The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of

Q: Wait, does the order of $a$ and $b$ matter for the mean deviation?

Great observation! Because the mean deviation is defined as the average of the absolute differences from the mean, the set of numbers remains the same regardless of whether $a=3, b=4$ or $a=4, b=3$. The sum of absolute deviations will be identical in both cases, so the order does not affect the final result.

Q: Why can't I just guess the values of $a$ and $b$?

While you might get lucky with small integers, JEE Advanced problems are designed to test your mastery of algebraic structures. Relying on guessing is risky and won't help you when the numbers are non-integers or complex. Always trust the algebraic path—it is your safety net.

Q: Is there a faster way to calculate the mean deviation without listing all the terms?

For a small set of five numbers, listing them is actually the most efficient and error-proof method. Trying to derive a generalized formula for mean deviation often leads to more complexity than simply calculating the absolute differences directly.

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

Both formulas are mathematically equivalent! However, the first one is often easier to use when you need to find the sum of squares of the variables, as it separates the variables from the mean constant, making the algebra much cleaner.

Q: I got $M = 2.4$. Is that correct?

Yes, $2.4$ is exactly $\frac{12}{5}$. Since the question asks for $25M$, you simply multiply $2.4$ by $25$ to get $60$. You are on the right track—keep that momentum going!

Question

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25 M$ is equal to:

Options:

(A) 60

(B) 55

(C) 50

(D) 45

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a statistics problem; we are performing a detective investigation.

We have a set of five numbers: $a, b, 8, 5, 10$. Two of them are shrouded in mystery, hidden behind the variables $a$ and $b$.

Nature—and mathematics—leaves clues. We are given the mean and the variance, and these two pieces of information are the keys to unlocking the entire set.

### The First Clue: The Mean

Imagine the mean as the 'center of gravity' of our data. We know the mean $\bar{x} = 6$.

By definition, the mean is the sum of all observations divided by the number of observations:

$$\bar{x} = \frac{a + b + 8 + 5 + 10}{5} = 6$$

Multiplying both sides by $5$, we get $a + b + 23 = 30$, which simplifies beautifully to $a + b = 7$. This is our first anchor.

We know the sum of our two mystery numbers is $7$. However, we need more information to isolate them individually.

### The Second Clue: The Variance

Now, we turn to the variance, $\sigma^2 = 6.8$. Variance measures the 'spread' of the data.

The most elegant way to handle this is the computational formula:

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

Substituting our known values, we have:

$$\sigma^2 = \frac{a^2 + b^2 + 8^2 + 5^2 + 10^2}{5} - 6^2 = 6.8$$

Expanding the squares, we get $64 + 25 + 100 = 189$. So, the equation becomes:

$$\frac{a^2 + b^2 + 189}{5} - 36 = 6.8$$

Adding $36$ to both sides gives $\frac{a^2 + b^2 + 189}{5} = 42.8$. Multiplying by $5$, we find $a^2 + b^2 + 189 = 214$, which leads us to the result: $a^2 + b^2 = 25$.

### The Algebraic Bridge

We now have $a + b = 7$ and $a^2 + b^2 = 25$. This is where the beauty of algebra shines.

We recall the identity $(a + b)^2 = a^2 + b^2 + 2ab$. Substituting our values, we get:

$$7^2 = 25 + 2ab \Rightarrow 49 = 25 + 2ab \Rightarrow 2ab = 24$$

This gives us $ab = 12$. Now, we have the sum ($7$) and the product ($12$) of two numbers.

These are the roots of the quadratic equation $t^2 - 7t + 12 = 0$. Factoring this, we get $(t-3)(t-4) = 0$. Thus, our mystery numbers are $3$ and $4$.

### The Final Act: Mean Deviation

With our set complete—$3, 4, 5, 8, 10$—we calculate the mean deviation $M$. The mean deviation is the average of the absolute differences from the mean:

$$M = \frac{1}{n} \sum |x_i - \bar{x}|$$

Calculating the deviations from $6$: 
$|3-6| = 3$, $|4-6| = 2$, $|5-6| = 1$, $|8-6| = 2$, and $|10-6| = 4$.

Summing these up: $3 + 2 + 1 + 2 + 4 = 12$. Therefore, $M = \frac{12}{5} = 2.4$.

The final step is to find $25M$. Since $M = 2.4$, then $25 	imes 2.4 = 60$.

The final answer is **60**.

Animated Solution for Mathematics - Statistics: The mean of the numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.8$ . If $M$ is the mean deviation of the numbers about the mean, then $25 M$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Data Set

Applying the Mean Formula

Finding the Sum $a + b$

Applying the Variance Formula

Simplifying the Variance Equation

Finding the Sum of Squares

Using Algebraic Identity

Solving for $a$ and $b$

The Mean Deviation Formula

Calculating Absolute Deviations

Finding the Value of $M$

Final Calculation for $25 M$

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The First Clue

The Second Clue

The Algebraic Bridge

The Final Act

Similar Questions

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $2, 3, 5, 10, 11, 13, 15, 21$ , then the mean deviation about the median of all the 10 observations is

Let the mean of the data $x Frequency (f) 14324528 7 α 98$ be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ ^{2}}$ is equal to _______.

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :

If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^{2}$ is equal to:

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$ , then the variance of the first four observations in equal to

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1, ∣ x - y ∣$ is :

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :