MCQ 11 Mark
The standard deviation of first 10 natural numbers is:
AnswerWe know that the standard deviation of first n natural number is $\sqrt{\frac{\text{n}^2-1}{12}}.$
$\therefore$ Standard deviation of first 10 natural numbers
$=\sqrt{\frac{10^2-1}{12}}$
$=\sqrt{\frac{99}{12}}$
$=\sqrt{8.25}$
$=2.87$
Hence, the correct answer is option (d).
View full question & answer→MCQ 21 Mark
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is:
- ✓
- B
- C
- D
$\frac{\text{s}}{\text{k}}$
AnswerThe given observations are a, b, c, d, e.
$\text{Mean}=\text{m}=\frac{\text{a+b+c+d+e}}{5}$
$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$
Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
Now, consider the observations a + k, b + k, c + k, d + k, e + k.
New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$
$=\frac{\text{a+b+c+d+e+5k}}{5}$
$=\frac{5\text{m}+5\text{k}}{5}$
$=\text{m}+\text{k}$
$\therefore$ New standard deviation
$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$
$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}}$ [Using (1)]
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
$=\text{s}$
Hence, the correct answer is option (a).
View full question & answer→MCQ 31 Mark
If n = 10, $\overline{\text{X}}=12$ and $\sum\text{x}_\text{i}^2=1530,$ then the coefficient of variation is:
AnswerStandard deviation is expressed in the following manner:
$\sigma=\sqrt{\frac{1}{\text{n}}\sum_\text{i}\text{x}_\text{i}^2-(\overline{\text{X}})^2}$
$=\sqrt{\frac{1530}{10}-(12)^2}$
$=\sqrt9$
$=3$
$\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$=\frac{3}{12}\times100$
$=25%$
View full question & answer→MCQ 41 Mark
Let $x_1, x_2, ..., x_n $ be n observations. Let $y_i = ax_i + by_i + b$ for $i = 1, 2, 3, ..., n,$ where $a$ and $b$ are constants. If the mean of $x_i's$ is $48$ and their standard deviation is $12,$ the mean of $y_i's \ 55$ and standard deviation of $y_i's$ is $15,$ the values of $a$ and $b$ are:
- ✓
$a = 1.25, b = -5$
- B
$a = -1.25, b = 5$
- C
$a = 2.5, b = -5$
- D
$a = 2.5, b = 5$
AnswerCorrect option: A. $a = 1.25, b = -5$
It is given that $y_i = ax_i + b$ for $i = 1, 2, 3, ..., n$, where a and b are constants.
$\overline{\text{x}_\text{i}}=48$ and $\sigma_{\text{x}_\text{i}}=12$
$\overline{\text{y}_\text{i}}=55$ and $\sigma_{\text{y}_\text{i}}=15$
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\sum(\text{ax}_\text{i}+\text{b})}{\text{n}}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\text{a}\frac{\sum\text{x}_\text{i}}{\text{n}}+\frac{\sum\text{b}}{\text{n}}$
$\Rightarrow\overline{\text{y}_\text{i}}=\text{a}\overline{\text{x}_\text{i}}+\text{b}$
$\Rightarrow55=48\text{a}+\text{b}\ ...(1)$
Now,
Standard deviation of $y_i =$ Standard deviation of $ax_i + b$
$\Rightarrow\sigma_{\text{y}_\text{i}}=\text{a}\times\sigma_{\text{x}_\text{i}}$
$\Rightarrow15=12\text{a}$
$\Rightarrow\text{a}=\frac{15}{12}=1.25$
Putting $a = 1.25$ in $(1)$, we get
$b = 55 - 48 × 1.25 = 55 - 60 = -5$
Thus, the values of $a$ and $b$ are $1.25$ and $-5,$ respectively.
