[5 marks sum]

Question 15 Marks

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:

Marks(more than)	90	80	70	60	50	40	30	20	10	0
No. of students	6	13	22	34	48	60	70	78	80	80

Answer

Given data is cumulative data , so draw the ogive as it is .

Marks (more than)	No. of students (f)
0	80
10	80
20	78
30	70
40	60
50	48
60	34
70	22
80	13
90	6

Take a graph paper and draw both the axes.
On the x-axis, take a scale of $1 cm =10$ to represent the marks (more than).
On the $y$ - axis, take a scale of $1 cm =10$ to represent the no. of students.
Now, plot the points $(0,80),(10,80),(20,78),(30,70)$, $(40,60),(50,48),(60,34),(70,22),(80,13),(90,6)$
Join them by a smooth curve to get the ogive.

No. of terms $=80$
$ \therefore \text { Median }=\frac{40+41}{2}=40.5^{\text {th }} \text { term } $
Through mark of 40.5 on y-axis draw a line parallel to $x$-axis which meets the curve at $A$. From $A$, draw a perpendicular to $x$-axis which meets it at $B$.
The value of $B$ is the median which is 55 .
Lower Quartile $\left(Q_1\right)=\frac{n}{4}=\frac{80}{4}=20^{\text {th }}$ term
Through mark of 20 on $y$-axis draw a line parallel to xaxis which meets the curve at P. From P, draw a perpendicular to $x$-axis which meets it at $Q$.
The value of $Q$ is the lower quartile which is 71 .
Upper Quartile $\left(Q_3\right)=\frac{n \times 3}{4}=\frac{80 \times 3}{4}=60^{\text {th }}$ term
Through mark of 60 on $y$-axis draw a line parallel to $x$ axis which meets the curve at $R$. From $R$, draw a perpendicular to $x$-axis which meets it at $S$.
The value of $S$ is the upper Quartile which is 40 .

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Question 25 Marks

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:

Age( in yrs)	Under 10	Under 20	Under 30	Under 40	Under 50	Under 60
No. of males	6	10	25	32	43	50

Answer

Given data is cumulative data , so draw the ogive as it is .

Age (in yrs) Under	No. of males (f)
10	6
20	10
30	25
40	32
50	43
60	50

Take a graph paper and draw both the axes.
On the x-axis, take a scale of $1 cm =10$ to represe the Age (in yrs) under.
On the y-axis, take a scale of $1 cm =10$ to represe the no. of males.
Now , plot the points $(10,6),(20,10),(30,25),(40,3$ $(50,43),(60,5)$.
Join them by a smooth curve to get the ogive.

$ \text { No. of terms }=50$
$\therefore \text { Median }=\frac{25+26}{2}=25.5^{\text {th }} \text { term } $
Through mark of 25.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $A$. From $A$, draw a perpendicular to $x$-axis which meets it at $B$.
The value of $B$ is the median which is 30 .
Lower Quartile $\left( Q _1\right)=\frac{n}{4}=\frac{50}{4}=12.5^{\text {th }}$ term
Through mark of 12.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $P$. From a perpendicular to $x$-axis which meets it at $Q$.
The value of $Q$ is the lower quartile which is 22 .
Upper Quartile $\left( Q _3\right)=\frac{n \times 3}{4}=\frac{50 \times 3}{4}=37.5^{\text {th }}$ term
Through mark of 37.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $R$. From $R$, draw a perpendicular to $x$-axis which meets it at $S$.
The value of $S$ is the upper quartile which is 44 .

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Question 35 Marks

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:

Marks (less than)	10	20	30	40	50	60	70	80
No. of students	5	15	30	54	72	86	94	100

Answer

Given data is a less than cumulative data , so draw the ogive as it is .

Marks (less than)	No. of students (f)
10	5
20	15
30	30
40	54
50	72
60	86
70	94
80	100

Take a graph paper and draw both the axes.
On the $x$-axis, take a scale of $1 cm =10$ to represent marks less than.
On the y-axis, take a scale of $1 cm =20$ to represent the number of students.
Now , plot the points $(10,5),(20,15),(30,30),(40,54)$, $(50,72),(60,86),(70,94),(80,100)$.
join them by a smooth curve to get the ogive.

No. of terms $=100$
$ \therefore \text { Median }=\frac{50+51}{2}=50.5^{\text {th }} \text { term } $
Through mark of 50.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $A$. From $A$, draw a perpendicular to $x$-axis which meets is at $B$.
The value of $B$ is the median which is 38 .
Lower Quartile $\left(Q_1\right)=\frac{n}{4}=\frac{100}{4}=25^{\text {th }}$ term
Through mark of 25 on $y$-axis draw a line parallel to $x-$ axis which meets the curve at $P$.From $P$, draw a perpendicular to $x$-axis which meets it at $Q$.
The value of $Q$ is the lower Quartile which is 28 .
Upper Quartile $\left(Q_3\right)=\frac{n \times 3}{4}=\frac{100 \times 3}{4}=75^{\text {th }}$ term
Through mark of 75 on $y$-axis draw a line parallel to $x-$ axis which meets the curve at R. From R, draw a perpendicular to $x$-axis which meets it at $S$.
The value of $S$ is the upper quartile which is 51 .

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Question 45 Marks

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:

Marks	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of boys	10	12	14	12	9	7	6

Answer

Marks	No. of boys (f)	Cumulative Frequency
30-40	10	10
40-50	12	22
50-60	14	36
60-70	12	48
70-80	9	57
80-90	7	64
90-100	6	70

Take a graph paper and draw both the axes.
On the $x$-axis, take a scale of $1 cm =20$ to represent the marks.
On the y-axis, take a scale of $1 cm =10$ to represent the number of boys.
Now , plot the points $(40,10),(50,22),(60,36),(70,48)$, $(80,57),(90,64),(100,70)$
Join them by a smooth curve to get the ogive.

