Short Answer Type Questions
Find the mean deviation about the mean for the data: $4, 7, 8, 9, 10, 12, 13, 17$.
Mean $\bar{x} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} = \frac{80}{8} = 10$.
Compute deviations $|x_i - \bar{x}|$: $|4 - 10| = 6, |7 - 10| = 3, |8 - 10| = 2, |9 - 10| = 1, |10 - 10| = 0, |12 - 10| = 2, |13 - 10| = 3, |17 - 10| = 7$.
Sum $\sum |x_i - \bar{x}| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24$.
$MD(\bar{x}) = \frac{24}{8} = 3$.
$3$
Find the mean deviation about the median for the observations: $39, 72, 48, 41, 43, 40, 45, 54, 70, 62$.
Arrange data in ascending order: $39, 40, 41, 43, 45, 48, 54, 62, 70, 72$.
Here $n = 10$ (even). Median is the average of $5^{\text{th}}$ and $6^{\text{th}}$ terms:
$M = \frac{45 + 48}{2} = 46.5$.
Compute $|x_i - M|$: $|39 - 46.5| = 7.5, |40 - 46.5| = 6.5, |41 - 46.5| = 5.5, |43 - 46.5| = 3.5, |45 - 46.5| = 1.5, |48 - 46.5| = 1.5, |54 - 46.5| = 7.5, |62 - 46.5| = 15.5, |70 - 46.5| = 23.5, |72 - 46.5| = 25.5$.
Sum $\sum |x_i - M| = 7.5 + 6.5 + 5.5 + 3.5 + 1.5 + 1.5 + 7.5 + 15.5 + 23.5 + 25.5 = 98$.
$MD(M) = \frac{98}{10} = 9.8$.
$9.8$
Find the variance and standard deviation of the numbers: $6, 7, 10, 12, 13, 4, 8, 12$.
Number of observations $n = 8$.
Sum $\sum x_i = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72$.
Mean $\bar{x} = \frac{72}{8} = 9$.
Sum of squares of deviations $\sum (x_i - \bar{x})^2$:
$(6 - 9)^2 + (7 - 9)^2 + (10 - 9)^2 + (12 - 9)^2 + (13 - 9)^2 + (4 - 9)^2 + (8 - 9)^2 + (12 - 9)^2$
$= 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74$.
Variance $\sigma^2 = \frac{74}{8} = 9.25$.
Standard deviation $\sigma = \sqrt{9.25} \approx 3.04$.
Variance $= 9.25$, Standard Deviation $\approx 3.04$
The mean and standard deviation of $20$ observations are found to be $10$ and $2$, respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.
Given $n = 20, \bar{x} = 10, \sigma = 2$.
Incorrect sum $\sum x_i = 20 \times 10 = 200$.
Omitting the wrong observation $8$: New $n' = 19$, Correct sum $= 200 - 8 = 192$.
Correct mean $\bar{x}' = \frac{192}{19} \approx 10.11$.
Given $\sigma^2 = 4 \implies \frac{\sum x_i^2}{20} - 10^2 = 4 \implies \frac{\sum x_i^2}{20} = 104 \implies \sum x_i^2 = 2080$.
Correct sum of squares $= 2080 - 8^2 = 2080 - 64 = 2016$.
Correct variance $\sigma'^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 = \frac{2016 \times 19 - 36864}{361} = \frac{38304 - 36864}{361} = \frac{1440}{361} \approx 3.99$.
Correct standard deviation $\sigma' = \sqrt{\frac{1440}{361}} = \frac{12\sqrt{10}}{19} \approx 1.997$.
Correct Mean $= 10.11$, Correct S.D. $\approx 2.00$
Find the mean deviation about the mean for the discrete frequency distribution: $x_i = [5, 10, 15, 20, 25]$, $f_i = [7, 4, 6, 3, 5]$.
$N = \sum f_i = 7 + 4 + 6 + 3 + 5 = 25$.
$\sum f_i x_i = 5(7) + 10(4) + 15(6) + 20(3) + 25(5) = 35 + 40 + 90 + 60 + 125 = 350$.
Mean $\bar{x} = \frac{350}{25} = 14$.
Absolute deviations $|x_i - \bar{x}|$: $|5 - 14| = 9, |10 - 14| = 4, |15 - 14| = 1, |20 - 14| = 6, |25 - 14| = 11$.
$\sum f_i |x_i - \bar{x}| = 7(9) + 4(4) + 6(1) + 3(6) + 5(11) = 63 + 16 + 6 + 18 + 55 = 158$.
