Question 1: Use a graph paper for this question. The daily pocket expenses of $200$ students in a school are given below;

 Pocket expenses (in Rs.) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 Number of students (frequency) 10 14 28 42 50 30 14 12

Draw a histogram representing the above distribution and estimate the mode from the graph. [2014]

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Question 2: 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 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 if the pass percentage was 35    [2014]

 Marks No. of Students Cumulative Frequency (c.f) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 8 7 12 17 23 14 9 6 5 4 3 10 22 39 62 76 85 91 96 100

On the graph paper, we plot the following points:

$(10, 3), (20, 10), (30, 22), (40,39), (50, 62), (60,76), (70,85), \\ (80, 91), (90, 96),(100, 100)$

(i) Medium $= (\frac{n}{2})^{th} term = \frac{100}{2}=50^{th} term$

From the graph $50^{th} term=44$

(ii) Lower quartile $= (\frac{n}{4})^{th} term = \frac{100}{4}=25^{th} term$

From the graph $25^{th} term =32$

(iii) The number of students who obtained more than $85\%$ marks in test $= 100-94 = 6$ students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 $= 30$ students.

Question 3:  The marks obtained by 30 students in a class assessment of 5 marks is given below:

 Marks 0 1 2 3 4 5 No. of Students 1 3 6 10 5 5

Calculate the mean, median and mode of the above distribution.    [2015]

Below table

 $x$$x$ $f$$f$ $fx$$fx$ $cf$$cf$ 0 1 0 1 1 3 3 4 2 6 12 10 3 10 30 20 4 5 20 25 5 5 25 30 $\Sigma f=30\ \ \ \ \ \ \ \Sigma fx=90$$\Sigma f=30\ \ \ \ \ \ \ \Sigma fx=90$

Mean $= \frac{\Sigma fx}{\Sigma f}=\frac{90}{30} = 3$

Median $= size \ of \left(\frac{N}{2}\right)th\ {obs}^n$

$= size \ of \left(\frac{30}{2}\right)th\ {obs}^n$

$= Size \ of 15 {}^{th} \ obs{}^{n} = 3$

Mode = 3 marks (as highest frequency is 10)

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Question 4:  Calculate the mean of the following distribution.    [2015]

 Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16

 Class Mid Value $x$$x$ $f$$f$ $fx$$fx$ 0-10 5 8 40 10-20 15 5 75 20-30 25 12 300 30-40 35 35 1225 40-50 45 24 1080 50-60 55 16 880 Total $\Sigma f=100$$\Sigma f=100$ $\Sigma fx=3600$$\Sigma fx=3600$

Mean $= \bar{x} = \frac{\mathcal{E}fx}{ \mathcal{E}f} = \frac{3600}{100}$ $= 36$

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Question 5: The weight of 50 workers is given below:

 Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on  one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight.     [2015]

Below the table

 Weight $f$$f$ $cf$$cf$ 50-60 4 4 60-70 7 11 70-80 11 22 80-90 14 36 90-100 6 42 100-110 5 47 110-120 3 50 Total 50

(i)   Lower quartile= $\frac{N^{th}}{4}{ \ obs}^n$

$= \frac{{50}^{th}}{4}{ \ obs}^n = 12.5th { \ obs}^n = 72 \ kg$

Upper quartile = $\frac{3N}{4}\ th\ {obs}^n = \frac{150}{4}th\ {obs}^n = 37.5th { \ obs}^n = 92 \ kg$

(ii)  No. of over weight workers $= 50-39 = 11$

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Question 6: The number $6,8,10,12,13$ and $x$ are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of $x$    [2014]

