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# Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18A

### Solution 3

We form the table as below:

 Class Frequency fi Class mark xi (fi × xi) 10-30 15 20 300 30-50 18 40 720 50-70 25 60 1500 70-90 10 80 800 90-110 2 100 200 𝛴fi = 70 𝛴(fi × xi) = 3520

Therefore, mean

### Solution 4

We form the table as below:

 Class Frequency fi Class mark xi (fi × xi) 0-20 6 10 60 20-40 8 30 240 40-60 10 50 500 60-80 12 70 840 80-100 6 90 540 100-120 5 110 550 120-140 3 130 390 𝛴fi = 50 𝛴(fi × xi) = 3120

Therefore, mean

### Solution 5

We form the table as below:

 Class Frequency fi Class mark xi (fi × xi) 0-6 10 3 30 6-12 11 9 99 12-18 7 15 105 18-24 4 21 84 24-30 4 27 108 30-36 3 33 99 36-42 1 39 39 𝛴fi = 40 𝛴(fi × xi) = 564

Therefore, mean number of days

### Solution 6

We form the table as below:

 Class Frequency fi Class mark xi (fi × xi) 500-700 6 600 3600 700-900 8 800 6400 900-1100 10 1000 10000 1100-1300 9 1200 10800 1300-1500 7 1400 9800 𝛴fi = 40 𝛴(fi × xi) = 40600

Therefore, mean daily expenses

### Solution 7

We form the table as below:

 Class Frequency fi Class mark xi (fi × xi) 64-68 6 66 396 68-72 8 70 560 72-76 10 74 740 76-80 12 78 936 80-84 3 82 246 84-88 1 86 86 𝛴fi = 40 𝛴(fi × xi) = 2964

Therefore, mean heartbeats per minute

### Solution 8

Here, class size, h = 10.

Let the assumed mean be A = 35.

For calculating the mean age, we prepare the following table:

 Class Frequency fi Class mark xi (fi × ui) 0-10 105 5 -3 -315 10-20 222 15 -2 -444 20-30 220 25 -1 -220 30-40 138 35 0 0 40-50 102 45 1 102 50-60 113 55 2 226 60-70 100 65 3 300 𝛴fi = 1000 𝛴(fi × ui) = -351

Therefore,

Thus, the mean age of persons visiting the marketing centre on that day is 31.5 years.

### Solution 9

 Class Frequency fi Class mark xi (fi × xi) 0-20 12 10 120 20-40 15 30 450 40-60 32 50 1600 60-80 x 70 70x 80-100 13 90 1170 𝛴fi = 72 + x 𝛴(fi × xi) = 3340 + 70x

Arithmetic mean

### Solution 10

 Class Frequency fi Class mark xi (fi × xi) 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 𝛴fi = 40 + f 𝛴(fi × xi) = 704 + 20f

Arithmetic mean

### Solution 14

We have

 Class Frequency Mid Value 0 - 20 7 10 70 20 - 40 30 30 40 - 60 12 50 600 60 - 80 =18 - 70 1260 - 70 80 - 100 8 90 720 100 - 120 5 110 550 = 50

### Solution 15

 Class Frequency fi Class mark xi (fi × xi) 0-80 20 40 800 80-160 25 120 3000 160-240 x 200 200x 240-320 y 280 280y 320-400 10 360 3600 𝛴fi = 55 + x + y 𝛴(fi × xi) = 7400 + 200x + 280y

Now,

Also, arithmetic mean = 188

### Solution 17

We have, Let A = 25 be the assumed mean.

 Marks Frequency Mid value Deviation 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 12 18 27 20 17 6 5 15 25 = A 35 45 55 -20 -10 0 10 20 30 -240 -180 0 200 340 180 = 100 = 300

Hence, mean marks per student = 28

### Solution 18

Let the assumed mean be 150, h = 20

 Marks Frequency Mid value Deviation di = - 150 di 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 10 20 30 15 5 110 130 150=A 170 190 -40 -20 0 20 40 -400 -400 0 300 200 = 80 di=-300

Hence, Mean = 146.25

### Solution 19

Let A = 50 be the assumed mean, we have

 Marks Frequency Mid value Deviation 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 20 35 52 44 38 31 10 30 50 = A 70 90 110 -40 -20 0 20 40 60 -800 -700 0 880 1520 1860 = 220

### Solution 25

Let h = 20 and assume mean = 550, we prepare the table given below:

 Age Frequency Mid value 500 - 520 520 - 540 540 - 560 560 - 580 580 - 600 600 - 620 14 9 5 4 3 5 510 530 550 = A 570 590 610 -2 -1 0 1 2 3 -27 -9 0 4 6 15 = 40

Thus, A = 550, h = 20, and = 40,

Hence, the mean of the frequency distribution is 544.