$MD(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$.
$6.32$
Find the mean deviation about the median for the data: $x_i = [10, 15, 20, 25, 30, 35]$, $f_i = [3, 2, 5, 8, 4, 3]$.
Compute cumulative frequencies ($cf$):
$x_i$: $10 (cf=3), 15 (cf=5), 20 (cf=10), 25 (cf=18), 30 (cf=22), 35 (cf=25)$.
$N = 25$. $\frac{N + 1}{2} = 13^{\text{th}}$ observation.
The observation corresponding to cumulative frequency $\ge 13$ is $x_i = 25$, so Median $M = 25$.
Absolute deviations $|x_i - M|$: $|10 - 25| = 15, |15 - 25| = 10, |20 - 25| = 5, |25 - 25| = 0, |30 - 25| = 5, |35 - 25| = 10$.
$\sum f_i |x_i - M| = 3(15) + 2(10) + 5(5) + 8(0) + 4(5) + 3(10) = 45 + 20 + 25 + 0 + 20 + 30 = 140$.
$MD(M) = \frac{140}{25} = 5.6$.
$5.6$
Two manufacturing plants $A$ and $B$ produce steel tubes. For Plant $A$, mean length $\bar{x}_A = 50\text{ cm}$, $\sigma_A = 5\text{ cm}$; for Plant $B$, $\bar{x}_B = 60\text{ cm}$, $\sigma_B = 6\text{ cm}$. Compare their variability.
Coefficient of Variation for Plant $A$: $\text{C.V.}_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{5}{50} \times 100 = 10\%$.
Coefficient of Variation for Plant $B$: $\text{C.V.}_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{6}{60} \times 100 = 10\%$.
Since $\text{C.V.}_A = \text{C.V.}_B = 10\%$, both manufacturing plants exhibit equal consistency and identical variability in production.
Both plants have equal consistency ($ ext{C.V.} = 10\%$)
The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three of the observations are $1, 2,$ and $6$, find the other two observations.
Let the other two observations be $x$ and $y$.
Mean: $\frac{1 + 2 + 6 + x + y}{5} = 4.4 \implies 9 + x + y = 22 \implies x + y = 13$.
Variance: $\frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2 = 8.24$.
$\frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36 = 8.24 \implies \frac{41 + x^2 + y^2}{5} = 27.60$.
$41 + x^2 + y^2 = 138 \implies x^2 + y^2 = 97$.
$(x + y)^2 = x^2 + y^2 + 2xy \implies 13^2 = 97 + 2xy \implies 169 = 97 + 2xy \implies 2xy = 72 \implies xy = 36$.
Equation: $t^2 - 13t + 36 = 0 \implies (t - 4)(t - 9) = 0 \implies t = 4, 9$.
$4$ and $9$
Find the standard deviation for the first $10$ multiples of $3$.
The first $10$ multiples of $3$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$.
These can be written as $3 \times (1, 2, 3, \dots, 10)$.
Variance of first $n = 10$ natural numbers is $\frac{n^2 - 1}{12} = \frac{10^2 - 1}{12} = \frac{99}{12} = \frac{33}{4} = 8.25$.
Multiplying each number by $3$ scales standard deviation by $3$:
$\sigma = 3 \times \sqrt{8.25} = 3 \times \frac{\sqrt{33}}{2} = \frac{3\sqrt{33}}{2} \approx 8.62$.
$\frac{3\sqrt{33}}{2} \approx 8.62$
The sum and sum of squares corresponding to $10$ observations are given by $\sum (x_i - 5) = 3$ and $\sum (x_i - 5)^2 = 43$. Find the mean and standard deviation.
Let $u_i = x_i - 5$, with $n = 10$.
$\bar{u} = \frac{\sum u_i}{n} = \frac{3}{10} = 0.3$.
Mean $\bar{x} = 5 + \bar{u} = 5 + 0.3 = 5.3$.
Variance $\sigma^2 = \frac{\sum u_i^2}{n} - (\bar{u})^2 = \frac{43}{10} - (0.3)^2 = 4.3 - 0.09 = 4.21$.
Standard deviation $\sigma = \sqrt{4.21} \approx 2.05$.
Mean $= 5.3$, Standard Deviation $\approx 2.05$
Find the coefficient of variation for the following observations: $35, 40, 42, 45, 48, 50$.
