16 questions · self-marked practice — reveal the answer and mark yourself.
The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be:
Solution:
Given that $\sigma_\text{c}=5$
We know that $\text{C}=\frac{5}{9}(\text{F}-32)$
$\Rightarrow\text{F}=\frac{9\text{C}}{5}+32$
$\therefore\ \sigma_\text{F}=\frac{9}{5}\sigma_\text{c}=\frac{9}{5}\times5=9$
$\therefore\ \sigma^2_{\text{F}}=(9)^2=81$
The following information relates to a sample of size 60 $\sum\text{x}^2=18000$ and $\sum\text{x}=960,$ then the variance is:
Solution:
We know that variance $(\sigma^2)\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2=300-256=44$
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:
Solution:
Here, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ 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, $\sigma_1-\sigma_2=15-15=0$
Mean deviation for n observations x1, x2, ...... , xn from their mean x is given by:
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})$
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
Solution:
$\text{MD}=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
Let x1, x2, ..., xn be n observations and $\bar{\text{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
$\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(\bar{\text{x}})^{-2}$
Solution:
The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
Standard deviations for first 10 natural numbers is:
Solution:
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}} $
$=\sqrt{8.25}=2.87$
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:
Solution:
$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$
$\text{M}_\text{e}=50$
| xi | di = |xi - Me| |
| 20 | 30 |
| 33 | 17 |
| 39 | 11 |
| 40 | 10 |
| 50 | 0 |
| 53 | 3 |
| 59 | 9 |
| 65 | 15 |
| 69 | 19 |
| N = 2 | $\sum\text{d}_\text{i}=114$ |
$\therefore\ \text{MD}=\frac{114}{9}=12.67$
Let x1, x 2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is:
Solution:
Here, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$
$\text{S}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\text{K}^2\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}.\text{S}$
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:
Solution:
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 $\sum\text{x}^2_\text{i}=2^2+3^2+4^2+5^2+\ .....\ +11^2=(1^2+2^2+3^2+\ .....\ +11^2)-(1^2)$
$=\frac{11\times12\times23}{6}-1=505$
$\therefore\ \text{s}^2=\frac{\sum\text{x}^2_\text{i}}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-(6.5)^2$
$=50.5-42.35$
$=8.25$
Let x1, x2, ... xn be n observations. Let wi = lxi + k for i = 1, 2, ... n, where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviation of wi's is 15, the values of l and k should be:
Solution:
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, $\bar{\text{w}}_\text{i}=\text{l}\bar{\text{x}}_\text{i}+\text{k}$ $\big[\text{where }\bar{\text{w}}_\text{i}\text{ is mean w}_\text{i}{'\text{s}}\text{ and }\bar{\text{x}}_\text{i}\text{ is mean of x}_\text{i}{'\text{s}}\big]$
$\Rightarrow55=\text{l}\times48+\text{k}\ ...(\text{i})$
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 × 48 = -5
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:
Solution:
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, $\text{variance}=(\text{SD})^2=\big(\sqrt{8.25}\big)^2=8.25$
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:
Solution:
Given 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, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$
$=\frac{(\text{a}+\text{b}+\text{c}+\text{d}+\text{e})+5\text{k}}{5}=\text{m}+\text{k}$
$\therefore\ \text{SD}=\sqrt{\frac{{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{k}\text{x}_\text{i})}}{\text{n}}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$
$=\text{s}$
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is:
Solution:
Observations are fiven by 3, 10, 10, 4, 7, 10, and 5
$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$
| 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 |
| Total | $\sum\text{d}_\text{i}=18$ |
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is:
$\sqrt{\frac{52}{7}}$
$\frac{52}{7}$
$\sqrt{6}$
$6$
Solution:
Given data are 6, 5, 9, 13, 12, 8 and 10
| xi | xi2 |
| 6 | 36 |
| 5 | 25 |
| 9 | 81 |
| 13 | 169 |
| 12 | 144 |
| 8 | 64 |
| 10 | 100 |
| $\sum\text{x}_\text{i}=63$ | $\sum\text{x}_\text{i}^2=619$ |
$\therefore\ \text{SD}=\sigma=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2}$
$=\sqrt{\frac{619}{7}-\Big(\frac{63}{7}\Big)^2}=\sqrt{\frac{4333-396}{49}}$
$=\sqrt{\frac{396}{49}}=\sqrt{\frac{52}{7}}$
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:
Solution:
The lines of 5 bulbs are given by
1357, 1090, 1666, 1494, 1623
$\therefore\ \text{Mean}=\frac{1357+1090+1666+1494+1623}{5}$
$\Rightarrow\bar{\text{x}}=\frac{7230}{5}=1446$
| xi | $\text{d}_\text{i}=|\text{x}_{\text{i}}-\bar{\text{x}}|$ |
| 1357 | 89 |
| 1090 | 356 |
| 1666 | 220 |
| 1494 | 48 |
| 1623 | 177 |
| Total | $\sum\text{d}_\text{i}=890$ |
$\therefore\ \text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{890}{5}=178$
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is:
Solution:
Here, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=5000$
$\therefore\ \text{SD}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$\Rightarrow\ 5=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{5000}{100}\Big)^2}$
$\Rightarrow25=\frac{\sum\text{x}_\text{i}^2}{100}=2525$
$\therefore\ {\sum\text{x}_\text{i}^2}=2525\times100=252500$