MCQ 11 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:
- ✓
$8.25$
- B
$6.5$
- C
$3.87$
- D
$2.87$
AnswerCorrect option: A. $8.25$
Since, the first $10$ positive integers are $1, 2, 3, 4, 5, 6, 7, 8, 9$ and $10.$
On multiplying each number by $-1,$ we get $-1, -2, -3, -4, -5, -6, -7, -8, -9, -10$ On adding $1$ in each number.
We get $0, -1, -2, -3, -4, -5, -6, -7, -8, -9.$
$\therefore\ \sum\text{x}_\text{i}=-\frac{9\times10}{2}=-45$
and $\sum\text{x}^2_\text{i}=0^2+(-1)^2+(-2)^2+\ ....\ +(9)^2=\frac{9\times10\times19}{6}=285$
$\text{SD}=\sqrt{\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2}$
$=\sqrt{\frac{285}{10}-\frac{2025}{100}}$
$=\sqrt{\frac{2850-2025}{100}}$
$=\sqrt{8.25}$
Now, $\text{variance}=(\text{SD})^2$
$=\big(\sqrt{8.25}\big)^2$
$=8.25$
View full question & answer→MCQ 21 Mark
Let $\ce{a, b, c, d, e}$ be the observations with mean $m$ and standard deviation $s.$ The standard deviation of the observations $\ce{a + k, b + k, c + k, d + k, e + k}$ is:
AnswerGiven observations are $a, b, c, d$ and $e.$
$\text{Mean}=\text{m}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}$
$\sum\text{x}_\text{i}={\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}={5}\text{m}$
Now, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$
$=\frac{(\text{a}+\text{b}+\text{c}+\text{d}+\text{e})+5\text{k}}{5}$
$=\text{m}+\text{k}$
$\therefore\ \text{SD}=\sqrt{\frac{{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{k}\text{x}_\text{i})}}{\text{n}}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$
$=\text{s}$
View full question & answer→MCQ 31 Mark
The standard deviation of some temperature data in $^\circ C$ is $5.$ If the data were converted into $^\circ F,$ the variance would be:
AnswerGiven that $\sigma_\text{c}=5$
We know that $\text{C}=\frac{5}{9}(\text{F}-32)$
$\Rightarrow\text{F}=\frac{9\text{C}}{5}+32$
$\therefore\ \sigma_\text{F}=\frac{9}{5}\sigma_\text{c}=\frac{9}{5}\times5=9$
$\therefore\ \sigma^2_{\text{F}}=(9)^2$
$=81$
View full question & answer→MCQ 41 Mark
The following information relates to a sample of size $60 \sum\text{x}^2=18000$ and $\sum\text{x}=960,$ then the variance is:
AnswerWe know that variance $(\sigma^2)\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2$
$=300-256$
$=44$
View full question & answer→MCQ 51 Mark
Mean deviation for n observations $x_1, x_2, ...... , x_n$ from their mean $x$ is given by:
- A
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})$
- ✓
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
- C
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
- D
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
AnswerCorrect option: B. $\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
$\text{MD}=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
View full question & answer→MCQ 61 Mark
Let $x_1, x_2, ..., x_n$ be n observations and $\bar{\text{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:
- A
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
- B
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
- ✓
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
- D
$\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(\bar{\text{x}})^{-2}$
