MATHS M.C.Q (1 Marks)

3. 252500

Solution:

Let $\overline{\text{x}}$ 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).

4. 2.87

Solution:

We 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).

1. s

Solution:

The 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).

3. 25%

Solution:

Standard deviation is expressed in the following manner:

$\sigma=\sqrt{\frac{1}{\text{n}}\sum_\text{i}\text{x}_\text{i}^2-(\overline{\text{X}})^2}$

$=\sqrt{\frac{1530}{10}-(12)^2}$

$=\sqrt9$

$=3$

$\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$

$=\frac{3}{12}\times100$

$=25%$

1. a = 1.25, b = -5

Solution:

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).

3. $\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$

Solution:

$\text{Y}=\frac{\text{aX+b}}{\text{c}}$

$\overline{\text{Y}}=\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\frac{\text{a}\sum\text{X}+\text{nb}}{\text{c}}}{\text{n}}$

$=\frac{\text{a}\sum\text{X}}{\text{nc}}+\frac{\text{nb}}{\text{nc}}$

$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$

$\text{Var}(\text{X})=\frac{\sum\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\sigma^2$

$\text{Var}(\text{Y})=\frac{\sum\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$

$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\Big)}{\text{n}}$

$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\Big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\big(\text{x}_1-\overline{\text{X}}\big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$

$\text{SD}(\sigma)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$

$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$

3. $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$

Solution:

Therefore are 2n + 1 terms.

⇒ 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 $=\frac{\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|}{\text{N}}$

$=\frac{\text{n}(\text{n}+1)\text{d}}{(2\text{n}+1)}$

1. 10%

Solution:

We 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 $\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$

$\Rightarrow\text{CV}=\frac{5}{50}\times100$

$=10\%$

3. $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$

Solution:

It is given that x₁, x₂, ...,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.

4. $\sqrt{\frac{52}{7}}$

Solution:

The given observations are 6, 5, 9, 13, 12, 8, 10.

Now,

$\sum\text{x}_\text{i}=$ 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63

$\sum\text{x}_\text{i}^2=$ 36 + 25 + 81 + 169 + 144 + 64 + 100 = 619

$\therefore$ Standard deviation of the observations, $\sigma$

$=\sqrt{\frac{1}{\text{N}}\sum\text{x}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{x}_\text{i}\Big)^2}$

$=\sqrt{\frac{1}{7}\times619-\Big(\frac{1}{7}\times63\Big)^2}$

$=\sqrt{\frac{619}{7}-81}$

$=\sqrt{\frac{619-567}{7}}$

$=\sqrt{\frac{52}{7}}$

Hence, the correct answer is option (d).

4. None of these

Solution:

Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$

$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$

$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$

$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$

$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$

$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$

$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$

$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$

$\sigma=\sqrt{\text{Variance}(\text{X})}$

$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$

3. 1.2

Solution:

$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$

$=\frac{25}{5}$

$=5$

Taking the absolute value of deviation of each term from the mean, we get:

$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$

$=\frac{2+1+0+1+2}{5}$

$=\frac{6}{5}$

$=1.2$

1. 8.6

Solution:

N = 10

$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$

$=\frac{494}{10}$

$=49.4$

Mean deviation from the mean $=\frac{88.8}{10}$

$= 8.88$

Disclaimer: No option is matching the answer.

1. 8.25

Solution:

The 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

$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$

$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$

$=50.5-42.25$

$=8.25$

1. $\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$

4. $\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$

1. Var (Y) = a² Var (X)

Solution:

$\text{Var}(\text{x})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}$ 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,

$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$

$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$

$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{ax}_\text{i}+\text{b}}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}+\frac{\text{nd}}{\text{n}}$

$=\text{a}\overline{\text{X}}+\text{b}$

$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big\{\text{ax}_\text{i}+\text{b}-\big(\text{a}\overline{\text{X}}+\text{b}\big)\Big\}^2}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{ax}_\text{i}-\text{a}\overline{\text{X}}\big)^2}{\text{n}}$

$=\text{a}^2\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\text{a}^2\text{Var}(\text{X})$

2. $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

Solution:

$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$

$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$

$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$

$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$

$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$

We know:

$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\sigma^2$

$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$

$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$

$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$

$\text{SD of Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$

$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$

$\text{ac}<0$

$\Rightarrow\text{a}<0\text{ or }\text{c}<0$

$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$

$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

4. 44

Solution:

Given $\sum\text{x}_\text{i}^2=18000,\ \sum\text{x}_\text{i}=960$ 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).

2. 2.57

Solution:

The 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).

1. 8.25

Solution:

The 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

$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$

$=\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2$

$=28.5-20.25$

$=8.25$

3. $\text{V}=\sigma^2$

Solution:

The variance is the square of the standard deviation.

4. 4

Solution:

If 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 $\text{S.D.}=\Big|-\frac{1}{2}\Big|\times8$

$=\frac{8}{2}$

$=4$

2. $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$

Solution:

The mean deviation for n observations x₁, x₂, ...,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 $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$