$n = 6$. Sum $\sum x_i = 35 + 40 + 42 + 45 + 48 + 50 = 260$.
Mean $\bar{x} = \frac{260}{6} = \frac{130}{3} \approx 43.33$.
Deviations $(x_i - \bar{x})$: $35 - 43.33 = -8.33, 40 - 43.33 = -3.33, 42 - 43.33 = -1.33, 45 - 43.33 = 1.67, 48 - 43.33 = 4.67, 50 - 43.33 = 6.67$.
Sum of squared deviations $\sum (x_i - \bar{x})^2 \approx 69.39 + 11.09 + 1.77 + 2.79 + 21.81 + 44.49 = 151.34$.
Variance $\sigma^2 = \frac{151.34}{6} \approx 25.22 \implies \sigma = \sqrt{25.22} \approx 5.02$.
$\text{C.V.} = \frac{5.02}{43.33} \times 100 \approx 11.59\%$.
$\approx 11.59\%$
Prove that the variance of the first $2n$ even natural numbers is $\frac{4(4n^2 - 1)}{12} = \frac{4n^2 - 1}{3}$.
The first $2n$ even natural numbers are $2, 4, 6, \dots, 4n$.
This can be written as $2(1, 2, 3, \dots, 2n)$.
The variance of the first $N = 2n$ natural numbers is $\frac{N^2 - 1}{12} = \frac{(2n)^2 - 1}{12} = \frac{4n^2 - 1}{12}$.
Since each number is multiplied by $2$, the variance is multiplied by $2^2 = 4$:
$\text{Variance} = 4 \times \frac{4n^2 - 1}{12} = \frac{4n^2 - 1}{3}$.
$\frac{4n^2 - 1}{3}$ (Hence proved)
Find the mean deviation about the mean for the discrete distribution: $x_i = [2, 4, 6, 8, 10]$, $f_i = [1, 3, 5, 3, 1]$.
$N = \sum f_i = 1 + 3 + 5 + 3 + 1 = 13$.
$\sum f_i x_i = 2(1) + 4(3) + 6(5) + 8(3) + 10(1) = 2 + 12 + 30 + 24 + 10 = 78$.
Mean $\bar{x} = \frac{78}{13} = 6$.
Deviations $|x_i - 6|$: $|2 - 6| = 4, |4 - 6| = 2, |6 - 6| = 0, |8 - 6| = 2, |10 - 6| = 4$.
$\sum f_i |x_i - \bar{x}| = 1(4) + 3(2) + 5(0) + 3(2) + 1(4) = 4 + 6 + 0 + 6 + 4 = 20$.
$MD(\bar{x}) = \frac{20}{13} \approx 1.54$.
$\frac{20}{13} \approx 1.54$
An analysis of monthly wages paid to workers in two firms $A$ and $B$ gave: Number of wage earners $n_A = 500, n_B = 600$; Mean monthly wage $\bar{x}_A = 186, \bar{x}_B = 175$; Variance $\sigma_A^2 = 81, \sigma_B^2 = 100$. (i) Which firm pays a larger total amount? (ii) Which firm shows greater variability in individual wages?
(i) Total monthly wage bill = $n \times \bar{x}$.
For firm $A$: $500 \times 186 = 93,000$.
For firm $B$: $600 \times 175 = 105,000$.
Firm $B$ pays a larger amount ($105,000 > 93,000$).
(ii) Variability is determined by Coefficient of Variation (C.V.):
$\sigma_A = \sqrt{81} = 9 \implies \text{C.V.}_A = \frac{9}{186} \times 100 \approx 4.84\%$.
$\sigma_B = \sqrt{100} = 10 \implies \text{C.V.}_B = \frac{10}{175} \times 100 \approx 5.71\%$.
Since $\text{C.V.}_B > \text{C.V.}_A$, firm $B$ shows greater variability in wages.
(i) Firm $B$, (ii) Firm $B$
The standard deviation of $15$ observations is $6$. If each observation is multiplied by $-2$ and then decreased by $3$, find the new variance and new standard deviation.
Let original observations be $x_i$ with $\sigma_x = 6$.
Transformed observations $y_i = -2x_i - 3$.
Under the linear transformation $y_i = ax_i + b$, $\sigma_y = |a|\sigma_x$.
Here $a = -2$, so $\sigma_y = |-2| \times 6 = 2 \times 6 = 12$.
New Variance $\sigma_y^2 = 12^2 = 144$.
New Variance $= 144$, New S.D. $= 12$