4 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip

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Question 14 Marks

For a certain distribution, $\Sigma x_i=20, \Sigma x_i^2=100$ and $\Sigma x_i^3$ $=1500$, and $n=10$, find coefficient of skewness.

Answer

Here, we calculate raw moments
$
\begin{array}{l}
\mu_1^{\prime}=\frac{\Sigma x_i}{n}=\frac{20}{10}=2 \\
\mu_2^{\prime}=\frac{\Sigma x_i^2}{n}=\frac{100}{10}=10 \\
\mu_3^{\prime}=\frac{\Sigma x_i^3}{n}=\frac{1500}{10}=150
\end{array}
$
Now, we calculate central moments
$\mu_2=\sigma^2=\mu_2^{\prime}-\left(\mu_1^{\prime}\right)^2$
$=10-(2)^2=10-4=6$
or, $\sigma=\sqrt{6}$
and $\mu_3=\mu_3^{\prime}-3 \mu_2^{\prime} \mu_1^{\prime}+2\left(\mu_1^{\prime}\right)^3$
$\begin{array}{l}=150-3(10)(2)+2(2)^3 \\ =150-60+16 \\ =106\end{array}$
Therefore,
Skewness, $S_k=\frac{\mu_3}{\sigma^3}=\frac{106}{(\sqrt{6})^3}$
$=\frac{106}{6 \sqrt{6}}=7.212$. (Approx.)

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Question 24 Marks

Find the first four moments about zero for the set of the number $1,3,5,7$.

Answer

The first four moment about zero means, first four raw moments about origin

X	X	$X^2$	$X^3$	$X^4$
1	1	1	1	1
3	3	9	27	81
5	5	25	125	625
7	7	49	343	2401
	$\Sigma X=16$	$\Sigma X^2=84$	$\Sigma X^3=496$	$\Sigma X^4=3108$

$\begin{array}{ll}\text { Here, } & n=4 \\ \therefore & \mu_1^{\prime}=\frac{\Sigma X}{n}=\frac{16}{4}=4\end{array}$
$\begin{array}{l}\mu_1^{\prime}=\frac{\Sigma X^2}{n}=\frac{84}{4}=21 \\ \mu_3^{\prime}=\frac{\Sigma X^3}{n}=\frac{496}{4}=124 \\ \mu_4^{\prime}=\frac{\Sigma X^4}{n}=\frac{3108}{4}=777\end{array}$

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Question 34 Marks

The first four raw moments of a distribution are $2,136,320$ and 40000 . Find out coefficient of skewness and kurtosis.

Answer

Given that, $\mu_1^{\prime}=2, \mu_2^{\prime}=136, \mu_3^{\prime}=320$ and $\mu_4^{\prime}=$ 40000
First of all we have to calculate the first four central moments
$\begin{aligned} \mu_1 & =\mu_1^{\prime}-\mu_1^{\prime}=0 \\ \mu_2 & =\mu_2^{\prime}-\left(\mu_1^{\prime}\right)^2=136-(2)^2=132 \\ \mu_3^{\prime} & =\mu_3^{\prime}-3 \mu_2^{\prime}+\left(2 \mu_1^{\prime}\right)^3 \\ & =320-3 \times 136 \times 2+2(2)^3 \\ & =320-816+16=-480 \\ \mu_4 & =\mu_4^{\prime}-4 \mu_1^{\prime} \mu_3^{\prime}+6 \mu_2^{\prime}\left(\mu_1^{\prime}\right)^2-3\left(\mu_1^{\prime}\right)^4 \\ & =40000-4 \times 2 \times 320+6 \times 136 \\ & \quad \times(2)^2-3 \times(2)^4 \\ & =40000-2560+3264-48 \\ & =40656\end{aligned}$
$\begin{aligned} \text { Skewness, } \beta_1 & =\frac{\mu_3^2}{\mu_2^3}=\frac{(-480)^2}{(132)^3}=0.10017 \\ & =0.1002 \text { (Approx.) } \\ \text { Kurtosis, } \beta_2 & =\frac{\mu_4}{\mu_2^2}=\frac{40656}{(132)^2}=2.333 . \text { (Approx.) }\end{aligned}$
The value of $\beta_2$ is less than 3 , hence the curve platykurtic.

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Question 44 Marks

Find the mean and standard deviation using short-cut method.

