3 Marks Question

Question 13 Marks

For a certain frequency distribution $\mu_1^{\prime}=2, \mu_2^{\prime}=18$ and $\mu_3^{\prime}=100$, find $S_k$.

Answer

Given, $\mu_1^{\prime}=2, \mu_2^{\prime}=18$ and $\mu_3^{\prime}=100$
Here, we need to calculate the central moments $\mu_2$ and $\mu_3$.
$
\begin{aligned}
\mu_2 & =\sigma^2=\mu_2^{\prime}-\left(\mu_1^{\prime}\right)^2 \\
& =18-(2)^2=18-4=14 \\
\mu_3 & =\mu_3^{\prime}-3 \mu_2^{\prime} \mu_1^{\prime}+2\left(\mu_1^{\prime}\right)^3 \\
& =100-3(18)(2)+2(2)^3 \\
& =100-108+16 \\
& =8
\end{aligned}
$
Now,
$
\begin{array}{l}
\sigma=\sqrt{\mu_2}=\sqrt{14}=3.74 \text { (Approx.) } \\
S_k=\frac{\mu_3}{\sigma^3}=\frac{8}{(3.74)^3}=0.1529 . \text { (Approx.) }
\end{array}
$

View full question & answer→

Question 23 Marks

For a frequency distribution the Bowley's coefficient of skewness is 1.2. If the sum of 1st and 3 rd quartile is 200 and median is 76 , then find the value of third quartile.

Answer

We have given, $S_k=1.2, Q_1+Q_2=200, Q_2=76$
$
\begin{aligned}
\because & S_k=\frac{Q_3+Q_1-2 Q_2}{Q_3-Q_1} \\
\Rightarrow & 1.2=\frac{(200-2 \times 76)}{Q_3-Q_1} \\
\Rightarrow & Q_3-Q_1=\frac{48}{1.2} \\
\Rightarrow & Q_3-Q_1=40
\end{aligned}
$
and it is given $Q_3+Q_1=200$
$
\therefore \quad Q_1=200-Q_3
$
From eq (i), we get
$
\begin{aligned}
Q_3-\left(200-Q_3\right) & =40 \\
\Rightarrow 2 Q_3 =240 \\
\Rightarrow Q_3 =120 .
\end{aligned}
$

View full question & answer→

Question 33 Marks

Co-efficient of variation of two distributions are 60 and 70, and their standard deviations are 21 and 16 respectively. What are their arithmetic mean?

Answer

Given, $C V_1=60, C V_2=70$
$\sigma_1=21, \sigma_2=16$
We know, $C V=\frac{\sigma}{\bar{x}} \times 100$
$\Rightarrow$ $\bar{x}=\frac{\sigma}{C V} \times 100$
$\therefore$ $\bar{x}_1=\frac{\sigma_1}{C V_1} \times 100$
$=\frac{21}{60} \times 100$
= 35
and, $\bar{x}_2=\frac{\sigma_2}{C V_2} \times 100$
$=\frac{16}{70} \times 100$
= 22.85 (Approx.)
$\therefore$ Their arithmetic mean are 35 and 22.85.

View full question & answer→

Question 43 Marks

Find the variance of first $n$ natural numbers.

Answer

Here, the variables are 1, 2, 3, ........, n
$\begin{aligned} \therefore \text { Mean, } \bar{x} & =\frac{\sum x_i}{n} \\ & =\frac{1+2+3+\ldots .+n}{n} \\ & =\frac{n(n+1)}{2 n} \\ & =\frac{n+1}{2} \\ \text { Variance } & =\frac{\sum x_i^2}{n}-(\bar{x})^2 \\ & =\frac{1^2+2^2+3^2+\ldots+n^2}{n}-\left(\frac{n+1}{2}\right)^2 \\ & =\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n+1}{2}\right)^2 \\ & =(n+1)\left\{\frac{4 n+2-3 n-3}{12}\right\} \\ & =\frac{(n+1)(n-1)}{12} \\ & =\frac{n^2-1}{12} .\end{aligned}$

View full question & answer→

Question 53 Marks

Find the mean and variance for the following data :
$
6,7,10,12,13,4,8,12 .
$

Answer

Here,
Mean$
x=\frac{6+7+10+12+13+4+8+12}{8}
$
$\begin{array}{l}=\frac{72}{8} \\ =9\end{array}$

$x_i$	$x_i-\bar{x}$	$\left(x_i-\bar{x}\right)^2$
6	-3	9
7	-2	4
10	-1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9

$\begin{aligned} \Sigma\left(x_i-\bar{x}\right)^2 & =9+4+1+9+16+25+1+9 \\ & =74 \\ \therefore \quad \text { Variance } & =\frac{\Sigma\left(x_i-\bar{x}\right)^2}{\Sigma f_i}=\frac{74}{8}=9.25 .\end{aligned}$

View full question & answer→

Question 63 Marks

Find the standard deviation for the following:
$20,22,27,30,31,32,35,40,45,48$

Answer

Here, $\bar{x}=\frac{\begin{array}{c}20+22+27+30+31 \\ +32+35+40+45+48\end{array}}{10}$
$\bar{x}=\frac{330}{10}=33$

Variable $x$	Deviation $x-\bar{x}$	$\begin{array}{l}\text { (Deviation) }^2 \\ =(x-\bar{x})^2\end{array}$
20	-13	169
22	-11	121
27	-6	36
30	-3	9
31	-2	4
32	1	1
35	2	4
40	7	49
45	12	144
48	15	225
n=10		$\Sigma(x-\bar{x})^2=762$

Hence,
$
\begin{aligned}
\text { Standard deviation } & =\sqrt{\frac{\Sigma(x-\bar{x})^2}{n}} \\
& =\sqrt{\frac{762}{10}}=\sqrt{76.2}=8.73 .
\end{aligned}
$

View full question & answer→

Question 73 Marks

Find the mean deviation about the median of the following distribution.