AnswerCorrect option: C. $\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
View full question & answer→MCQ 71 Mark
The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is:
AnswerCorrect option: B. $2.57$
$2.57$
Observations are fiven by $3, 10, 10, 4, 7, 10,$ and $5$
$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$
|
$\text{xi}$
|
$\text{d}_\text{i}=|\text{x}_\text{i}-\bar{\text{x}}|$
|
|
$3$
|
$4$
|
|
$10$
|
$3$
|
|
$10$
|
$3$
|
|
$4$
|
$3$
|
|
$7$
|
$0$
|
|
$10$
|
$3$
|
|
$5$
|
$2$
|
|
$\text{Total}$
|
$\sum\text{d}_\text{i}=18$
|
View full question & answer→MCQ 81 Mark
Coefficient of variation of two distributions are $50$ and $60,$ and their arithmetic means are $30$ and $25$ respectively. Difference of their standard deviation is:
AnswerHere, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ and $\bar{\text{x}}_2=25$
$\therefore\ \text{CV}_1=\frac{\sigma_1}{\bar{\text{x}}_1}\times100$
$\Rightarrow50=\frac{\sigma_1}{30}\times100$
$\therefore\ \sigma_1=\frac{30\times50}{100}=15$
and $\text{CV}_2=\frac{\sigma_2}{\bar{\text{x}}_2}\times100$
$\Rightarrow60=\frac{\sigma_2}{25}\times100$
$\therefore\ \sigma^2=\frac{60\times25}{100}=15$
Now, $\sigma_1-\sigma_2=15-15=0$
View full question & answer→MCQ 91 Mark
Standard deviations for first $10$ natural numbers is:
- A
$5.5$
- B
$3.87$
- C
$2.97$
- ✓
$2.87$
AnswerCorrect option: D. $2.87$
We know that $SD$ of first $n$ natural numbers $\sqrt{\frac{\text{n}^2-1}{12}}$
Here, $\text{n}=10$
$\therefore\ \text{SD}=\sqrt{\frac{(10)^2-1}{12}}$
$=\sqrt{\frac{99}{12}} $
$=\sqrt{8.25}$
$=2.87$
View full question & answer→MCQ 101 Mark
Let $x_1, x _2, x_3, x_4, x_5$ be the observations with mean m and standard deviation $s.$ The standard deviation of the observations $kx_1, kx_2, kx_3, kx_4, kx_5$ is:
Answer$ks$
Here, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$
$\text{S}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\text{K}^2\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}.\text{S}$
View full question & answer→MCQ 111 Mark
Following are the marks obtained by $9$ students in a mathematics test: $50, 69, 20, 33, 53, 39, 40, 65, 59$ The mean deviation from the median is:
- A
$9$
- B
$10.5$
- ✓
$12.67$
- D
$14.76$
AnswerCorrect option: C. $12.67$
$12.67$
$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$
$\text{M}_\text{e}=50$
| $\text{xi}$ |
$d_i = |x_i - M_e|$ |
| $20$ |
$30$ |
| $33$ |
$17$ |
| $39$ |
$11$ |
| $40$ |
$10$ |
| $50$ |
$0$ |
| $53$ |
$3$ |
| $59$ |
$9$ |
| $65$ |
$15$ |
| $69$ |
$19$ |
| $N = 2$ |
$\sum\text{d}_\text{i}=114$ |
$\therefore\ \text{MD}=\frac{114}{9}=12.67$ View full question & answer→MCQ 121 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:
- A
$6.5$
- B
$2.87$
- C
$3.87$
- ✓
$8.25$
AnswerCorrect option: D. $8.25$
Given numbers are $1, 2, 3,4, 5, 6, 7, 8, 9$ and $10$
If $1$ is added to each number, then observations will be $2, 3,4, 5, 6,7, 8, 9, 10$ and $11$
$\therefore\ \sum\text{x}_\text{i}=2+3+4+\ ....\ +11$
$=\frac{10}{2}\big[2\times2+9\times1\big]=5[4+9]=65$
and $\sum\text{x}^2_\text{i}=2^2+3^2+4^2+5^2+\ .....\ +11^2=(1^2+2^2+3^2+\ .....\ +11^2)-(1^2)$
$=\frac{11\times12\times23}{6}-1=505$
$\therefore\ \text{s}^2=\frac{\sum\text{x}^2_\text{i}}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-(6.5)^2$