$x_i$	60	61	62	63	64	65	66	67	68
$f_i$	2	1	12	29	25	12	10	4	5

Answer

Let $y_i=\frac{x_i-a}{i}=x_i-64[\because i=1$ and ' $a$ ' is assumed to be 64$]$

$x_i$	$f_i$	$y_i=x_i-64$	$f_i y_i$	$y_i^2$	$f_i y_i^2$
60	2	-4	-8	16	32
61	1	-3	-3	9	9
62	12	-2	-24	4	48
63	29	-1	-29	1	29
64	25	0	0	0	0
65	12	1	12	1	12
66	10	2	20	4	40
67	4	3	12	9	36
68	5	4	20	16	80
Total	100		0	60	286

Here,
$\quad \begin{aligned} \text { Mean, } \bar{x} & =64+\frac{\sum f_i y_i}{\sum f_i} \\ & =64+0 \\ & =64\end{aligned}$
Variance,
$
\begin{aligned}
\sigma^2 & =\frac{1}{\sum f_i}\left[\sum f_i y_i^2-n \bar{y}^2\right] \\
& =\frac{1}{100}[286]=2.86 \\
\sigma & =\sqrt{2.86}=1.69 . \text { (Approx.) }
\end{aligned}
$

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Question 54 Marks

If $\bar{x}$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, x_3, \ldots, x_n$, then prove that the mean and variance of the observations $a x_1, a x_2, \ldots$, $a x_n$ are $a \bar{x}$ and $a^2 \sigma^2$ respectively $(a \neq 0)$.

Answer

We know :
$
\text { Mean, } \bar{x}=\frac{x_1+x_2+\ldots+x_n}{n}
$
and, mean of $a x_1, a x_2, \ldots, a x_n$
$
\begin{array}{l}
=\frac{a x_1+a x_2+\ldots+a x_n}{n} \\
=a \cdot \frac{x_1+x_2+\ldots+x_n}{n} \\
=a \cdot \bar{x}
\end{array}
$
Also, Variance, $\quad \sigma^2=\frac{\sum\left(x_i-\bar{x}\right)^2}{n}$
$\begin{array}{l}\text { Variance of } a x_1, a x_2, a x_3, \ldots, a x_n \\ \qquad=\frac{\sum\left(a x_i-a \bar{x}\right)^2}{n}\end{array}$
$\begin{array}{l}=\frac{\left(a x_1-a \bar{x}\right)^2+\left(a x_1-a \bar{x}\right)^2+\ldots+\left(a x_n-a \bar{x}\right)^2}{n} \\ =\frac{a^2\left(x_1-\bar{x}\right)^2+a^2\left(x_1-\bar{x}\right)^2+\ldots+a^2\left(x_n-\bar{x}\right)^2}{n} \\ =a^2 \frac{\sum\left(x_1-\bar{x}\right)^2}{n} \\ =a^2 \sigma^2\end{array}$
$\therefore$ New mean $=a x_1$ and new variance $=a^2 \sigma^2$.

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Question 64 Marks

Find the mean, variance and standard deviation of first 10 multiples of 4 .

Answer

Here, the variables are $
4,8,12,16,20,24,28,32,36,40
$
Here, $N =10$,
So,
Mean, $\bar{x}=\frac{4+8+\ldots+40}{10}$
$\begin{array}{l}=\frac{4(1+2+\ldots 10)}{10} \\ =\frac{4}{10} \cdot \frac{10(10+1)}{2} \\ =2 \times 11 \\ =22\end{array}$

$x_i$	$x_i-\bar{x}$	$\left(x_i-\bar{x}\right)^2$
4	-18	324
8	-14	196
12	-10	100
16	-6	36
20	-2	4
24	2	4
28	6	36
32	10	100
36	14	196
40	18	324
Total	$\sum\left(x_i-\bar{x}\right)=0$	$\sum\left(x_i-\bar{x}\right)^2=1320$

We have $n=10, \sum\left(x_i-\bar{x}\right)^2=1320$
$\therefore$ Variance, $\sigma^2=\frac{1}{n} \sum\left(x_i-\bar{x}\right)^2=\frac{1}{10} \times 1,320=132$
and
$
\sigma=\sqrt{132}=11.49
$
$\therefore$ Mean $=22$, Variance $=132$ (Approx.) and Standard deviation $=11.49$ (Approx.).

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Question 74 Marks

Co-efficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25 respectively. Find the difference of their standard deviation.

