $4 d^{2} = 16$
$d^{2} = 4 \Rightarrow d = 2$
Note: $x_{1} < x_{2} < \dots < x_{7} \Rightarrow d > 0$

$Find Mean \overset{x}{ˉ} and x_{6}$

Mean $\overset{x}{ˉ} = 9 + 3 (2) = 15$
Sixth term $x_{6} = 9 + 5 (2) = 19$

$Final Calculation$

Required value: $\overset{x}{ˉ} + x_{6}$
$\overset{x}{ˉ} + x_{6} = 15 + 19 = 34$
Final Answer: 34

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are considering an Arithmetic Progression (A.P.) consisting of seven terms,

x_{1}, x_{2}, \dots, x_{7}

, with a common difference

d

. We are given that the first term is

x_{1} = 9

In any A.P. with an odd number of terms, the mean

\overset{x}{ˉ}

is equal to the middle term. For seven terms, the middle term is the fourth term,

x_{4}

Using the general formula for an A.P.,

x_{n} = x_{1} + (n - 1) d

, we can express the mean as:

\overset{x}{ˉ} = x_{4} = 9 + 3 d

The Symmetry of Deviations

To calculate the variance

σ^{2}

, we examine the deviation of each term

x_{k}

from the mean

\overset{x}{ˉ}

x_{k} - \overset{x}{ˉ} = (9 + (k - 1) d) - (9 + 3 d) = (k - 4) d

The set of deviations for

k = 1, 2, \dots, 7

is given by:

{- 3 d, - 2 d, - d, 0, d, 2 d, 3 d}

The variance is defined as the average of the squared deviations:

σ^{2} = \frac{1}{7} k = 1 \sum 7 (x_{k} - \overset{x}{ˉ})^{2}

Substituting our deviations, we obtain:

σ^{2} = \frac{d ^{2}}{7} ((- 3)^{2} + (- 2)^{2} + (- 1)^{2} + 0^{2} + 1^{2} + 2^{2} + 3^{2})

σ^{2} = \frac{d ^{2}}{7} (9 + 4 + 1 + 0 + 1 + 4 + 9) = \frac{28 d ^{2}}{7} = 4 d^{2}

Final Calculation

We are given that the standard deviation

σ = 4

, which implies the variance is

σ^{2} = 16

. Equating our derived expression to this value:

4 d^{2} = 16 \Rightarrow d^{2} = 4

Since the sequence is increasing, we take the positive root,

d = 2

. Now, we calculate the required values:

\overset{x}{ˉ} = 9 + 3 (2) = 15

x_{6} = 9 + 5 (2) = 19

The sum of the mean and the sixth term is:

\overset{x}{ˉ} + x_{6} = 15 + 19 = 34

Common Traps & FAQs

Wait, why is the mean just the middle term? Can't I use the standard formula?

You absolutely can use the standard formula, but in an Arithmetic Progression with an odd number of terms, the terms are perfectly symmetric about the middle term. Because the sum of deviations from the mean is always zero, the middle term must be the mean. It is a beautiful property that saves you a lot of calculation time!

Why can't we just use energy conservation or some other physics principle here?

This is a pure mathematics problem involving statistics and sequences. While physics often uses these concepts, here we rely on the algebraic properties of an A.P. and the definition of variance. Stick to the definitions, and the algebra will guide you to the answer.

Why is

d = 2

and not

- 2

? Both satisfy

d^{2} = 4

Great catch! The problem explicitly states that

x_{1} < x_{2} < \dots < x_{7}

. This means the sequence is strictly increasing. For an A.P. to be increasing, the common difference

d

must be positive. Thus, we reject

d = - 2

Animated Solution for Mathematics - Statistics: Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is 4 and the mean is $\overset{x}{ˉ}$ , then $\overset{x}{ˉ} + x_{6}$ is equal to :

Select Answer:

Visualized Solution

$x_{1}, x_{2}, \dots, x_{7} in A.P.$

$Common Difference d$

$Identify the Mean \overset{x}{ˉ}$

$Set up Variance σ^{2}$

$Calculate Deviations$

$Simplify the Variance$

$Solve for d$

$Find Mean \overset{x}{ˉ} and x_{6}$

$Final Calculation$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Symmetry of Deviations

Final Calculation

Similar Questions

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to

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:

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $x$ . If $\sum_{i = 1}^{10} (x_{i} - p) = 3$ and $\sum_{i = 1}^{10} (x_{i} - p)^{2} = 9$ where $0 \neq = p \in R$ , then the standard deviation of these observations is

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 :

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

If $\sum_{i = 1}^{n} (x_{i} - a) = n$ and $\sum_{i = 1}^{n} (x_{i} - a)^{2} = na$ , $(n, a > 1)$ then the standard deviation of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ is:

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} (x_{i} - 5)^{2} = 45$ , then the standard deviation of the 9 items $x_{1}, x_{2}, \dots, x_{9}$ 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 ______.

Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100$ . Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to :

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​,x2​,…,x7​ in A.P.

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identify the Mean xˉ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Set up Variance σ2

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find Mean xˉ and x6​

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

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Symmetry of Deviations

Final Calculation

Similar Questions

.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 :

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑i=1n​(xi​−a)=n and ∑i=1n​(xi​−a)2=na, (n,a>1) then the standard deviation of n observations x1​,x2​,…,xn​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑i=19​(xi​−5)=9 and ∑i=19​(xi​−5)2=45, then the standard deviation of the 9 items x1​,x2​,…,x9​ is :

$x_{1}, x_{2}, \dots, x_{7} in A.P.$

$Common Difference d$

$Identify the Mean \overset{x}{ˉ}$

$Set up Variance σ^{2}$

$Calculate Deviations$

$Simplify the Variance$

$Solve for d$

$Find Mean \overset{x}{ˉ} and x_{6}$

$Final Calculation$

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 :

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

If $\sum_{i = 1}^{n} (x_{i} - a) = n$ and $\sum_{i = 1}^{n} (x_{i} - a)^{2} = na$ , $(n, a > 1)$ then the standard deviation of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ is:

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} (x_{i} - 5)^{2} = 45$ , then the standard deviation of the 9 items $x_{1}, x_{2}, \dots, x_{9}$ is :