JEE Main 2022 (25 July Shift 2)

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Animated Solution for Mathematics - Statistics: If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is

Q: Why do we take the average of the 4th and 5th terms?

In statistics, when the number of observations $n$ is even, there is no single middle term. The median is defined as the arithmetic mean of the two central values, which in our case are the 4th and 5th terms.

Q: How can we be sure that $2k$ is less than the median?

The problem states the data is in ascending order. Since $2k$ is the 4th term and $12$ is the 5th term, and the median $M$ lies between them, it is mathematically guaranteed that $2k \le M \le 12$.

Select Answer:

Visualized Solution

Analyze the Given Data

Given observations in ascending order: $3, 5, 7, 2 k, 12, 16, 21, 24$
Total number of observations ( $n$ ) = $8$
Mean Deviation about Median ( $M D_{M}$ ) = $6$

Identify the Median Position

Since $n = 8$ (even), the median $M$ is the average of the $(\frac{n}{2})^{t h}$ and $(\frac{n}{2} + 1)^{t h}$ terms.
Middle terms are the $4^{t h}$ term ( $2 k$ ) and the $5^{t h}$ term ( $12$ ).

Express Median $M$ in terms of $k$

Median $M = \frac{2 k + 12}{2}$
Simplifying, $M = k + 6$

The Mean Deviation Formula

Formula: $M D_{M} = \frac{\sum _{i = 1}^{n} ∣ x _{i} - M ∣}{n} = 6$
Substituting $n = 8$ : $\sum_{i = 1}^{8} ∣ x_{i} - M ∣ = 6 \times 8 = 48$

Expanding the Summation

Expanding the sum: $∣3 - M ∣ + ∣5 - M ∣ + ∣7 - M ∣ + ∣2 k - M ∣ + ∣12 - M ∣ + ∣16 - M ∣ + ∣21 - M ∣ + ∣24 - M ∣ = 48$

Removing Modulus: First 4 Terms

Since $3, 5, 7, 2 k \leq M$ , then $∣ x_{i} - M ∣ = M - x_{i}$ for $i = 1, 2, 3, 4$ .
Sum of first 4 deviations: $(M - 3) + (M - 5) + (M - 7) + (M - 2 k)$

Removing Modulus: Last 4 Terms

Since $12, 16, 21, 24 \geq M$ , then $∣ x_{i} - M ∣ = x_{i} - M$ for $i = 5, 6, 7, 8$ .
Sum of last 4 deviations: $(12 - M) + (16 - M) + (21 - M) + (24 - M)$

Simplifying the Equation

Total Sum: $(4 M - (3 + 5 + 7 + 2 k)) + ((12 + 16 + 21 + 24) - 4 M) = 48$
The $M$ terms cancel out: $(12 + 16 + 21 + 24) - (3 + 5 + 7 + 2 k) = 48$

Arithmetic Calculation

Sum of last 4 terms: $12 + 16 + 21 + 24 = 73$
Sum of first 3 terms: $3 + 5 + 7 = 15$
Equation: $73 - (15 + 2 k) = 48$

Solving for $2 k$

$73 - 15 - 2 k = 48$
$58 - 2 k = 48$
$2 k = 58 - 48 = 10$

Final Median Calculation

Substitute $2 k = 10$ into the median formula:
$M = \frac{2 k + 12}{2}$
$M = \frac{10 + 12}{2} = \frac{22}{2}$
Median ( $M$ ) = 11

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The Sigma Insight: Mean Deviation

Solution Diagram

Analyzing the Setup

We are given a set of eight numbers:

3, 5, 7, 2 k, 12, 16, 21, 24

. Since the number of observations is

n = 8

, the median

M

is the average of the two middle terms.