Arrange numbers in ascending order are $6,8,10,12,13, x.$

Mean $= \frac{6+8+10+12+13+x}{6}=\frac{49+x}{6}$

No. of terms $(n) = 6 \ (even)$

Median $= \frac{ (\frac{n}{2})^{th} \ term + (\frac{n}{2}+1)^{th} \ term }{2}$

Median $= \frac{ (\frac{6}{2})^{th} \ term + (\frac{6}{2}+1)^{th} \ term }{2} = \frac{3^{rd}+4^{th}}{2} = \frac{10+12}{2} = \frac{22}{2}$ $= 11$

According to given condition

$\frac{49+x}{6}=11 \Rightarrow 49+x=66$

or $x=17$

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Question 8:  Calculate the mean of the distribution given below using the short cut method;     [2014]

 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

Table as follows

 $Marks \\ (C.I)$$Marks \\ (C.I)$ $f$$f$ Mean Value $x$$x$ $A=45.5$$A=45.5$ $d=x-A$$d=x-A$ $f \times d$$f \times d$ 11-20 21-30 31-40 41-50 51-60 61-70 71-80 2 6 10 12 9 7 4 15-5 25-5 35-5 45-5 55-5 65-5 75-5 -30 -20 -10 0 10 20 30 -60 -120 -100 0 90 140 120 $\Sigma f=50$$\Sigma f=50$ $\Sigma fd=70$$\Sigma fd=70$

$Mean = A + \frac{ \Sigma fd}{\Sigma f} = 45.5+\frac{70}{50} = 45.5 + 1.4 = 46.9$

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Question 8: The median of the following observation $11,12,14 (x-2), (x+4), (x+9), 32, 38, 47$ arranged in ascending order is $24$. Find the value of $x$ and hence find the mean.     [2013]

Given observation are $11,12,14, (x-2), (x+4), (x+9), 32, 38, 47$ and mediam $=24$

Since $n=9$ which is odd, therefore

$Median=\ \frac{n+1}{2}th\ term =\frac{9+1}{2}th\ term$

$24=5th\ term$

$x+4=24$

$\Rightarrow x=24-4 \Rightarrow x=20$

Therefore, $11,12,14, (20-2), (20+4), (20+9),32,38,47$

$=11,12,14,18,24,29,32,38,47$

Now, Mean $=\frac{\sum{x}}{n}$

$=\ \frac{11+12+14+18+24+29+32+38+47}{9} =\frac{225}{9}$ $=25$

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Question 9: Draw a histogram from the following frequency distribution and find the made from the graph:   [2013]

 Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5

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Question 10: Find the mean of the following distribution by step deviation method:   [2013]

 Class Internal 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 10 6 8 12 5 9

 C.I Frequency $(f_i)$$(f_i)$ Mid-Value $(x)$$(x)$ $d_i = \frac{x-a}{h}$$d_i = \frac{x-a}{h}$ $(f_i. d_i)$$(f_i. d_i)$ 20-30 30-40 40-50 50-60 60-70 70-80 10 6 8 12 5 9 25 35 45 55 65 75 -2 -1 0 1 2 3 -20 -6 0 12 10 27 $\Sigma f_i = 50$$\Sigma f_i = 50$ $\Sigma f_i.d_i = 23$$\Sigma f_i.d_i = 23$

Here, $a=45$ and $h=10$

Mean $= a + \frac{\Sigma f_i.d_i }{\Sigma f_i} = 45 + \frac{23}{50} . 10 = 49.6$

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Question 11: The mark obtained by 120 students in a 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 Students 5 9 16 22 26 18 11 6 4 3

Draw an ogive for the given distribution on a graph sheet; Using suitable scale for ogive to estimate the following:

(i) The mediam

(ii) The number of students who obtained more than 75% marks in the test.