Answer

Given,
$
\begin{aligned}
C . V_1 & =50, C . V_{\cdot}=60 \\
\bar{x}_1 & =30, \bar{x}_2=25
\end{aligned}
$
We know :
$
\begin{array}{l}
\text { C.V. }=\frac{\sigma}{\bar{x}} \times 100 \\
\Rightarrow \quad \bar{x}=\frac{\sigma}{C . V .} \times 100 \\
\therefore \quad \bar{x}_1=\frac{\sigma_1}{C . V_{\cdot}} \times 100 \\
\Rightarrow \quad 30=\frac{\sigma_1}{50} \times 100 \\
\Rightarrow \quad \sigma_1=15 \\
\text { and } \quad \bar{x}_2=\frac{\sigma_2}{C . V_{\cdot 2}} \times 100 \\
\Rightarrow \quad 25=\frac{\sigma_2}{60} \times 100 \\
\Rightarrow \quad \sigma_2=25 \times \frac{3}{5} \\
\Rightarrow \quad \sigma_2=15 \\
\Rightarrow \text { Required difference }=\sigma_1-\sigma_2 \\
\therefore \quad=15-15 \\
=0 \text {. }
\end{array}
$

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Question 84 Marks

The mean and SD for $a, b$ and 2 are 3 and $2 \sqrt{3}$, respectively. Find value of $a b$.

Answer

Given,
$
\begin{aligned}
\text { mean } & =3 \\
\text { S.D. } & =2 \sqrt{3}
\end{aligned}
$
Given numbers are: $a, b$, and 2
$
\begin{array}{ll}
\therefore & \text { mean }=\frac{a+b+2}{3} \\
\Rightarrow & a+b+2=9
\end{array}
$
$
a+b=7 \qquad\ldots(i)
$
$
\begin{aligned}
\text { Standard Deviation } & =\sqrt{\frac{\Sigma(x-\text { mean })^2}{n}} \quad[\because n=3] \\
2 \sqrt{3} & =\sqrt{\frac{\Sigma(x-\text { mean })^2}{3}}
\end{aligned}
$
Squaring both sides, we get
$
\begin{aligned}
12 & =\frac{\sqrt{\Sigma(x-\text { mean })^2}}{3} \\
\Rightarrow \quad 36 & =\sqrt{\Sigma(x-\text { mean })^2}
\end{aligned}
$
Now,
$
\Sigma(x-\text { mean })^2=(a-3)^2+(b-3)^2+(2-3)^2
$
$
\begin{array}{l}
\Rightarrow \quad 36=\left(a^2+9-6 a\right)+\left(b^2+9-6 b\right)+1 \\
\Rightarrow \quad 36=a^2+b^2-6(a+b)+19 \\
\text { Using } \quad(a+b)^2=a^2+b^2+2 a b \\
\Rightarrow \quad 36=(a+b)^2-2 a b-6(a+b)+19 \\
\Rightarrow \quad 36=(7)^2-2 a b-6(7)+19 \\
\Rightarrow \quad 2 a b=49-42+19-36 \\
\Rightarrow \quad 2 a b=68-78 \\
\Rightarrow \quad a b=-5 .
\end{array}
$

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Question 94 Marks

Find the mean deviation about the mean for the following data :

Answer

Let us calculate Mean deviation.

Here, $\qquad$ Mean $(x)=$ assumed mean $+\frac{\Sigma f_i d_i}{\Sigma f_i} \times h$
$\begin{array}{l}=350+\frac{4}{50} \times 100 \\ =358\end{array}$
$\therefore$ Mean deviation about mean $(x)$,
$
\text { M.D. }(\bar{x})=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{7896}{50}=157.92 .
$

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Question 104 Marks

Find the mean deviation about the median for the following data :

$x_i$	5	7	9	10	12	15
$f_i$	8	6	2	2	2	6

Answer

$x_i$	$f_i$	$c.f.$	$\left\|x_i-M\right\|=\left\|x_i-7\right\|$	$f_i\left\|x_i-7\right\|$
5	8	8	2	16
7	6	14	0	0
9	2	16	2	4
10	2	18	3	6
12	2	20	5	10
15	6	26	8	48
	$\Sigma f=26$			$\Sigma f_i\left\|x_i-M\right\|=84$

Here, $N=26$, which is even.
Median observation are $\left(\frac{26}{2}\right)^{\text {th }}$ and $\left(\frac{26}{2}+1\right)^{\text {th }}$ observations
i.e., $13^{\text {th }}$ and $14^{\text {th }}$ Observations
i.e., 7 and 7.
Median $=\frac{7+7}{2}=7$
$\therefore$ Mean deviation about median $(M)$
$
\begin{aligned}
\text { M.D. }(M) & =\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i} \\
& =\frac{84}{26} \\
& =3.23 . \text { (Approx.) }
\end{aligned}
$

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Question 114 Marks

If $\bar{x}$ is the mean and M.D $\bar{x})$ is mean deviation from mean then find the number of observations lying between $\bar{x}$-M.D. $(\bar{x})$ and $\bar{x}+$ M.D. $(\bar{x})$. Use the data $22,24,30,27,29,31,25,28,41,42$.

