[5 marks sum]

Question 15 Marks

Use graph paper for this question.
The table given below shows the monthly wages of some factory workers.
(i) Using the table, calculate the cumulative frequency of workers.
(ii) Draw the cumulative frequency curve.
Use 2 cm = ₹500, starting the origin at ₹6,500 on X-axis, and 2 cm = 100 worker at they Y-axis.
(iii) Use your graph to write down the median wages in ₹.

Wages in ₹ (CLass)	No. of workers (frequency)	Cumulative frequency f(x)
6500 - 7000	10	-
7000 - 7500	18	-
7500 - 8000	22	-
8000 - 8500	25	-
8500 - 9000	17	-
9000 - 9500	10	-
9500 - 10000	8	-

Answer

(i)

(ii) Plot the points (6500, 0), (7000, 10), (7500, 28), (8000, 75), (9000, 92), (9500, 102), (10000, 110) and join them by a free hand curve.

(iii) Here, N = 110
To find the median we shall construct a horizontal line at cumulative frequency
$=\frac{ N }{2}=\frac{110}{2}=55$,
intersecting the ogive at (8100, 55)
Hence, median wages = ₹8100.

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Question 25 Marks

The mark of 200 students in a test were recorded as follows:

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

Draw the cumulative frequency table.
Draw an ogive and use it to find:
(i) The median
(ii) The number of students who scored more than 35% marks.

Answer

The given frequency distribution is discontinuous, to convert it into continuous distribution.
Adjustment factor $=\frac{20-19}{2}=0.5$.
Cumulative (continuous) frequency tab;e for the given data is :

Take 1 cm along X-axis = 10% marks and 1 cm along Y-axis = 25 students.
Plot the point $(19·5, 7), (29·5 - 18), (39·5 - 38), (49·5 - 141), (69·5 - 178), (9·5 - 193), (89·5 - 200)$ and $(9·5 - 0)$ join these points by a free hand drawing.
The required ogive is drawn in the figure given below:

(i) To find the median: Let A be a point on Y-axis representing frequency
$=\frac{1}{2}\left(\left(\frac{ n ^{\text {th }}}{2} \text { term }\right)+\left(\frac{ n }{2}+1\right)^{\text {th }} \text { term }\right)$
$=\frac{1}{2}(100+101)$
$= 100.5.$
Through A draw a horizontal line to meet the ogive at P. Through P draw a vertical line to meet X-axis at M. the abscissae of point M represents $52\%.$
$\therefore $ The required median $= 52\%.$
(ii) Let the point B on X-axis represent 35% marks. Through B draw a vertical line to meet the ogive at Q. Through Q draw a horizontal line to meet Y-axis at C. The ordinate of the point C represents 28 students on Y-axis.
$\therefore $ The number of students who scored more than 35% marks = total no. of students - no. of students who scored ≤35%
$= 200 - 8$
$= 172.$

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Question 35 Marks

Attempt this question on graph paper. Marks obtained by $200$ students in examination are given below:

Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$	$80 - 90$	$90 - 100$
No. of students	$5$	$10$	$14$	$21$	$25$	$34$	$36$	$27$	$16$	$12$

Draw an ogive for the given distribution taking $2$ cm $= 10$ makrs on one axis and $2\ cm = 20$ students on the other axis.
From the graph find:
(i) the median
(ii) the upper quartile
(iii) number of student scoring above $65$ marks.
(iv) If to students qualify for merit scholarship, find the minimum marks required to qualify.

