The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations ar

Q: Why can't we just use the definition of variance $\frac{\sum (x_i - \bar{x})^2}{n}$?

You absolutely can! However, that definition requires you to know all the values first. Since we are missing two values, the computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is much more efficient because it allows us to isolate the sum of the squares of the unknowns without knowing the unknowns themselves.

Q: How do I know if the quadratic equation will have real roots?

In these JEE problems, the data is carefully constructed to yield real, often integer, solutions. If you find yourself with a negative discriminant, it is a sign to re-check your arithmetic in the sum of squares calculation. Trust the process!

Q: Why is the median the average of the 5th and 6th terms?

For any set of $n$ observations, if $n$ is even, the median is defined as the arithmetic mean of the two middle terms. For $n=10$, the middle positions are $\frac{n}{2} = 5$ and $\frac{n}{2} + 1 = 6$. Thus, we average the 5th and 6th terms after sorting the data.

Question

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

Options:

(A) 4

(B) 7

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a set of $n = 10$ observations with a mean $\bar{x} = 9$ and a variance $\sigma^2 = 34.2$. The 8 known observations are $2, 3, 5, 10, 11, 13, 15, 21$. Let the two missing numbers be $x$ and $y$.

Using the mean formula $\bar{x} = \frac{\sum x_i}{n}$, we find the total sum:
$$9 = \frac{\sum x_i}{10} \Rightarrow \sum x_i = 90$$

The sum of the 8 known values is $2 + 3 + 5 + 10 + 11 + 13 + 15 + 21 = 80$. Therefore, the sum of the missing numbers is:
$$x + y = 90 - 80 = 10$$

### The Algebraic Bridge

Next, we utilize the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ to find the sum of the squares of the observations. Substituting the given values:
$$34.2 = \frac{\sum x_i^2}{10} - 9^2$$

$$34.2 = \frac{\sum x_i^2}{10} - 81 \Rightarrow \frac{\sum x_i^2}{10} = 115.2$$

Thus, the total sum of squares is $\sum x_i^2 = 1152$. The sum of squares of the 8 known values is:
$$2^2 + 3^2 + 5^2 + 10^2 + 11^2 + 13^2 + 15^2 + 21^2 = 1094$$

Subtracting this from the total sum of squares gives:
$$x^2 + y^2 = 1152 - 1094 = 58$$

Using the identity $(x + y)^2 = x^2 + y^2 + 2xy$, we substitute our known values:
$$10^2 = 58 + 2xy \Rightarrow 100 - 58 = 2xy \Rightarrow xy = 21$$

### The Reveal

We now have the system $x + y = 10$ and $xy = 21$. These are the roots of the quadratic equation $t^2 - 10t + 21 = 0$.

Factoring the quadratic:
$$(t - 3)(t - 7) = 0$$
The missing numbers are $3$ and $7$.

Arranging all 10 observations in ascending order, we get:
$$2, 3, 3, 5, 7, 10, 11, 13, 15, 21$$

Since $n = 10$ is even, the median $M$ is the average of the 5th and 6th terms:
$$M = \frac{7 + 10}{2} = 8.5$$

### Final Calculation

The mean deviation about the median is defined as $\frac{\sum |x_i - M|}{n}$. We calculate the absolute deviations from $8.5$:

$|2 - 8.5| = 6.5$
$|3 - 8.5| = 5.5$
$|3 - 8.5| = 5.5$
$|5 - 8.5| = 3.5$
$|7 - 8.5| = 1.5$
$|10 - 8.5| = 1.5$
$|11 - 8.5| = 2.5$
$|13 - 8.5| = 4.5$
$|15 - 8.5| = 6.5$
$|21 - 8.5| = 12.5$

Summing these deviations:
$$6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5 = 50$$

Dividing by $n = 10$, the mean deviation is:
$$	ext{Mean Deviation} = \frac{50}{10} = \mathbf{5}$$

Animated Solution for Mathematics - Statistics: 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

Select Answer:

Visualized Solution

Decoding the Given Data

Sum of Unknowns $x + y$

Applying the Variance Formula

Total Sum of Squares $\sum x_{i}^{2}$

Sum of Squares of Unknowns $x^{2} + y^{2}$

Finding the Product $x y$

Solving for $x$ and $y$

Arranging Data & Finding Median

Absolute Deviations $∣ x_{i} - M ∣$

Final Mean Deviation

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Algebraic Bridge

The Reveal

Final Calculation

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