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 51 Mark
If the standard deviation of a variable X is $\sigma,$ then the standard deviation of variable $\frac{\text{aX+b}}{\text{c}}$ is:
- A
$\text{a}\ \sigma$
- B
$\frac{\text{a}}{\text{c}}\sigma$
- ✓
$\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
- D
$\frac{\text{a}\sigma+\text{b}}{\text{c}}$
AnswerCorrect option: C. $\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{Y}=\frac{\text{aX+b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\frac{\text{a}\sum\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\text{a}\sum\text{X}}{\text{nc}}+\frac{\text{nb}}{\text{nc}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
$\text{Var}(\text{X})=\frac{\sum\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\Big)}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\Big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\big(\text{x}_1-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD}(\sigma)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
View full question & answer→MCQ 61 Mark
The mean deviation of the series $a, a + d, a + 2d, ..., a + 2n$ from its mean is:
- A
$\frac{(\text{n}+1)\text{d}}{2\text{n}+1}$
- B
$\frac{\text{n}\text{d}}{2\text{n}+1}$
- ✓
$\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
- D
$\frac{(2\text{n}+1)\text{d}}{\text{n}(\text{n}+1)}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
| $x_i$ |
$\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\big|\text{x}_\text{i}-(\text{a}+\text{nd})\big|$ |
| $a$ |
$nd$ |
| $a + d$ |
$(n - 1)d$ |
| $a + 2d$ |
$(n - 2)d$ |
| $a + 3d$ |
$(n - 3)d$ |
| : |
: |
| : |
: |
| $a + (n + 1)d$ |
$d$ |
| $a + nd$ |
$0$ |
| $a + (n + 1)d$ |
$d$ |
| : |
: |
| : |
: |
| $a + 2nd$ |
$nd$ |
| $\sum\text{x}_\text{i}=(2\text{n}+1)(\text{a}+\text{nd})$ |
$\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\text{n}(\text{n}+1)\text{d}$ |
Therefore are $2n + 1$ terms.
$\Rightarrow N = 2n + 1$
$\sum\text{x}_\text{i}=\text{a}+\text{a}+\text{d}+\text{a}+2\text{d}+\text{a}+3\text{d}+...+\text{a}+2\text{nd}$
$=(2\text{n}+1)\text{a}+\text{d}(1+2+3+...+2\text{n} [a + a + a + ...(2n + 1)$times $= (2n + 1)a])$
$=(2\text{n}+1)\text{a}+\frac{2\text{n}(2\text{n}+1)\text{d}}{2}$
$\Big[$Sum of the first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2},$ but here we are considering$\Big]$
$=(2\text{n}+1)\text{a}+(2\text{n}+1)\text{nd}$
$=(2\text{n}+1)(\text{a}+\text{nd})$
$\overline{\text{X}}=\frac{(2\text{n}+1)(\text{a}+\text{nd})}{(2\text{n}+1)}$
$=\text{a}+\text{nd}$
$\sum\big|\text{x}_\text{i}-\overline{\text{X}}\big|=\text{nd}+(\text{n}-1)\text{d}+(\text{n}-2)\text{d}$
$+...+\text{d}+0+\text{d}+2\text{d}+3\text{d}+...+\text{nd}$
$=\text{d}(\text{n}+(\text{n}-1)+(\text{n}-2)+...+1)$
$\ +0+\text{d}(1+2+3+....+\text{n})$
$=\frac{\text{dn}(\text{n}+1)}{2}+\frac{\text{dn}(\text{n}+1)}{2}$
$\Big\{\because1+2+3+....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big\}$
$=\text{n}(\text{n}+1)\text{d}$
Mean deviation about the mean $=\frac{\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|}{\text{N}}$
$=\frac{\text{n}(\text{n}+1)\text{d}}{(2\text{n}+1)}$ View full question & answer→MCQ 71 Mark
The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is:
AnswerWe have:
$\overline{\text{X}}=50,\ \text{n}=10$
$\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2=250$
$\therefore\text{SD}=\sqrt{\text{Variance of X}}$
$=\sqrt{\frac{\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}}$
$=\sqrt{\frac{250}{10}}$
$=5$
Using $\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$\Rightarrow\text{CV}=\frac{5}{50}\times100$
$=10\%$
View full question & answer→MCQ 81 Mark
let $x_1, x_2, ...,x_n$ be $n$ observations and $\overline{\text{X}}$ be their arithmetic mean. The standard deviation is given by:
- A
$\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- B
$\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- ✓
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
- D
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
AnswerCorrect option: C. $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
It is given that $x_1, x_2, ...,x_n$ be $n$ observations and $\overline{\text{X}}$ be their arithmetic mean.
The standard deviation is given observations is $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}.$
Also, $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}=\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
Hence, the correct answers are options $(c)$ and $(d).$
Disclaimer: For option $(c)$ to be the only correct answer, option $(d)$ should be different from the given value.
View full question & answer→MCQ 91 Mark
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is:
AnswerLet $\overline{\text{x}}$ and $\sigma$ be the mean and standard deviation of 100 observations, respectively.
$\therefore\overline{\text{x}}=50,\ \sigma=5$ and n = 100
$\text{Mean},\ \overline{\text{x}}=50$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{100}=50$
$\Rightarrow\sum\text{x}_\text{i}=5000\ ...(1)$
Now,
Standard deviation, $\sigma=5$
$\Rightarrow\sqrt{\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{\sum\text{x}_\text{i}}{100}\Big)^2}=5$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{5000}{100}\Big)^2=25$ [From (1)]
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}=25+2500=2525$
$\Rightarrow{\sum\text{x}_\text{i}^2}=252500$
Thus, the sum of all squares of all the observations is 252500.