No. of terms $=70$
$ \therefore \text { Median }=\frac{35+36}{2}=35.5^{\text {th }} \text { term } $
Through mark of 35.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $A$. From $A$, draw a perpendicular to $x$-axis which meets is at $B$.
The value of $B$ is the median which is 60 .
Lower Quartile $\left( Q _1\right)=\frac{n}{4}=\frac{70}{4}=17.5^{\text {th }}$ term
Through mark of 17.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $P$. From $P$, draw a perpendicular to $x$-axis which meets it at $Q$.
The value of $Q$ is the lower quartile which is 47.5
Upper Quartile $\left(Q_3\right)=\frac{n \times 3}{4}=\frac{70 \times 3}{4}=52.5^{\text {th }}$ term
Through mark of 52.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $R$,
draw a perpendicular to $x$-axis which meets it at $S$.
The value of $S$ is the upper quartile which is 74.5 .

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Question 55 Marks

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	4	12	21	18	15	7	3

Answer

We first construct the cumulative frequency table of the given distribution.

Class Interval	Frequency (f)	Cumulative Frequency
0-10	4	4
10-20	12	16
20-30	21	37
30-40	18	55
40-50	15	70
50-60	7	77
60-70	3	80

Take a graph paper and draw both the axes.
On the $x$-axis, take a scale of $1 cm =10$ to represent the class intervals.
On the $y$-axis, take a scale of $1 cm =10$ to represents the frequency .
Now , plot the points $(10,4),(20,16),(30,37),(40,55)$, $ (50,70),(60,77),(70,80) $
Join them by a smooth curve to get the ogive.
No. of terms $=n=80$
$\therefore$ Median $=\frac{40+41}{2}=40.5^{\text {th }}$ term

Through mark of 40.5 on $y$-axis draw a line parallel to $x$-axis which meets the curve at $A$. From $A$, draw a perpendicular to $x$-axis which meets it at $B$.
The value of $B$ is the median which is 32 .
Lower Quartile $\left(Q_1\right)=\frac{n}{4}=\frac{80}{4}=20^{\text {th }}$ term
Through mark of 20 on $y$-axis draw a line parallel to $x$ axis which meets the curve at P. From P, draw a perpendicular to $x$-axis which meets it at $Q$.
The value of $Q$ is the lower quartile which is 23 .
Upper Quartile $\left(Q_3\right)=\frac{n \times 3}{4}=\frac{80 \times 3}{4}=60^{\text {th }}$ term
Through mark of 60 on $y$-axis draw a line parallel to $x$ axis which meets the curve at R. From R, draw a perpendicular to -axis which meets it at $S$.
The value of $S$ is the upper quartile which is 43.5 .

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Question 65 Marks

Find the mean of the following frequency distribution by the step deviation method :

$\text{Class}$	$100-110 $	$110-120 $	$120-130 $	$130-140 $	$140-150 $
$\text{Frequency}$	$15$	$18$	$32$	$25$	$10$

Answer

$\text{Class Interval}$	$x_i$	$f_i$	$A = 35.5 d = x - A$	$f_id$
$100-110$	$105$	$15$	$-2$	$-30$
$110-120$	$115$	$18$	$-1$	$-18$
$210-130$	$A=125$	$32$	$ 0$	$0$
$310-140$	$135$	$25$	$1$	$25$
$140-150$	$145$	$10$	$2$	$20$
$\text{Total}$		$100$		$-3$

$ A =125 \text { and } h _{ i }=10$
$\bar{x}=A+ h \times \frac{\Sigma f_i u }{\Sigma f_i}$
$\bar{x}=125+10 \times \frac{-3}{100}$
$\bar{x}=125-0.3$
$\bar{x}=124.70$
$\therefore \text { Mean }=124.70$

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Question 75 Marks

The marks of 200 students in a test is given below :

Marks%	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89
No. of Students	7	11	20	46	57	37	15	7

Draw an ogive and find
(i) the median
(ii) the number of students who scored more than 35% marks

Answer

We construct cumulative frequency table of the given distribution :

Marks	No.of students (f)	Cumulative Frequency
9.5-19.5	7	7
19.5-29.5	11	18
29.5-39.5	20	38
39.5-49.5	46	84
49.5-59.5	57	141
59.5-69.5	37	178
69.5-79.5	15	193
79.5-89.5	7	200

Take a graph paper and draw both the axes.
On the x -axis , take a scale of 1cm=10 to represent the marks.
On the y - axis , take a scale of 1 cm =50 to represent the no. of students .
Now, plot the points (19.5,7) ,(29.5,18) ,(39.5,38) ,( 49.5,84) ,(59.5,141) ,(69.5,178) ,(79.5,193) ,(89.5,200).
Join them by a smooth curve to get the ogive.

(i) No. of terms = 200 $\therefore$ Median $=\frac{100+101}{2}=100.5^{\text {th }}$ term Through mark of 100.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B. The value of B is the median which is 52. (ii) From marks % = 35 draw a line parallel to y-axis and meet the curve at R. From R, Draw a perpendicular on y-axis which meets it at S. The difference of the value obtained when subtracted from 200 gives the number of students who scored more than 35%. ⇒ 200 - 23 = 172 172 students scored more than 35 %

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Mathematics [5 marks sum]

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