Objective Type Questions (MCQs)
The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is
Calculate the arithmetic mean $\bar{x}$:
$\bar{x} = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7$.
Calculate the absolute deviations $|x_i - \bar{x}|$:
$|3 - 7| = 4, |10 - 7| = 3, |10 - 7| = 3, |4 - 7| = 3, |7 - 7| = 0, |10 - 7| = 3, |5 - 7| = 2$.
Sum of absolute deviations $\sum |x_i - \bar{x}| = 4 + 3 + 3 + 3 + 0 + 3 + 2 = 18$.
Mean Deviation $MD(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{18}{7} \approx 2.57$.
(b)
The variance of the first $n$ natural numbers is
Mean of first $n$ natural numbers $\bar{x} = \frac{n(n + 1)}{2n} = \frac{n + 1}{2}$.
Sum of squares of first $n$ natural numbers $\sum x_i^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4}$.
$\sigma^2 = \frac{n + 1}{12}[2(2n + 1) - 3(n + 1)] = \frac{n + 1}{12}[4n + 2 - 3n - 3] = \frac{(n + 1)(n - 1)}{12} = \frac{n^2 - 1}{12}$.
(a)
If the standard deviation of numbers $2, 3, 2x, 11$ is $3.5$, then the variance of the numbers is
Standard deviation $\sigma = 3.5$.
By definition, Variance $= \sigma^2$.
Variance $= (3.5)^2 = 12.25$.
(c)
If the variance of a set of observations is $16$ and each observation is multiplied by $3$, the variance of the resulting observations is
Let initial observations have variance $\sigma^2 = 16$.
If each observation $x_i$ is transformed into $y_i = c \cdot x_i$, then $\sigma_y^2 = c^2 \sigma_x^2$.
Here $c = 3$, so new variance $= 3^2 \times 16 = 9 \times 16 = 144$.
(c)
If each observation in a data set is increased by $5$, then the standard deviation
Standard deviation is independent of the change of origin.
If $y_i = x_i + a$, then $\bar{y} = \bar{x} + a$, so $(y_i - \bar{y}) = (x_i + a - (\bar{x} + a)) = x_i - \bar{x}$.
Hence, the deviations from the mean remain identical.
Thus, the standard deviation remains unchanged.
(c)
The mean of $100$ observations is $50$ and their standard deviation is $5$. The sum of all squares of the observations is
Given $n = 100$, $\bar{x} = 50$, and $\sigma = 5 \implies \sigma^2 = 25$.
Using the variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$25 = \frac{\sum x_i^2}{100} - (50)^2 = \frac{\sum x_i^2}{100} - 2500$.
$\frac{\sum x_i^2}{100} = 2525 \implies \sum x_i^2 = 252,500$.
(c)
The coefficient of variation for a distribution with mean $40$ and standard deviation $8$ is
Formula for Coefficient of Variation: $\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$.
Given $\sigma = 8$ and $\bar{x} = 40$.
$\text{C.V.} = \frac{8}{40} \times 100 = \frac{1}{5} \times 100 = 20\%$.
(a)
For two data sets $A$ and $B$, $\text{C.V.}_A = 40\%$ and $\text{C.V.}_B = 55\%$. Which data set is more consistent?
The series having a smaller coefficient of variation is said to be more consistent or less variable.
Here $\text{C.V.}_A = 40\% < \text{C.V.}_B = 55\%$.
Therefore, series $A$ is more consistent than series $B$.
(a)
The mean deviation about the median for the observations $11, 3, 8, 7, 5, 14, 10, 2, 9$ is
Arrange data in ascending order: $2, 3, 5, 7, 8, 9, 10, 11, 14$.
Number of observations $n = 9$ (odd).
$\text{Median } M = \left(\frac{9 + 1}{2}\right)^{\text{th}} \text{ observation} = 5^{\text{th}} \text{ observation} = 8$.