Hence, the correct answer is option (c).
View full question & answer→MCQ 101 Mark
The standard deviation of the observations $6, 5, 9, 13, 12, 8, 10$ is:
- A
$6$
- B
$\sqrt6$
- C
$\frac{52}{7}$
- ✓
$\sqrt{\frac{52}{6}}$
AnswerCorrect option: D. $\sqrt{\frac{52}{6}}$
View full question & answer→MCQ 111 Mark
The standard deviation of the data:
| $x$ |
$1$
|
$a$
|
$a^2$
|
$....$ |
$a^n$
|
| $f$ |
$^nC_0$
|
$^nC_1$
|
$^nC_2$
|
$....$ |
$^nC_2$
|
is, - A
$\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- B
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- C
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
- ✓
Answer
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
$x_i^2$
|
$f_ix_i^2$
|
|
$1$
|
$^nC_0$
|
$^nC_0$
|
$1$
|
$1$
|
|
$a$
|
$^nC_1$
|
$a^nC_1$
|
$a^2$
|
$a^{2\ n}C_1$
|
|
$a$
|
$^nC_2$
|
$a^{2\ n}C_2$
|
$a^4$
|
$a^{4\ n}C_2$
|
|
$a$
|
$^nC_3$
|
$a^{3\ n}C_3$
|
$a^6$
|
$a^{6\ n}C_3$
|
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
|
$a^n$
|
$^nC_n$
|
$a^{n\ n}C_n$
|
$a^{2n}$
|
$a^{2n\ n}C_n$
|
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}=2^\text{n}$
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}=(1+\text{a})^\text{n}$
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$ |
Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$
$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$
$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$
$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$
$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$
$\sigma=\sqrt{\text{Variance}(\text{X})}$
$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$ View full question & answer→MCQ 121 Mark
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is:
Answer$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$
$=\frac{25}{5}$
$=5$
Taking the absolute value of deviation of each term from the mean, we get:
$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$
$=\frac{2+1+0+1+2}{5}$
$=\frac{6}{5}$
$=1.2$
View full question & answer→MCQ 131 Mark
A batsman scores runs in $10$ innings as $38, 70, 48, 34, 42, 55, 63, 46, 54$ and $44.$ The mean deviation about mean is:
Answer$N = 10$
$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$
$=\frac{494}{10}$
$=49.4$
|
$x_i$
|
$\text{d}_\text{i}=\big|\text{x}_\text{i}-49.4\big|$
|
|
$34$
|
$15.4$
|
|
$38$
|
$11.4$
|
|
$42$
|
$7.4$
|
|
$44$
|
$5.4$
|
|
$46$
|
$3.4$
|
|
$48$
|
$1.4$
|
|
$54$
|
$4.6$
|
|
$55$
|
$5.6$
|
|
$63$
|
$13.6$
|
|
$70$
|
$20.6$
|
|
|
$\sum\limits^{\text{n}}_{\text{i}=}\text{d}_\text{i}=88.8$
|
Mean deviation from the mean $=\frac{88.8}{10}$
$= 8.88$
Disclaimer: No option is matching the answer. View full question & answer→MCQ 141 Mark
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 so obtained is:
AnswerThe given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
If 1 is added to each number, then the new numbers obtained are
2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Now,
$\sum\text{x}_\text{i}=$ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65
$\sum\text{x}_\text{i}^2=$ 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505
$\therefore$ Variance of the numbers so obtained
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-42.25$
$=8.25$
View full question & answer→MCQ 151 Mark
For a frequency distribution standard deviation is computed by applying the formula:
- ✓
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
- B
$\sigma=\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$
- C
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
- D
$\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$
AnswerCorrect option: A. $\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
View full question & answer→MCQ 161 Mark
For a frequency distribution mean deviation from mean is computed by:
- A
$\text{M.D.}=\frac{\sum\text{f}}{\sum\text{f}\ |\text{d}|}$
- B
$\text{M.D.}=\frac{\sum\text{d}}{\sum\text{f}}$
- C
$\text{M.D.}=\frac{\sum\text{fd}}{\sum\text{f}}$
- ✓
$\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$
AnswerCorrect option: D. $\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$
$\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$
View full question & answer→MCQ 171 Mark
Let $x_1, x_2, ..., x_n $ be values taken by a variable $X$ and $y_1, y_2, ..., y_n$ be the values taken by a variable $Y$ such that $y_i = ax_i + b, i = 1, 2,..., n$. Then,
- ✓
Var $(Y) = a^2$ Var $(X)$
- B
Var $(X) = a^2$ Var $(Y)$
- C
Var $(X) =$ Var $(X) + b$
- D
AnswerCorrect option: A. Var $(Y) = a^2$ Var $(X)$
$\text{Var}(\text{x})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}$ where Mean $\Big(\overline{\text{X}}\Big)=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$ and $\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