Answer

Let us first arrange the data in ascending order :
$22,24,25,27,28,29,30,31,41,42$
Mean,
$
\begin{aligned}
\bar{x} & =\frac{22+24+25+27+28+29+30+31+41+42}{10} \\
& =\frac{299}{10} \\
& =29.9
\end{aligned}
$
and
Mean deviation from mean,
M.D. $(\bar{x})=\frac{\sum_{i=1}^{10}\left|x_i-\bar{x}\right|}{10}$
$\begin{aligned} & |22-29.9|+|24-29.9|+|25-29.9|+|27-29.9| \\ = & \frac{+|28-29.9|+|29-29.9|+|30-29.9|+|31-29.9|+|41-29.9|+|42-29.9|}{10} \\ = & \frac{7.9+5.9+4.9+2.9+1.9+0.9+0.1+1.1+11.1+12.1}{10}\end{aligned}$
$\begin{array}{l}=\frac{48.8}{10} \\ =4.88 \\ =4.9 \text { (approx.) }\end{array}$
$
\begin{array}{ll}
\therefore & \bar{x}-\text { M.D. }(\bar{x})=29.9-4.9=25 \\
\text { and } & \bar{x}+\text { M.D. }(\bar{x})=29.9+4.9=34.8
\end{array}
$
Observations lying between 25 and 34.8 are 27, 28, 29, 30, 31 .

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Question 124 Marks

Find the quartile deviation and its coefficient for the marks obtained by 10 students :
$
56,48,65,35,42,75,82,60,55
$

Answer

After arranging the marks in an ascending order of magnitude, we get
$
35,42,48,50,55,56,60,65,75,82
$
First quartile $\left( Q _1\right)=\left(\frac{n+1}{4}\right)^{\text {th }}$ observation
$
\begin{array}{l}
=\left(\frac{10+1}{4}\right)^{\text {th }} \text { observation } \\
=2.75^{\text {th }} \text { observation } \\
=2^{\text {nd }} \text { observation }+0.75
\end{array}
$
$\times$ (difference between the third and the $2^{\text {nd }}$ observation)
$
\begin{array}{l}
=42+0.75 \times(48-42) \\
=46.50
\end{array}
$
Third quartile $\left(Q_3\right)=\frac{3(n+1)}{4}$ th observation
$=3\left(\frac{10+1}{4}\right)^{\text {th }}$ observation
$=8.25^{\text {th }}$ observation
$=65+0.25 \times$ (difference between $9^{\text {th }}$ and $8^{\text {th }}$ observation)
$\begin{aligned} & =65+0.25 \times 10 \\ & =65+2.5=67.50 \\ \text { Quartile Deviation } & =\frac{Q_3-Q_1}{2} \\ & =\frac{67.50-46.50}{2} \\ & =10.50\end{aligned}$
Coefficient of quartile deviation
$
\begin{array}{l}
=\frac{Q_3-Q_1}{Q_3+Q_1} \times 100 \\
=\frac{67.50-46.50}{67.50+46.50} \times 100 \\
=18.42
\end{array}
$

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Question 134 Marks

Find the percentage of workers who earned more than ₹ 10200 from the following data :

Income in ₹	7000 - 8000	8000 - 9000	9000 - 10000	10000 - 11000	11000 - 12000
No. of workers	4	8	9	6	3

Answer

Let $x \%$ workers earn upto ₹ 10200
$\therefore P_x$= ₹ 10200, n=30, i=1000
Setting the data in tabular form, we have

Class interval	Frequency	Cumulative frequency
7000 - 8000	4	4
8000 - 9000	8	12
9000 - 10,000	9	21
10,000 - 11,000	6	$27 \leftarrow P_x$
11,000 - 12,000	3	30

Obviously, $P_x$ lies in the interval 10000-11000.
$\therefore$ $10,200=10,000+\frac{\frac{30 \times x}{100}-21}{6} \times 1000$
$\Rightarrow$ $200 \times 6=300 x-21000$
$\Rightarrow$ $300 x=22200 \Rightarrow x=74 \%$
$\Rightarrow$ $74 \%$ workers earn upto ₹ 10,200.
Hence, 26% workers earn more than ₹10,200.