Answer

(i) Let A be the point on y-axis representing frequency
Here, n (no. of student) $= 200$ (even)
$\text { Median }=\left(\frac{ n }{2}\right)^{\text {th }} \text { term }$
$=\left(\frac{200}{2}\right)^{\text {th }} \text { term }$
$=100^{\text {th }} \text { term }$
From the graph $100$th term $= 57.5$
(ii) Upper quartile = $\frac{3 n}{4}$
$=\frac{3 \times 200^{\text {th }}}{4} \text { term }$
$=\frac{600}{4}=150^{\text {th }} \text { term }$
From graph $150^{th}$ term
The upper quartile $= 72$
(iii) No. of students scoring above $65$ marks
$\Rightarrow$ Total No. of students - No. of students scoring $\leq 65$ marks
$\Rightarrow 200 - 126$
$\Rightarrow 74$ (approx.)
(iv) From the above diagram, we observe the students from $191$ to $200$ qualify for merit scholarship.
$\therefore$ The student who qualifies for merit scholarship scores more than $91$ marks.
$\therefore$ The minimum marks required to qualify for merit scholarship $= 92$ (approx.).

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Question 45 Marks

The marks obtained by $200$ students in an examination are given below :

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
No.of students	$5$	$10$	$11$	$20$	$27$	$38$	$40$	$29$	$14$	$6$

Using a graph paper, draw an Ogive for the above distribution. Use your Ogive to estimate:
(i) the median;
(ii) the lower quartile;
(iii) the number of students who obtained more than $80\%$ marks in the examination and
(iv) the number of students who did not pass, if the pass percentage was $35.$
Use the scale as $2 cm = 10$ marks on one axis and $2 cm = 20$ students on the other axis.

Answer

We construct cumulative frequency table of the given distribution :

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 $1 cm=50$ to represent the no. of students.
Now, plot the points $(10 ,5) ,(20 ,15) ,(30 ,26) ,( 40 ,46) ,(50 ,73) ,(60 ,111), (70 ,151) ,(80 ,180) ,(90 ,194) ,(100 ,200) .$
Join them by a smooth curve to get the ogive.

(i) No. of terms = 200
∴ Median = $\frac{100+101}{2}$ = $100.5^th$ 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 $58.5.$
(ii) Lower Quartile ($Q_1$) = $\frac{n}{4}=\frac{200}{4}=50^{\text {th }}$ term
Through mark of 50 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 41 .
$(iii)$ From marks $=80$ 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 $80 \%$.
$\Rightarrow 200-180=20$
$20$ students scored more than $80 \%$

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Question 55 Marks

The marks obtained by $100$ students in a Mathematics test are given below:

Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$	$80 - 90$	$90 - 100$
No. of Students	$3$	$7$	$12$	$17$	$23$	$14$	$9$	$6$	$5$	$4$

Draw an ogive for the given distribution on a graph sheet.
(Use a scale of $2 cm = 10$ units on both axis).
use the ogive to estimate the :
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was $35.$

Answer

(i) Median $=\left(\frac{ n }{2}\right)^{\text {th }}$ observation
$=\left(\frac{100}{2}\right)^{\text {th }}$ observation
$= 50^{th}$ observation
$= 45.$
(ii) Lower Quartile $\left( Q _1\right)=\left(\frac{ N }{2}\right)^{ th }$ observation
$=\left(\frac{100}{4}\right)^{\text {th }}$ observation
$= 25^{th}$ observation
$= 32.$
(iii) Number of students who obtained more than $85\%$ marks
$= (100 - 94)$
$= 6.$
(iv) Number of students who did not pass if passing $\%$ of marks is $35$
$= 30.$

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Question 65 Marks

The daily wages of $160$ workers in a building project are given below:

Wages in ₹	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
No. of Workers	$12$	$20$	$30$	$38$	$24$	$16$	$12$	$8$

Using a graph paper, draw in Ogive for the above distribution.
Use your Ogive to estimate :
(i) the median wage of the workers.
(ii) the upper quartile wage of the workers
(iii) the lower quartile wages of the workers
(iv) the percentage of workers who earn more than ₹ $45$ a day.