### Solution 26

The given series is an inclusive series, making it an exclusive series, we have

 Class Frequency Mid value 24.5 - 29.5 29.5 - 34.5 34.5 - 39.5 39.5 - 44.5 44.5 - 49.5 49.5 - 54.5 54.5 - 59.5 4 14 22 16 6 5 3 27 32 37 42 = A 47 52 57 -3 -2 -1 0 1 2 3 -12 -28 -22 0 6 10 9 = 70

Thus, A = 42, h = 5, = 70 and

Hence, Mean = 39.36 years

### Solution 27

The given series is an inclusive series making it an exclusive series,we get

 class Frequency Mid value 4.5 - 14.5 14.5 - 24.5 24.5 - 34.5 34.5 - 44.5 44.5 - 54.5 54.5 - 64.5 6 11 21 23 14 5 9.5 19.5 29.5=A 39.5 49.5 59.5 -2 -1 0 1 2 3 -12 -11 0 23 28 15 = 80

Thus, A = 29.5, h = 10, = 80 and

Hence, Mean = 34.87 years

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18B

### Solution 2

We prepare the frequency table, given below

 Marks No. of students C.F. 0 - 7 7 - 14 14 - 21 21 - 28 28 - 35 35 - 42 42 - 49 3 4 7 11 0 16 9 3 7 14 25 25 41 50 N = = 50

Now,

The cumulative frequency is 25 and corresponding class is 21 - 28.

Thus, the median class is 21 - 28

l = 21, h = 7, f = 11, c = C.F. preceding class 21 - 28 is 14 and  = 25

Hence the median is 28.

### Solution 3

We prepare the frequency table given below:

 Daily wages Frequency C.F. 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 40 32 48 22 8 40 72 120 142 150 N = = 150

Now, N = 150, therefore

The cumulative frequency just greater than 75 is 120 and corresponding class is 200 - 300.

Thus, the median class is 200 - 300

l = 200, h = 100, f = 48

c = C.F. preceding median class = 72 and

Hence the median of daily wages is Rs. 206.25.

Hence the median is 28.

### Solution 4

We prepare the frequency table, given below:

 Class Frequency C.F 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35-40 40 - 45 5 6 15 10 5 4 2 2 5 11 26 36 41 45 47 49 = 49

Now, N = 49

The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 - 20.

Thus, the median class is 15 - 20

l = 15, h = 5, f = 15

c = CF preceding median class = 11 and

Median of frequency distribution is 19.5

### Solution 5

We prepare the cumulative frequency table as given below:

 Consumption Frequency C.F 65 - 85 85 - 105 105 - 125 125 - 145 145 - 165 165 - 185 185 - 205 4 5 13 20 14 7 4 4 9 22 42 56 63 67 N = = 67

Now, N = 67

The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 - 145.

Thus, the median class is 125 - 145

l = 125, h = 20, and c = CF preceding the median class = 22,  = 33.5

Hence median of electricity consumed is 136.5

### Solution 6

Frequency table is given below:

 Height Frequency C.F 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 6 10 18 22 20 15 6 3 6 16 34 56 76 91 97 100 N = =100

N = 100,

The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 - 155

Thus, the median class is 150 - 155

l = 150, h = 5, f = 22, c = C.F.preceding median class = 34

Hence, Median = 153.64

### Solution 7

The frequency table is given below. Let the missing frequency be x.

 Class Frequency C.F 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 5 25 x 18 7 5 30 30 + x 48 + x 55 + x

Median = 24 Median class is 20 - 30

l = 20, h = 10, f = x, c = C.F. preceding median class = 30

Hence, the missing frequency is 25.