Marks obtained	10	11	12	14	15
No. of students	2	3	8	3	4

Answer

$x_i$	$f_i$	$c . f$	$\left\|x_i-M\right\|$	$f_i\left\|x_i-M\right\|$
10	2	2	2	4
11	3	5	1	3
12	8	13	0	0
14	3	16	2	6
15	4	20	3	12
	N = 20			25

Here $N =20$
$
\therefore \quad \text { Median }=12
$
Hence, Mean deviation about Median (M),
$
\begin{aligned}
\text { M.D. }(M) & =\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i} \\
& =\frac{25}{20}=1.25 .
\end{aligned}
$

View full question & answer→

Question 83 Marks

Find the mean deviation about the mean of the distribution.

Size	20	21	22	23	24
Frequency	6	4	5	1	4

Answer

$x_i$	$f_i$	$f_i x_i$	$\left\|x_i-\bar{x}\right\|$	$f_i\left\|x_i-\bar{x}\right\|$
20	6	120	1.65	9.90
21	4	84	0.65	2.60
22	5	110	0.35	1.75
23	1	23	1.35	1.35
24	4	96	2.35	9.40
	20	433		25

$
\therefore \quad \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{433}{20}=21.65
$
$\therefore$ Required mean deviation about mean $(\bar{x})$ i.e.,
$
\text { M.D. }(\bar{x})=\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i}=\frac{25}{20}=1.25 .
$

View full question & answer→

Question 93 Marks

Find the mean deviation from mean and coefficient of deviation for the following series:
Marks : $17,18,25,20,26,28,38,30,22$

Answer

Arranging the marks in ascending order, the series obtained is :
$
17,18,20,22,25,26,28,30,38
$
Here, number of terms $n=9$ is odd, so that
Median, $M_d=$ Values of $\left(\frac{9+1}{2}\right)^{\text {th }}$ term
$
=\text { Value of } 5^{\text {th }} \text { term }
$
$
=25
$
Hence, Mean deviation
$
=\frac{\Sigma\left|x-M_d\right|}{n}
$
$
\begin{aligned}
& |17-25|+|18-25|+|20-25|+ \\
& |22-25|+|25-25|+|26-25|+ \\
= & \frac{|28-25|+|30-25|+|38-25|}{9} \\
= & \frac{8+7+5+3+0+1+3+5+13}{9} \\
= & \frac{45}{9}=5
\end{aligned}
$
Also, coefficient of deviation
$
\begin{array}{l}
=\frac{\text { Mean deviation }}{\text { Median }} \times 100 \\
=\frac{5}{25} \times 100=20.
\end{array}
$

View full question & answer→

Question 103 Marks

Find the mean about median for the following profits (in ₹ 1000 ) of a firm during a week
82,56,75,70,52,80,68

Answer

The profits in thousand rupees is denoted by $x$. Arranging the values of $x$ in an ascending order, we get
52,56,68,70,75,80,82
Therefore, Median, $M_d=70$, thus, median profit = ₹ 70,000
Computation of Mean Deviation

$x_i$	$d_i=\left\|x_i-M_d\right\|$
52 56 68 70 75 80 82	18 14 2 0 5 10 12
Total	$\Sigma d_i=61$

Mean Deviation =$\frac{\Sigma d_i}{n}$= ₹ $\frac{61}{7} \times 1000$
= ₹ $8714.28$

View full question & answer→

Question 113 Marks

What is the range and its coefficient for the following distribution of weight?

Weights in kgs	50 - 54	55 - 59	60 - 64	65 - 69	70 - 74
No. of students	12	18	23	10	3

Answer

The lowest class boundary is 49.50 kgs and the highest class boundary is 74.50 kgs
Thus, L= 74.50 and S= 49.50
$\therefore \quad$ Range $=74.50-49.50=25 kgs$
Also, coefficient of range
$\begin{array}{l}=\frac{L-S}{L+S} \times 100 \\ =\frac{74.50-49.50}{74.50+49.50} \times 100 \\ =\frac{25}{124} \times 100=0.2016 \times 100\end{array}$
$=20.16$

View full question & answer→

Question 123 Marks

The value attendance of 9 students in a test were $13,17,20,5,3,3,18,15$ and 20 . Find the first and three quartiles.

Answer

First we have to arrange the data in ascending order :
3,3,5,13,15,17,18,20,20
Here, n=9.
Lower quartile or first quartile, $Q_1=\frac{n+1}{4}$ th value
$
\begin{array}{l}
=\frac{10}{4} \text { th value } \\
=2.5 \text { th value }
\end{array}
$
$\begin{aligned} \therefore \quad Q_1 & =\frac{2^{n d} \text { value }+3^{r d} \text { value }}{2} \\ & =\frac{3+5}{2}=4 .\end{aligned}$
Upper quartile or third quartile,
$
Q_3=\frac{3(n+1)}{4} \text { th value }
$
$=\frac{30}{4}$ th value $=7.5$ th value
$\begin{aligned} \therefore \quad Q_3 & =\frac{7 \text { th value }+8 \text { th value }}{2} \\ & =\frac{18+20}{2}=19 .\end{aligned}$

View full question & answer→

Applied Maths 3 Marks Question

Test yourself on this topic