(iii) The number of students who did not pass the test if minimum marks required to pass is 40.   [2013]

 Marks No. of students Cumulative Frequency 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 5 9 16 22 26 18 11 6 4 3 5 14 30 52 78 96 107 113 117 120 N=120

On the graph paper we plot the following points:

$(10,5), (20,14), (30,30), (40,52), (50,78), (60,96), (70,107), (80,113), (90,117), (100,120)$

(i) Mediam $= (\frac{n}{2})^{th} \ term = \frac{120}{2} =60^{th} \ term$

From the graph 60th term $=42$

(ii) The number of students who obtained more than $75\%$ marks in test $= 120-110 = 10$

(iii) The number of students who did not pass the test if the minimum pass marks $40=52$

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Question 12: 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 No. of Students 6 A 16 13 B

If the mean of the distribution is $7.2$ find $a \ and \ b$.    [2012]

Given, the total number of students $= 40$

Therefore $6 + a+ 16+ 13+ b = 40$

$\Rightarrow a + b = 5$ … … … (i)

Given mean $(\overline{x}) = \frac {\Sigma fx}{\Sigma f}$ $= 7.2$

Therefore $7.2 =$ $\frac{5 \times 6 + 6 \times a +7 \times 16 +8 \times 13 +9 \times b}{40}$

$246 + 6a + 9 b = 288$

$6a + 9 b = 42$

$2a + 3b = 14$… … … (ii)

Solving (i) and (ii) we get $a = 1$ and $b = 4$

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Question 13: The following distribution represents the height of 160 students of a school

 Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 Students 12 20 30 38 24 16 12 8

Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm=20 students on the other axis. Using the graph, determine:

(i) The medium height

(ii) The interquartile range

(iii) The number of students whose height is above 172 cm.     [2012]

Following table:

 Height F c.f. 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 12 20 30 38 24 16 12 8 12 32 62 100 124 140 152 160

(i) Mean $= 157.3$

(ii) Interquartile range $= Q_3 - Q_1 = 163.5 - 152 = 11.5$

(iii) No. of students above $172 \ cm = 160-144 = 16$

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Question 14: Find the mode and medium of the following frequency distribution:    [2012]

 X 10 11 12 13 14 15 f 1 4 7 5 9 3

Mode is the value of the highest frequency.

Therefore Mode $= 14$

For Median, first write the data in ascending order as follows:

$10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, \\ 13, 13, 13, 13,13, 14, 14, 14,14, 14, 14, 14, 14, 14,15, 15, 15$.

Since Median is the middle most value.

Median $= (\frac{N+1}{2})^{th} observation = (\frac{29+1}{2})^{th} observation = 15^{th} \ observation = 13$

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Question 15: A mathematics aptitude test of 50 students was recorded as follows:

 Marks 50-60 60-70 70-80 80-90 90-100 No. of students 4 8 14 19 5

Draw a histogram for the above data using a graph paper and locate the mode.   [2011]

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Question 16: Using step division method, calculate the mean marks of the following distribution: State the modal class:     [2011]

 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

 C.I $f$$f$ $x$$x$ $d=x-67.5$$d=x-67.5$ $u$$u$ $f.u.$$f.u.$ 50-55 5 52.5 -15 -3 -15 55-60 20 57.5 -10 -2 -40 60-65 10 62.5 -5 -1 -10 65-70 10 67.5 0 0 0 70-75 9 72.5 5 1 9 75-80 6 77.5 10 2 12 80-85 12 82.5 15 3 36 85-90 8 87.5 20 4 32 $\Sigma f = 80$$\Sigma f = 80$ $\Sigma fu = 24$$\Sigma fu = 24$

$A.M = 67.5$

(i) $\bar{x} = A.M. + \frac{\Sigma fu}{\Sigma f} \times i = 67.5 + \frac{24}{80} \times 5 = 69$

(ii) Modal class is $55-60$ (class with highest freq.)

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Question 17: 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 11 10 20 28 37 40 29 14 6

Draw an ogive for the given distribution taking $2 \ cm=10$ marks on one axis and $2 \ cm=20$ students on the other axis. Using the graph, determine.

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is $40$.