Assuming the data is ordered, the 4th and 5th terms are

2 k

and

12

. Therefore, the median is defined as:

M = \frac{2 k + 12}{2} = k + 6

The Master Equation

The mean deviation about the median is given as

6

. The formula for mean deviation is:

M D_{M} = \frac{1}{n} i = 1 \sum n ∣ x_{i} - M ∣ = 6

Multiplying by

n = 8

, we find that the sum of absolute deviations is

48

i = 1 \sum 8 ∣ x_{i} - M ∣ = 48

Expanding the Sum

Expanding the summation for our specific set of numbers, we have:

∣3 - M ∣ + ∣5 - M ∣ + ∣7 - M ∣ + ∣2 k - M ∣ + ∣12 - M ∣ + ∣16 - M ∣ + ∣21 - M ∣ + ∣24 - M ∣ = 48

Because the data is sorted, the first four terms are less than or equal to

M

, and the last four are greater than or equal to

M

. We can remove the absolute value bars by changing the signs accordingly:

(M - 3) + (M - 5) + (M - 7) + (M - 2 k) + (12 - M) + (16 - M) + (21 - M) + (24 - M) = 48

Final Calculation

Observe the symmetry: the

M

terms cancel out completely. This leaves us with the following expression:

(12 + 16 + 21 + 24) - (3 + 5 + 7 + 2 k) = 48

Calculating the sums, we obtain:

73 - (15 + 2 k) = 48

58 - 2 k = 48

Solving for

2 k

, we find

2 k = 10

. Substituting this value back into our median formula:

M = \frac{10 + 12}{2} = 11

The final median of the set is 11.

Common Traps & FAQs

Why do we take the average of the 4th and 5th terms?

In statistics, when the number of observations

n

is even, there is no single middle term. The median is defined as the arithmetic mean of the two central values, which in our case are the 4th and 5th terms.

How can we be sure that

2 k

is less than the median?

The problem states the data is in ascending order. Since

2 k

is the 4th term and

12

is the 5th term, and the median

M

lies between them, it is mathematically guaranteed that

2 k \leq M \leq 12

Why do the

M

terms cancel out in the summation?

This is a beautiful property of the median! Because we have an equal number of terms (4) on either side of the median, the sum of deviations becomes

(4 M - \sum_{l e f t} x_{i}) + (\sum_{r i g h t} x_{i} - 4 M)

. The

4 M

and

- 4 M

terms subtract to zero, leaving only the constants.

Animated Solution for Mathematics - Statistics: If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is

Select Answer:

Visualized Solution

Analyze the Given Data

Identify the Median Position

Express Median $M$ in terms of $k$

The Mean Deviation Formula

Expanding the Summation

Removing Modulus: First 4 Terms

Removing Modulus: Last 4 Terms

Simplifying the Equation

Arithmetic Calculation

Solving for $2 k$

Final Median Calculation

The Sigma Insight: Mean Deviation

Analyzing the Setup

The Master Equation

Expanding the Sum

Final Calculation

Similar Questions

If the mean deviation about the median of the numbers a, 2a,.......,50a is 50, then | a| equals

Consider the given data with frequency distribution $x_{i}$ : 3, 8, 11, 10, 5, 4 $f_{i}$ : 5, 2, 3, 2, 4, 4 Match each entry in List-I to the correct entries in List-II.

List-I

List-II

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $2$ respectively. Then the mean deviation about the mode of these observations is :

Let $α, β \in R$ . Let the mean and the variance of 6 observations $- 3, 4, 7, - 6, α, β$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to:

Animated Solution for Mathematics - Statistics: If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is

Select Answer:

Visualized Solution

Analyze the Given Data

Identify the Median Position

Express Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

The Mean Deviation Formula

Expanding the Summation

Removing Modulus: First 4 Terms

Removing Modulus: Last 4 Terms

Simplifying the Equation

Arithmetic Calculation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2k

Final Median Calculation

The Sigma Insight: Mean Deviation

Analyzing the Setup

The Master Equation

Expanding the Sum

Final Calculation

Similar Questions

If the mean deviation about the median of the numbers a, 2a,.......,50a is 50, then | a| equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the given data with frequency distribution xi​: 3, 8, 11, 10, 5, 4 fi​: 5, 2, 3, 2, 4, 4 Match each entry in List-I to the correct entries in List-II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β∈R. Let the mean and the variance of 6 observations −3,4,7,−6,α,β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to:

Express Median $M$ in terms of $k$

Solving for $2 k$

Consider the given data with frequency distribution $x_{i}$ : 3, 8, 11, 10, 5, 4 $f_{i}$ : 5, 2, 3, 2, 4, 4 Match each entry in List-I to the correct entries in List-II.

Let $α, β \in R$ . Let the mean and the variance of 6 observations $- 3, 4, 7, - 6, α, β$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :