Descriptive Statistics Test 21

In the given graph, The straight line represents the growth of the Plant $$B$$.

In the given graph, The straight line represents the growth of the Plant $$B$$.

In the given graph, the straight line represents the actual temperature and the dotted line represents the forecast temperature.

In the given graph, The dotted line represents the growth of the Plant $$A$$.

In the given graph, the straight line represents the actual temperature.

The cumulative frequency distribution is as given below

Marks Groups (class)	No. of Students	Cumulative frequency
5-10	5	5
10-15	6	11
15-20	15	26
20-25	10	36
25-30	5	41
30-35	4	45
35-40	2	47
40-45	1	48
$$N = 48$$

We have $$N = 48 \quad \Rightarrow \displaystyle\frac{3N}{4} = 36$$
The cumulative frequency just greater than $$\displaystyle\frac{3N}{4}$$ is $$41$$. The corresponding class is $$25-30$$. It is the upper quartile class.

$$\therefore \quad l = 25, \space f = 5, \space h = 5, \space F = 36$$

$$\therefore \quad Q_3 = l + \displaystyle\frac{\displaystyle\frac{3N}{4} - F}{f}\times h = 25 + \displaystyle\frac{36-36}{5}\times 5 = 25$$

Arranging the wages in ascending order, we obtain the following table:

S. No	Wages (Rs.)	S. No.	Wages (Rs.)
1	120	16	400
2	170	17	420
3	210	18	440
4	240	19	440
5	260	20	450
6	270	21	470
7	300	22	480
8	320	23	480
9	330	24	490
10	330	25	500
11	330	26	520
12	330	27	550
13	350	28	580
14	370	29	620
15	380	30	680

We have, $$ n = 30$$
Lower Quartile:
$$\quad Q_1 =$$ Value of $$\left(\displaystyle\frac{n+1}{4}\right)^{th}$$ observation
$$\Rightarrow\quad Q_1 =$$ Value of $$\left(\displaystyle\frac{30+1}{4}\right)^{th}$$ observation
$$\Rightarrow\quad Q_1 =$$ Value of $$\left(7.75\right)^{th}$$ observation
$$\Rightarrow\quad

Q_1 =$$ Value of $$\left(7\right)^{th}$$ observation +

$$\displaystyle\frac{3}{4}$$(Value of $$8^{th}$$ observation - Value of

$$7^{th}$$ observation)
$$\Rightarrow \quad Q_1 = 300 + \displaystyle\frac{3}{4}(320 - 300) = 315$$

Middle Quartile (Median):
$$Q_2$$

= A.M. of $$\left(\displaystyle\frac{n}{2}\right)^{th}$$ and

$$\left(\displaystyle\frac{n}{2} + 1\right)^{th}$$ observation
$$\Rightarrow

\quad Q_2 = \displaystyle\frac{\mbox{value of } 15^{th} \mbox{

observation} + \mbox{value of } 16^{th} \mbox{ observation}}{2}$$
$$\Rightarrow \quad Q_2 = \displaystyle\frac{380 + 400}{2} = 390$$

Upper Quartile:
$$\quad Q_3 = $$ Value of $$3\left(\displaystyle\frac{n+1}{4}\right)^{th}$$ observation
$$\Rightarrow\quad Q_3 = $$ Value of $$3\left(\displaystyle\frac{30+1}{4}\right)^{th}$$ observation
$$\Rightarrow\quad Q_3 = $$ Value of $$23.25^{th}$$ observation
$$\Rightarrow\quad

Q_3 = $$ Value of $$23^{rd}$$ observation +

$$\displaystyle\frac{1}{4}($$ Value of $$24^{th}$$ Observation - Value

of $$23^{rd}$$ observation $$)$$
$$\Rightarrow\quad Q_3 = 480 + \displaystyle\frac{1}{4}(490 - 480) = 482.50$$
$$\therefore \quad Q_1+Q_2+Q_3 = 315+390+482.50 = 1187.50$$

Quartile Deviation ($$ { Q }_{ D } $$) means the semi variation between the upper quartile or third quartile ($$ { Q }_{ 3 } $$) and lower quartile or first quartile ($$ { Q }_{ 1 } $$) in a distribution.
Formula = $$ { Q }_{ D }=\cfrac { { Q }_{ 3 }-{ Q }_{ 1 } }{ 2 } $$
Given, $$ { Q }_{ D } $$$$ = 10$$  and  $$ { Q }_{ 3 } $$ $$= 35$$
  $$ 10=\cfrac { 35-{ Q }_{ 1 } }{ 2 }  $$ 
  $$ { Q }_{ 1 } $$ $$= 15$$

The cumulative frequency table is as given below

Age in years	No. of students	Cumulative frequency
20	3	3
30	61	64
40	132	196
50	153	349
60	140	489
70	51	540
81	3	543
N $$= $$543

Lower Quartile: We have, $$\displaystyle\frac{N}{4} = \displaystyle\frac{543}{4} = 135.75$$
Cumulative frequency just greater than $$N/4$$ is $$196$$ and the corresponding value of the variable is $$40$$.
$$\therefore\quad Q_1 = $$ Lower Quartile = $$40$$ years
Middle Quartile (Medium):
We have, $$\displaystyle\frac{N}{2} = \displaystyle\frac{543}{2} = 271.5$$
Cumulative frequency just greater than $$N/2$$ is $$349$$ and the corresponding value of the variable is $$50$$.
$$\therefore\quad Q_2 = $$ Middle quartile = $$50$$ years
Upper Quartile:
We have $$\displaystyle\frac{3N}{4} = \displaystyle\frac{3\times543}{4} = 407.25$$
Cumulative frequency just greater than $$407.25$$ is $$489$$ and the corresponding value of the variable is $$60$$.
$$\therefore\quad Q_3= $$ Upper quartile  $$=60$$ years
$$\therefore\quad Q_1 + Q_3 - Q_2 = 40+60-50 = 50 = Q_2$$

The cumulative frequency of the table is a given below:

Weekly income ( in Rs.)	No. of workers	Cumulative frequency
58	2	2
59	3	5
60	6	11
61	15	26
62	10	36
63	5	41
64	4	45
65	3	48
66	1	49
N = 49

Lower Quartile: 
We have $$\displaystyle\frac{N}{4} = \displaystyle\frac{49}{4} = 12.25$$
The cumulative frequency just greater than $$\displaystyle\frac{N}{4}$$ is $$26$$ and the corresponding value of the variable is $$61$$
$$\therefore\quad Q_1 = Rs.\space 61$$

Middle Quartile (median): 
We have $$\displaystyle\frac{N}{2} = \displaystyle\frac{49}{2} = 24.5$$
The cumulative frequency just greater than $$\displaystyle\frac{N}{2}$$ is $$26$$ and the corresponding value of the variable is $$61$$
$$\therefore\quad Q_2 = Rs.\space 61$$

Upper Quartile:
We have $$\displaystyle\frac{3N}{4} = \displaystyle\frac{3\times49}{4} = 36.75$$
The cumulative frequency just greater than $$\displaystyle\frac{3N}{4}$$ is $$41$$ and the corresponding value of the variable is $$63$$
$$\therefore\quad Q_3 = Rs.\space 63$$

$$\therefore\quad \displaystyle\frac{Q_1+Q_3}{2} = \displaystyle\frac{61+63}{2} = 62$$