(iii) If scoring $85$ and more marks is considered as grade one, find the number of students who secured grade on in the examination;     [2011]

 Class Interval Frequency Cumulative Frequency 0-10 5 5 10-20 11 16 20-30 10 20 30-40 20 46 40-50 28 74 50-60 37 111 60-70 40 151 70-80 29 180 80-90 14 194 90-100 6 200

(i) $n = 200$ (even)

Median $= (\frac{n}{2})^{th}$ observation $= (\frac{200}{2})^{th}$ observation $= 100^{th}$ observation $= 57$

(ii) Number of student who failed $= 46$

(iii) Number of students who secured grade one  $200 - 1888 = 12$

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Question 18:  The distribution given below shows the marks obtained by $25$ students in an aptitude test. Find the mean, median and mode of the distribution.     [2010]

 Marks obtained 5 6 7 8 9 10 No. of students 3 9 6 4 2 1

 $x_i$$x_i$ $f_i$$f_i$ $x_if_i$$x_if_i$ $cf$$cf$ 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32 22 9 2 18 24 10 1 10 25 $\Sigma = 25$$\Sigma = 25$ $\Sigma x_if_i = 171$$\Sigma x_if_i = 171$

Mean = $\frac{\Sigma x_if_i}{N} = \frac{171}{25}$ $= 6.84$

Since $N = 25$ is odd,

Median = $(\frac{n+1}{2})^2 \ term = 13^{th} \ term = 7$

Mode $= 6$ (maximum frequency)

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Question 19: The mean of the following distribution is 52 and the frequency of class interval 30-40 is $f$. Find $f$     [2010]

 Class Interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 5 3 $f$$f$ 7 2 6 13

 Interval Frequency $(f_i)$$(f_i)$ $x-i$$x-i$ $d_i = x_i - A$$d_i = x_i - A$ $f_id_i$$f_id_i$ 10-20 5 15 -30 -150 20-30 3 25 -20 -60 30-40 F 35 -10 -10f 40-50 7 45 (A) 0 0 50-60 2 55 10 20 60-70 6 65 20 120 70-80 13 75 30 390 $36+f$$36+f$ $\Sigma f_id_i= 320 - 10f$$\Sigma f_id_i= 320 - 10f$

$Mean = A +$ $\frac{\Sigma f_id_i}{N}$

$\Rightarrow 52 = 45 +$ $\frac{320-10f}{36+f}$

$\Rightarrow 7 =$ $\frac{320-10f}{36+f}$

$\Rightarrow 252 + 7f = 320 - 10f$

$\Rightarrow 17f = 68$

$\Rightarrow f = 4$

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Question 20: The monthly income of a group of 320 employees in a company is given below:

 Monthly Income No. of Employees 6000-7000 20 7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5

Draw an ogive of the given. distribution on a graph sheet taking $2 \ cm = Rs. \ 1000$ on one axis and $2 \ cm = 50 \ employees$ on the other axis. From the graph determine:

(i) the median wage

(ii) the number of employees whose income is below $Rs. \ 8500$

(iii) If the salary of a senior employee is above $Rs. \ 11500$ find the number of senior employees in the company.

(iv) The upper quartile     [2010]

 Monthly Income No. of Employees Cumulative Frequency 6000-7000 20 20 7000-8000 45 65 8000-9000 65 130 9000-10000 95 225 10000-11000 60 285 11000-12000 30 315 12000-13000 5 320

Here n (no. of employees) $= 320$ (even)

(i) Median $= \frac{1}{2} \{ \frac{n}{2} + (\frac{n}{2} + 1) \} = \frac{1}{2} (160+161) = 160.5$

Required median $= Rs. \ 9800$ (from graph)

(ii) Number of employees whose income is below $Rs. \ 8500 = 95$ approx

(iii) Number of senior employees in the company $= 320 - 305 = 15$

(iv) Upper Quartile $= (\frac{3n}{4})^{th} = \frac{3 \times 320}{4} = 240^{th}$term

Upper Quartile $= Rs. \ 10200$

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