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Question 144 Marks

Following is the distribution of marks obtained by 50 students in a class test :

Marks more then	0	10	20	30	40	50
No. of students	50	42	38	28	16	3

(i) If $60 \%$ of the students pass the test, find the minimum marks obtained by a pass student.
(ii) Find the percentage of students who scored more than 33 marks.

Answer

First we prepare the following cumulative frequency table :

Here, $n=50$
(i) As $60 \%$ students pass the test, it means that $40 \%$ students get less than minimum passing marks. minimum passing marks are $P_{40}$.
Now, $P_{40}=\frac{40 n}{100}$ th value
$=\frac{40 \times 50}{100}$ th value
$=20$ th value, which lies in class $20-30$
$\therefore \quad P_{40}=l+\left(\frac{\frac{40 n}{100}-c}{f}\right) \times i$
$=20+\frac{20-12}{10} \times 10=28$.
Hence, minimum passing marks are 28 .
(ii) Let us assume that $r \%$ students score less than 33 marks than $P_r=33$. As 33 lies in the class $30-40$, we get
$P_{ r }=l+\frac{\frac{r n}{100}-c}{f} \times i$
$\Rightarrow \quad 33=30+\frac{\frac{r \times 50}{100}-22}{12} \times 10$
$\Rightarrow \quad \frac{r}{2}-22=\frac{3 \times 12}{10}=3.6 \Rightarrow \frac{r}{2}=25.6 \Rightarrow r=51.2$
Thus, $51.2 \%$ students scored less than 33 marks.
Hence, $100-51.2$ i.e., $48.8 \%$ students scored more than 33 marks.

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Question 154 Marks

Find the value of $a$ and $b$ from the following data :

Given that the third decile is 11.

Answer

We construct the cumulative frequency table as under :

Marks	0-5	5-10	10-15	15-20	20-25
No. of students	7	a	25	30	b
Cumulative frequency	7	7+ a	32 + a	62 + a	62 + a + b

As the total number of students is 100,
$62+a+b=100 \Rightarrow a+b=38$
Third decide, $D_3$ is $\frac{3}{10}$ of $100^{\text {th }}$ i.e., $30^{\text {th }}$ value. Given third decile is 11 , so it lies in the class $10-15$.
Using formula,
$D_3=l+\frac{\frac{3}{10} n-c}{f} \times i$, we get
$\Rightarrow \quad 11=10+\frac{30-(7+a)}{25} \times 5$
$\Rightarrow \quad 1=\frac{23-a}{5} \Rightarrow 5=23-a \Rightarrow a=18$
From eq. (i), $b=38-a=38-18=20$
Hence, $a=18$ and $b=20$.

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Question 164 Marks

The heights (to nearest cm ) of 60 students of a certain school are given in the following frequency distribution table:

Find the (i) median (ii) lower quartile (iii) upper quartile (iv) inter quartile range.

Answer

The given variates are already in ascending order. We construct the cumulative frequency table as under :

Variate (Height)	Frequency (No. of students)	Cumulative frequency
151	6	6
152	4	10
153	11	21
154	9	30
155	16	46
156	12	58
157	2	60

(i)
$\begin{aligned} \text { Median } & =\frac{n+1}{2} \text { th item } \\ & =\frac{60+1}{2} \text { th item } \\ & =30.5 \text { th item }\end{aligned}$
$\begin{array}{l}=\frac{30 th \text { item }+31 \text { th item }}{2} \\ =\frac{154+155}{2}=154.5 cm .\end{array}$
(ii)
$\begin{aligned} Q_1 & =\frac{n+1}{4} \text { th value } \\ & =\frac{60+1}{4} \text { th value } \\ & =15.25 \text { th value }\end{aligned}$
$=153 cm$ (as both 15th and 16th value are 153)
(iii)
$\begin{aligned} Q_3 & =\frac{3(n+1)}{4} \text { th value } \\ & =\frac{3(60+1)}{4} \text { th value }\end{aligned}$
$\begin{array}{l}=45.75 th \text { value } \\ =155 cm \text { (as both 45th and 46th } \\ \quad \text { values are } 155)\end{array}$
(iv)
Interquartile range $=Q_3-Q_1=155-153=2 cm$

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