Answer

(i) Median $=\left(\frac{ n }{2}\right)^{\text {th }}$ observation
$=\left(\frac{160}{2}\right)^{\text {th }}$ observation
$= 80^{th}$ observation
Now the $80^{th}$ position in the ogive diagram represent the median wage of workers by the graph $= 35.$
(ii) Upper quartile = $\left(\frac{3 n}{4}\right)^{\text {th }}$ observation
$=\left(\frac{3 \times 160}{4}\right)^{\text {th }} \text { observation }$
$=120^{\text {th }} \text { observation }$
(iii) Lower quartile $=\left(\frac{ n }{4}\right)^{\text {th }}$ observation
$=\left(\frac{160}{4}\right)^{\text {th }}$ observation
$40^{\text {th }}$ observation
The $40^{\text {th }}$ observation in the ogive diagram represents wage of the workers $Q_1=23.5$
(iv) The percentage of workers earmore than ₹45
$=\frac{160-115}{100} \times 100$
$=\frac{45 \times 10}{16}$
$= 28.25\%.$

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Question 75 Marks

Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

Marks	5	6	7	8	9
Number of Students	6	a	16	13	b

If the mean of the distribution is 7.2, find a and b.

Answer

Marks (x)	Number of students (f)	fx
5	6	30
6	a	6a
7	16	112
8	13	104
9	b	9b
	`sumf = 34 + a + b`	`sumfx = 246 + 6a + 9b`

It is given that the number of students is 40.
∴ 35 + a + b = 40
⇒ a + b – 5 = 0 ....(1)
"Mean" = (sum fx)/(sum f)
=> (246 + 6a + 9b)/(35 + a + b) = 7.2`
=> 246 + 6a + 9b = 7.2(35 + a + b)`
=> 246 + 6a + 9b = 252 + 7.2a + 7.2b`
=> 0 = 252 - 246 + 7.2a - 6a + 7.2b - 9b`
=> 6 + 1.2a - 1.8b = 0`
=> 10 + 2a - 3b = 0` .... (2)
Solving equations (1) and (2), we have
5a - 5 = 0
=> a = 1
From (1), we have b = 4
Hence, the values of a and b are 1 and 4 respectively.

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Question 85 Marks

For the following frequency distribution find:
(i) Lower quartile
(ii) Upper quartile
(iii) Inter quartile range
(iv) Semi-inter quartile range.

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$y$	$3$	$5$	$9$	$15$	$20$	$16$	$10$	$2$

Answer

(i) Lower quartile
($Q_1$) = The value of $\left(\frac{ n }{4}\right)^{\text {th }}$ observation
$=$ The value of (804)th observation
$=$ The value of $20^{\text {th }}$ observation $Q _1=4$.
(ii) Upper quartile
$\left(Q_3\right)=$ The value of $(3 n 4)$ th observation
$=$ The value of $(3 \times 804)$ th observation
$=$ The value of $60^{\text {th }}$ observation
$\therefore Q_3 = 6.$
(iii) Inter quartile range
$= Q_3 - Q_1$
$= 6 - 4$
$= 2.$
(iv) Semi-quartile range
$=\frac{Q_3-Q_1}{2}$
$=\frac{2}{2}$
$= 1.s$

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Question 95 Marks

Calculate the mean of the distribution given below using the short cut method.

Marks	$11 - 20$	$21 - 30$	$31 - 40$	$41 - 50$	$51 - 60$	$61 - 70$	$71 - 80$
No. of students	$2$	$6$	$10$	$12$	$9$	$7$	$4$

Answer

$\therefore \text { Mean }=A+\frac{\sum f d}{\sum f}$
$ =45.5+\frac{70}{50}$
$ =45.5+\frac{7}{5} $
$ =\frac{227.5+7}{5}$
$=\frac{234.5}{5} $
$ =46.9$

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Question 105 Marks

Using step-deviation method, calculate the mean marks of the following distribution

Class Interval	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$	$80 - 85$	$85 - 90$
Frequency	$5$	$20$	$10$	$10$	$9$	$6$	$12$	$8$

Answer

$\therefore \text { Mean } X = A +\frac{\sum f \cdot u}{\sum f} \times i \quad .[ i =\text { length of C.I. }] $
$ =67.5+\frac{24}{80} \times 5$
$ =67.5+1.5 $
$ =69 .$

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Question 115 Marks

A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:

Life time (in hrs)	Number of tubes
300 - 399	14
400 - 499	46
500 - 599	58
600 - 699	76
700 - 799	68
800 - 899	62
900 - 999	48
1000 - 1099	22
1100 - 1199	6

Answer

Here, the class-intervals are formed by exclusive method. If we make the series an inclusive one the mid-values remain same. So, there is no need to convert the series.
Let the assumed mean be A = 749·5 and h = 100.