We have,
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{ax}_\text{i}+\text{b}}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}+\frac{\text{nd}}{\text{n}}$
$=\text{a}\overline{\text{X}}+\text{b}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big\{\text{ax}_\text{i}+\text{b}-\big(\text{a}\overline{\text{X}}+\text{b}\big)\Big\}^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{ax}_\text{i}-\text{a}\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\text{Var}(\text{X})$
View full question & answer→MCQ 181 Mark
If two variates X and Y are connected by the relation $\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}},$ where a, b, c are constants such that ac < 0, then
- A
$\sigma\text{Y}=\frac{\text{a}}{\text{c}}\sigma\text{X}$
- ✓
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
- C
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}+\text{b}$
- D
AnswerCorrect option: B. $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
We know:
$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD of Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{ac}<0$
$\Rightarrow\text{a}<0\text{ or }\text{c}<0$
$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$
$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
View full question & answer→MCQ 191 Mark
If for a sample of size 60, we have the following information $\sum\text{x}_\text{i}^2=18000$ and $\sum\text{x}_\text{i}=960$ then the variance is:
AnswerGiven $\sum\text{x}_\text{i}^2=18000,\ \sum\text{x}_\text{i}=960$ and n = 60
$\therefore$ Variance
$=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\bigg(\frac{\sum\text{x}_\text{i}}{\text{n}}\bigg)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2$
$=300-256$
$=44$
Hence, the correct answer is option (d).
View full question & answer→MCQ 201 Mark
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is:
AnswerThe given observations are 3, 10, 10, 4, 7, 10, 5.
$\therefore\text{Mean},\ \overline{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
Now,
Mean deviation from mean, MD
$=\frac{\sum|\text{x}_\text{i}-7|}{7}$
$=\frac{|3-7|+|10-7|+|10-7|+|4-7|+|7-7|+|10-7|+|5-7|}{7}$
$=\frac{4+3+3+3+0+3+2}{7}$
$=\frac{18}{7}$
$=2.57$
Hence, the correct answer is (b).
View full question & answer→MCQ 211 Mark
The mean deviation from the median is:
- A
Equal to that measured from another value.
- B
Maximum if all observations are positive.
- C
Greater than that measured from any other value.
- ✓
Less than that measured from any other value.
AnswerCorrect option: D. Less than that measured from any other value.
In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.
View full question & answer→MCQ 221 Mark
Consider the first 10 positive integers. If we multiply each number by -1 and then add 1 to each number, the variance of the numbers so obtained is:
AnswerThe first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Multiplying each number by −1, we get
-1, -2, -3, -4, -5, -6, -7, -8, -9, -10
Adding 1 to each of these numbers, we get
0, -1, -2, -3, -4, -5, -6, -7, -8, -9
Now,
$\sum\text{x}_\text{i}=$ 0 + (-1) + (-2) + (-3) + (-4) + (-5) + (-6) + (-7) + (-8) + (-9) = -45
$\sum\text{x}_\text{i}^2=$ 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285
$\therefore$ Variance of the obtained numbers
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2$
$=28.5-20.25$
$=8.25$
View full question & answer→MCQ 231 Mark
If v is the variance and σ is the standard deviation, then:
AnswerCorrect option: C. $\text{V}=\sigma^2$
The variance is the square of the standard deviation.
View full question & answer→MCQ 241 Mark
If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be:
AnswerIf a set of observations, with SD σσ, are multiplied with a non-zero real number a, then SD of the new observations will be $|\text{a}|\sigma.$
Dividing the set of observations by -2 is same as multiplying the observations by $\frac{1}{-2}.$
New $\text{S.D.}=\Big|-\frac{1}{2}\Big|\times8$
$=\frac{8}{2}$
$=4$
View full question & answer→MCQ 251 Mark
The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is given by:
- A
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
- ✓
$\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
- C
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- D
$\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
AnswerCorrect option: B. $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$
Disclaimer: There is some printing error in option $(b)$ given in the question. The answer would be option $(b)$ if it given as $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$
View full question & answer→