$=50.5-42.35$
$=8.25$
View full question & answer→MCQ 131 Mark
The standard deviation of the data $6, 5, 9, 13, 12, 8, 10$ is:
- ✓
$\sqrt{\frac{52}{7}}$
- B
$\frac{52}{7}$
- C
$\sqrt{6}$
- D
$6$
AnswerCorrect option: A. $\sqrt{\frac{52}{7}}$
$\sqrt{\frac{52}{7}}$
Given data are $6, 5, 9, 13, 12, 8$ and $10$
| $x_i$ |
$x_i^2$ |
| $6$ |
$36$ |
| $5$ |
$25$ |
| $9$ |
$81$ |
| $13$ |
$169$ |
| $12$ |
$144$ |
| $8$ |
$64$ |
| $10$ |
$100$ |
| $\sum\text{x}_\text{i}=63$ |
$\sum\text{x}_\text{i}^2=619$ |
$\therefore\ \text{SD}=\sigma=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2}$
$=\sqrt{\frac{619}{7}-\Big(\frac{63}{7}\Big)^2}=\sqrt{\frac{4333-396}{49}}$
$=\sqrt{\frac{396}{49}}=\sqrt{\frac{52}{7}}$ View full question & answer→MCQ 141 Mark
Let $x_1, x_2, ... x_n$ be n observations. Let $w_i = lx_i + k for i = 1, 2, ... n,$ where l and k are constants. If the mean of $\text{x}_\text{i}{'\text{s}}$ is $48$ and their standard deviation is $12,$ the mean of $\text{w}_\text{i}{'\text{s}}$ is $55$ and standard deviation of $\text{w}_\text{i}{'\text{s}}$ is $15,$ the values of l and $k$ should be:
- ✓
$l = 1.25, k = -5$
- B
$l = -1.25, k = 5$
- C
$l = 2.5, k = -5$
- D
$l = 2.5, k = 5$
AnswerCorrect option: A. $l = 1.25, k = -5$
$l = 1.25, k = -5$
Given, $\text{w}_\text{i}=\text{lx}_\text{i}+\text{k},\ \bar{\text{x}}_\text{i}=48,\text{ sx}_\text{i}=12,\text{ w}_\text{i}=55$ and $\text{sw}_\text{i}=15$
Then, $\bar{\text{w}}_\text{i}=\text{l}\bar{\text{x}}_\text{i}+\text{k}$
$\big[$ where $\bar{ w_ i}$ is mean $w_ i{' s}$ and $\bar{ x_ i}$ is mean of $x_ i{' s}\big]$
$\Rightarrow55=\text{l}\times48+\text{k}\ ...(\text{i})$
Now, $\text{SD of w}_\text{i}=\text{l}(\text{SD of x}_\text{i})$
$\Rightarrow15=\text{l}\times12$
$\Rightarrow\text{l}=\frac{15}{12}=12.5$
From Eq. $(i)$ we get $k = 55 - 1.25 \times 48 = -5$
View full question & answer→MCQ 151 Mark
When tested, the lives $($in hours$)$ of $5$ bulbs were noted as follows: $1357, 1090, 1666, 1494, 1623$ The mean deviations $($in hours$)$ from their mean is:
Answer$178$
The lines of $5$ bulbs are given by
$1357, 1090, 1666, 1494, 1623$
$\therefore\ \text{Mean}=\frac{1357+1090+1666+1494+1623}{5}$
$\Rightarrow\bar{\text{x}}=\frac{7230}{5}=1446$
| $x_i$ |
$\text{d}_\text{i}=|\text{x}_{\text{i}}-\bar{\text{x}}|$ |
| $1357$ |
$89$ |
| $1090$ |
$356$ |
| $1666$ |
$220$ |
| $1494$ |
$48$ |
| $1623$ |
$177$ |
| $Total$ |
$\sum\text{d}_\text{i}=890$ |
$\therefore\ \text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{890}{5}=178$ View full question & answer→MCQ 161 Mark
The mean of $100$ observations is $50$ and their standard deviation is $5$. The sum of all squares of all the observations is:
- A
$50000$
- B
$250000$
- ✓
$252500$
- D
$255000$
AnswerCorrect option: C. $252500$
Here, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=5000$
$\therefore\ \text{SD}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$\Rightarrow\ 5=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{5000}{100}\Big)^2}$
$\Rightarrow25=\frac{\sum\text{x}_\text{i}^2}{100}=2525$
$\therefore\ {\sum\text{x}_\text{i}^2}=2525\times100$
$=252500$
View full question & answer→