Absolute deviations $|x_i - M|$: $|2 - 8| = 6, |3 - 8| = 5, |5 - 8| = 3, |7 - 8| = 1, |8 - 8| = 0, |9 - 8| = 1, |10 - 8| = 2, |11 - 8| = 3, |14 - 8| = 6$.
Sum $\sum |x_i - M| = 6 + 5 + 3 + 1 + 0 + 1 + 2 + 3 + 6 = 27$.
$MD(M) = \frac{\sum |x_i - M|}{n} = \frac{27}{9} = 3.0$.
(c)
The range of the data $12, 25, 15, 18, 35, 29, 40, 10, 22$ is
Maximum observation $X_{\max} = 40$.
Minimum observation $X_{\min} = 10$.
$\text{Range} = X_{\max} - X_{\min} = 40 - 10 = 30$.
(c)
If the standard deviation of $x_1, x_2, \dots, x_n$ is $\sigma$, then the standard deviation of $ax_1 + b, ax_2 + b, \dots, ax_n + b$ is
Let $y_i = ax_i + b$. Then $\bar{y} = a\bar{x} + b$.
$y_i - \bar{y} = (ax_i + b) - (a\bar{x} + b) = a(x_i - \bar{x})$.
Variance $\sigma_y^2 = \frac{1}{n}\sum (y_i - \bar{y})^2 = \frac{1}{n}\sum a^2(x_i - \bar{x})^2 = a^2 \sigma_x^2$.
Standard deviation $\sigma_y = \sqrt{a^2\sigma_x^2} = |a|\sigma$.
(b)
The mean and variance of $7$ observations are $8$ and $16$, respectively. If five of the observations are $2, 4, 10, 12, 14$, then the remaining two observations are
Let the remaining two observations be $x$ and $y$.
Sum of $7$ observations: $2 + 4 + 10 + 12 + 14 + x + y = 7 \times 8 = 56 \implies 42 + x + y = 56 \implies x + y = 14$.
Sum of squares: $\sigma^2 = \frac{\sum x_i^2}{7} - (\bar{x})^2 \implies 16 = \frac{\sum x_i^2}{7} - 64 \implies \sum x_i^2 = 7 \times 80 = 560$.
Sum of squares of five known observations: $2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 4 + 16 + 100 + 144 + 196 = 460$.
Therefore, $x^2 + y^2 = 560 - 460 = 100$.
$(x + y)^2 = x^2 + y^2 + 2xy \implies 14^2 = 100 + 2xy \implies 196 = 100 + 2xy \implies 2xy = 96 \implies xy = 48$.
Quadratic equation: $t^2 - 14t + 48 = 0 \implies (t - 6)(t - 8) = 0 \implies t = 6, 8$.
(a)
Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. If $1$ is added to each number, the variance of the numbers will
Variance is independent of change of origin.
Adding a constant to each observation shifts the distribution as a whole without altering the spread or dispersion.
Hence, the variance remains identical.
(c)
If $\sum_{i=1}^{10} (x_i - 5) = 10$ and $\sum_{i=1}^{10} (x_i - 5)^2 = 40$, then the standard deviation of $x_1, x_2, \dots, x_{10}$ is
Let $d_i = x_i - 5$.
Then $\sigma_x = \sigma_d = \sqrt{\frac{\sum d_i^2}{n} - \left(\frac{\sum d_i}{n}\right)^2}$.
Substitute $n = 10$, $\sum d_i = 10$, and $\sum d_i^2 = 40$:
$\sigma = \sqrt{\frac{40}{10} - \left(\frac{10}{10}\right)^2} = \sqrt{4 - 1^2} = \sqrt{3}$.
(a)
The algebraic sum of the deviations of a set of $n$ values from their arithmetic mean is always
By definition, $\bar{x} = \frac{\sum x_i}{n} \implies \sum x_i = n\bar{x}$.
Sum of deviations from mean: $\sum (x_i - \bar{x}) = \sum x_i - \sum \bar{x} = n\bar{x} - n\bar{x} = 0$.
Therefore, the algebraic sum of deviations from the arithmetic mean is always zero.
(c)