Here,
$N = 400, A = 749·5, h = 100$ and $\sum f_i \mu_i=-138$
$\therefore X = A +\frac{ h }{ N } \sum f_i \mu_i$
$\Rightarrow X =749 \cdot 5+100 \times\left(\frac{-138}{400}\right)$
$=749 \cdot 5-\frac{138}{4}$
$= 749.5 - 34.5$
$= 715.$
Hence, the average life time of a tube is $715$ hours.

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Question 125 Marks

The weights of $50$ apples were recorded as given below. Calculate the mean weight, to the nearest gram. by the Step Deviation Method.

Weights in grams	No. of apples
$80 - 55$	$5$
$85 - 90$	$8$
$90 - 95$	$10$
$95 - 100$	$12$
$100 - 105$	$8$
$105 - 110$	$4$
$110 - 115$	$3$

Answer

$A =97.5, \sum f_i=50, \sum f_i u_i=-16, h =5 . $
$\therefore \text { Mean } \overline{ X }= A +\frac{\sum f_i u_i}{\sum f_i} \times h$
$ =90 \cdot 5+\frac{-16}{50} \times 5$
$= 95.9.$

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Question 135 Marks

Helping the step deviation method find the arithmetic mean of the distribution:

Variable (x)	5	10	15	20	25	30	35	40	45	50
Frequency(f)	20	43	75	67	72	45	39	9	8	6

Answer

Let the assumed Mean be A = 25 and h = 5.

We have,
N = 384, A = 25, h = 5 and $\sum f_i u_i=-214$
Mean $\overline{( X )}= A + h \left(\frac{1}{ N } \sum f_i u_i\right)$
$=25+5 \times\left(\frac{-214}{384}\right)$
= 25 - 2.786
= 22.214.

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Question 145 Marks

The average score of girls in class $X$ examination in school is $67$ and that of boys is $63$. The average score for the whole class is $64.5$. Find the percentage of girls and boys in the class.

Answer

Let the number of boys and girls in the class be $n_1$ and $n_2$ respectively.
We have
$\bar{X}_1$ = Average score of girls $= 67$
$\bar{X}_2$= Average score of boys $= 63$
$\bar{X}$= Average score of the whole class $= 64.5$
$\therefore \overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2}{ n _1+ n _2}$
$\Rightarrow 64 \cdot 5=\frac{67 n _1+63 n _2}{ n _1+ n _2}$
$\Rightarrow 64 \cdot 5 n _1+64 \cdot 5 n _2=67 n _1+63 n _2$
$\Rightarrow 2 \cdot 5 n _1=1 \cdot 5 n _2$
$\Rightarrow 25 n _1=15 n _2$
$\Rightarrow 5 n _1=3 n _2$
Total number of students in the class $= n _1+ n _2$
$\therefore \text { Percentage of girls }=\frac{n_1}{n_1+n_2} \times 100$
$=\frac{n_1}{n_1+\frac{5 n_1}{3}} \times 100, \quad \ldots\left[\because 5 n_1=3 n_2\right]$
$=\frac{3 n_1}{3 n_1+5 n_1} \times 100$
$=\frac{3}{8} \times 100$
$= 37·5\%$
and
Percentage of boys = $\frac{ n _2}{ n _1+ n _2} \times 100$
$=\frac{ n _2}{\frac{3 n _2}{5}+ n _2} \times 100$
$=\frac{5 n _2}{3 n _2+5 n _2} \times 100$
$= 62.5\%$.
Hence, there are $37.5\%$ girls and $62.5\%$ boys in the class.

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Mathematics [5 marks sum]

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