6 Marks Question

Question 16 Marks

Calculate arithmetic mean with the help of following data using step deviation method.

Marks (Less than)	10	20	30	40	50	60
Number of Students	3	10	20	25	28	30

Answer

The given data is less than type hence, first of all convert the less than cumulative frequency series into an ordinary series and then calculate the value of arithmetic mean. For the calculation of Arithmetic Mean let A =25

Calculation of Arithmetic Mean using step deviation method

Marks	Frequency (f)	Mid-Value (m) m=(L1+L2)/2	dm=m-A (A=25)	$\begin{array}{l}d^{\prime} m=\frac{d m}{c} \\ ( c = 1 0 ) \end{array}$	fd'm
0-10	3	5	-20	-2	-6
10-20	10-3=7	15	-10	-1	-7
20-30	20-10=10	25	0	0	0
30-40	25-20=5	35	+10	+1	+5
40-50	28-25=3	45	+20	+2	+6
50-60	30-28=2	55	+30	+3	+6
	$\Sigma f=30$				$\Sigma f d^{\prime} m=+4$

Here,
$
\begin{array}{l}
A=25, \Sigma f=30, \Sigma f d^{\prime} m=+4, c=10 \\
\text { Now, } \bar{X}=A+\frac{\Sigma f d^{\prime} m}{\Sigma f} \times c=25+\frac{4}{30} \times 10 \\
=25+1.33=26.33
\end{array}
$
Therefore,arithmetic mean of the given data is 26.33

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Question 26 Marks

Calculate the upper and lower quartiles for the following frequency distribution.

Class Interval	Frequency (f)
13-25	6
25-37	11
37-49	23
49-61	7
61-73	3
Total	50

Answer

Class Interval	Frequency (f)	Cumulative Frequency (cf)
13-25	6	6
25-37	11	17
37-49	23	40
49-61	7	47
61-73	3	50
	$n=\Sigma f=50$

Calculation of Upper and Lower Quartiles

Lower Quartile	Upper Quartile
$\begin{array}{l}\text { Lower Quartile number }\left(q_1\right) \\ =\text { Size of }\left(\frac{n}{4}\right) \text { th item }\end{array}$	$\begin{array}{l}\text { Upper Quartile number }\left(q_3\right) \\ =\text { Size of } 3\left(\frac{n}{4}\right) \text { th item }\end{array}$
$=\left(\frac{50}{4}\right)$ th item $=12.5$ th item cf just greater than 12.5 is 17 and the corresponding class is 25-37.So, $1_1=25$$cf =6, f =11$ and $c =12$ $\begin{array}{l}\therefore Q_1=l_1+\frac{\frac{n}{4}-c f}{f} \times c \\=25+\frac{12.5-6}{11} \times 12=25+\frac{6.5}{11} \\\times 12=25+7.09 \\\Rightarrow Q_1=32.09\end{array}$	$=$ Size of $3\left(\frac{50}{4}\right)$ th item $=37.5$ th item cf just greater than 37.5 is 40 and corresponding class is $37-$ 49. So, $1_1=37$, cf=17 $\begin{array}{l} f=23 \\ c=12 \\ \therefore Q_3=l_1+\frac{\frac{3 m}{4}-c f}{f} \times c \\=37+\frac{37.5-17}{23} \times 12=37 \\+\frac{20.5}{23} \times 12 \\=37+10.70 \Rightarrow Q_3=47.70\end{array}$

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Question 36 Marks

From the data given below, calculate Karl Pearson’s coefficient of correlation between density of population and death rate by step deviation method.

Region	Area(in sq km)	Population	Death
A	200	40000	480
B	150	75000	1200
C	120	72000	80
D	80	20000	280

Answer

Regio n	Density(X)	dx(X A), A = 500	$\begin{array}{c} d x \left(\frac{d x}{c_1}\right), \\ c _1= 5 0 \end{array}$	$dx ^{\prime 2}$	Death Rate(Y)	dy(Y A), A = 16	$\begin{array}{c} d y ^{\prime}\left(\frac{d y}{c_2}\right) \\ , c _2=1\end{array}$	$d y^{\prime 2}$	dx'dy'
A	200	-300	-6	36	12	-4	-4	16	24
B	500	0	0	0	16	0	0	0	0
C	600	100	2	4	15	-1	-1	1	-2
D	250	-250	-5	25	14	-2	-2	4	10
			$\Sigma dx ^{\prime}=-9$	$\begin{array}{c}\Sigma dx x^{\prime 2} =65\end{array}$			$\Sigma dy ^{\prime}=-7$	$\begin{array}{l} \Sigma dy ^{\prime 2} =21\end{array}$	$\begin{array}{l}\Sigma dx ^{\prime} dy ^{\prime} =32\end{array}$

Density is calculated as $\frac{\text { population }}{\text { area }}$
Death Rate is calculated as $\frac{\text { death }}{\text { population }} \times 100$
Here, $\Sigma d x { }^{\prime}=-9, \Sigma dx ^{\prime 2}=65, \Sigma dy ^{\prime}=-7, \Sigma dy ^{\prime 2}=21, \Sigma dx ^{\prime} dy ^{\prime}=32$
Now, $r =\frac{\sum d x^{\prime} d y^{\prime}-\frac{\Sigma d x^{\prime} \times \Sigma d y^{\prime}}{n}}{\sqrt{\sum d x^n-\frac{\left(\Sigma d d^{\prime}\right)^2}{n}} \times \sqrt{\Sigma d y^a-\frac{(\Sigma d)^2}{n}}}$
$=\frac{32-\frac{(-9 \times-7)}{4}}{\sqrt{65-\frac{(-9)^2}{4}} \times \sqrt{21-\frac{(-7)^2}{4}}}$
$\begin{array}{l}
=\frac{32-15.75}{\sqrt{65-20.25} \times \sqrt{21-12.25}} \\
=\frac{16.25}{\sqrt{44.75} \times \sqrt{8.75}}=\frac{16.25}{6.69 \times 2.96}=\frac{16.25}{19.80}=0.82
\end{array}$
- Therefore, Karl Pearson's coefficient of correlation between density of population and death rate is 0.82 .
- Interpretation of r: There is a high degree of positive correlation between density of population and death rate.

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