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Question 16 Marks
Following table gives the distribution of companies according to the size of capital. Using step deviation method, find out the mean size of the capital of a company.
Capital (Rs. in lakh)Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30
Number of Companies202729384853
Answer
Capital (Rs. in lakh)Number of Companies (f)MidValue (m)dim(m−A) (A =12.5)$\begin{array}{l} d ^{\prime} m \left(\frac{ m }{ c }\right)  ( c = 5 )\end{array}$fd' m
0-5202.5-10-2$\left.\begin{array}{c}-40 \\ -7\end{array}\right]-47$
5-1027-20=77.5-5-1
10-1529-27=212.5000
15-2038-29=917.5+5+1$\left.\begin{array}{l}+9 \\ +20 \\ +15\end{array}\right]+44$
20-2548-38=1022.5+10+2
25-3053-48=527.5+15+3
 $\Sigma f=53$   $\Sigma f d^{\prime} m=-3$
Here,
$\begin{array}{l}
A=12.5, \Sigma f=53, \Sigma f d^{\prime} m=-3 \text { and } c=5 \\
\bar{X}=A+\frac{\Sigma f d^{\prime} m}{\Sigma f} \times c=12.5+\frac{(-3)}{53} \times 5=R s .12 .5-0.28=R s .12 .22 \\
\therefore \bar{X}=12.22
\end{array}$
Hence, mean size of capital is Rs. 12.22 lakhs
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Question 26 Marks
Give formula for:
a. Simple mean in individual series by short cut method
b. Weighted mean
c. Simple mean in continuous series by direct method
d. Simple mean in discrete series by short cut method
e. Combined Mean 
f. Simple mean in continuous series by step deviation method
Answer
(a) $\bar{x}=\frac{1}{n} A+\frac{\sum d}{N}$
(b) Weighted Mean $=\Sigma W X / \Sigma W$
(c) $\bar{x}=\frac{\sum_{i=1}^n f_i m_i}{\sum_{i=1}^n f_i}$
(d) $\bar{x}=A+\frac{\sum_{i=1}^n f_i d_i}{\sum_{i=1}^n f_i}$
(e) Combined Mean $\bar{x} 12=\frac{\bar{x}_1 N_1+\bar{x}_2 N_2}{N_1+N_2}$
(f) $\bar{x}=A+\frac{\sum d^{\prime}}{N} \times i$
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Question 36 Marks
From the following data, calculate coefficient of correlation between age and playing habits.
Age Group20 -3030 - 4040 - 5050 - 6060 - 70
Number of students2560402020
Number of Regular Players10301221
Answer
 First, we are required to calculate the percentage of regular players.
Calculation of Percentage of Regular Players:
Number of StudentsNumber of Regular PlayersPercentage of Regular Players
2510$\frac{10}{25} \times 100=40$
6030$\frac{30}{60} \times 100=50$
4012$\frac{12}{40} \times 100=30$
202$\frac{2}{20} \times 100=10$
201$\frac{1}{20} \times 100=5$
Denoting mid value of age as X and percentage of regular players as Y. 
Age GroupXdx(X - A), A = 45$\begin{array}{c}d x^{\prime}\left(\frac{d z}{a}\right),  c_1=10\end{array}$$dx ^{\prime 2}$Ydy(Y - A), A = 30$\begin{array}{c} dy ^{ \prime }\left(\frac{d y}{c_2}\right),  c _2=5\end{array}$$dy ^{\prime 2}$dx'dy'
20 -3025-20-24401024-4
30-4035-10-115020416-4
40-5045000300000
50-6055101110-20-416-4
60 -706520245-25-525-10
   $\Sigma d x^{\prime}=0$\begin{array}{c}\Sigma d x^2=10\end{array}  $\Sigma d y^{\prime}=-3$\begin{array}{c}\Sigma \Sigma d^{\prime 2}=61\end{array}\begin{array}{c}\Sigma dx^{\prime} dy y^{\prime}=-22\end{array}
Here, $n =5, \Sigma dx ^{\prime}=0, \Sigma dx ^{\prime 2}=10, \Sigma dy ^{\prime}=-3, \Sigma dx ^{\prime} dy ^{\prime}=-22 \Sigma dy ^{\prime 2}=61$
Now,Putting the values in the given formula:
$\begin{array}{l}r=\frac{\Sigma d x^{\prime} d y^{\prime}-\frac{\Sigma d x^{\prime} \times \Sigma d y^{\prime}}{n}}{\sqrt{\Sigma d x^{\prime 2}-\frac{\left(\Sigma d x^{\prime}\right)^2}{n}} \times \sqrt{\Sigma d d y^2-\frac{\left(\Sigma d y^{\prime}\right)^2}{n}}}= \\ =\frac{-22-\frac{0 \times-3}{5}}{\sqrt{10-\frac{(0)^2}{5}} \times \sqrt{61-\frac{(-3)^2}{5}}} \\ =\frac{-22}{\sqrt{10} \times \sqrt{61-1.8}}=\frac{-22}{\sqrt{10} \times \sqrt{59.2}} \\ =\frac{-22}{3.16 \times 7.69}=\frac{-22}{24.3}=-0.90\end{array}$
- Therefore, Karl Pearson's coefficient of correlation between age and playing habits is $=0.90$.
- Interpretation of $r$
1. It indicates that there is a high degree of negative correlation between age and playing habits.
2. It indicates that as age increases, the tendency to play decreases.
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6 Marks Question - Economics STD 11 Commerce Questions - Vidyadip