### Solution 10

Let be the frequencies of class intervals 0 - 10 and 40 - 50

Median is 32.5 which lies in 30 - 40, so the median class is 30 - 40

l = 30, h = 10, f = 12, N = 40 and

### Solution 11

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

 Class Frequency C.F 18.5 - 25.5 25.5 - 32.5 32.5 - 39.5 39.5 - 46.5 46.5 - 53.5 53.5 - 60.5 35 96 68 102 35 4 35 131 199 301 336 340 fi = N = 340

The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 - 39.5.

Median class is 32.5 - 39.5

l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131

Hence median is 36.5 years

### Solution 12

Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get

 Wages per day (in Rs) Frequency C.F 60.5 - 70.5 70.5 - 80.5 80.5 - 90.5 90.5 - 100.5 100.5 - 110.5 110.5 - 120.5 5 15 20 30 20 8 5 20 40 70 90 98 fi = N =98

The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 - 100.5.

median class is 90.5 - 100.5

l = 90.5, h = 10, f = 30, c = CF preceding median class = 40

Hence, Median = Rs 93.50

### Solution 13

The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get

 Marks Frequency C.F 0.5 - 5.5 5.5 - 10.5 10.5 - 15.5 15.5 - 20.5 20.5 - 25.5 25.5 - 30.5 30.5 - 35.5 35.5 - 40.5 40.5 - 45.5 7 10 16 32 24 16 11 5 2 7 17 33 65 89 105 116 121 123 fi = N =123

The cumulative frequency just greater than 61.5 is 65.

The corresponding median class is 15.5 - 20.5.

Then the median class is 15.5 - 20.5

l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33

Hence, Median = 19.95

### Solution 14

 Marks Frequency C.F 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 12 20 25 23 12 24 48 36 12 32 57 80 92 116 164 200 N =

The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 - 60.

Thus the median class is 50 - 60

l = 50, h = 10, f = 24, c = C.F. preceding median class = 92,  = 100

Hence, Median = 53.33

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18C

### Solution 4

As the class 26 - 30 has maximum frequency so it is modal class

Hence, mode = 28.5

### Solution 5

As the class 1500 - 2000 has maximum frequency, so it os modal class

Hence the average expenditure done by maximum number of workers = Rs. 1820

### Solution 6

As the class 5000 - 10000 has maximum frequency, so it is modal class

Hence, mode = Rs. 7727.27

### Solution 7

As the class 15 - 20 has maximum frequency so it is modal class.

Hence mode = 17.3 years

### Solution 8

As the class 85 - 95 has the maximum frequency it is modal class

Hence, mode = 85.71

### Solution 9

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

 Class Frequency 0.5 - 5.5 5.5- 10.5 10.5 - 15. 5 15.5 - 20.5 20.5 - 25. 5 25.5 - 30.5 30.5 - 35.5 35.5 - 40.5 40.5 - 45.5 45.5 - 50.5 3 8 13 18 28 20 13 8 6 3

As the class 20.5 - 25.5 has maximum frequency, so it is modal class

Hence, mode = 23.28

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18D

### Solution 1

Mean:

 Class Frequency fi Cumulative frequency Class mark xi (fi × xi) 0-10 4 4 5 20 10-20 4 8 15 60 20-30 7 15 25 175 30-40 10 25 35 350 40-50 12 37 45 540 50-60 8 45 55 440 60-70 5 50 65 325 𝛴fi = 50 𝛴(fi × xi) = 1910

Therefore, mean

Median:

Cumulative frequency just greater than 25 is 37 and the corresponding class is 40-50.

Thus, the median class = 30-40

Mode:

Mode = 3(Median) - 2(Mean)

= 3(40) - 2(38.2)

= 120 - 76.4

= 43.6

### Solution 5

Let the assumed mean A be 145.Class interval h = 10.

 Class Frequency Mid-Value C.F. 120-130 130-140 140-150 150-160 160-170 2 8 12 20 8 125 135 145=A 155 165 -2 -1 0 1 2 -4 -8 0 20 16 2 10 22 42 50 N = 50

(i)Mean

(ii)N = 50,

Cumulative frequency just after 25 is 42

Corresponding median class is 150 - 160

Cumulative frequency before median class, c = 22

Median class frequency f = 20

(iii)Mode = 3 median - 2 mean

= 3 151.5 - 2 149.8 = 454.5 - 299.6

= 154.9

Thus, Mean = 149.8, Median = 151.5, Mode = 154.9

### Solution 6

 Class Frequency Mid-value C.F. 100-120 120-140 140-160 160-180 180-200 12 14 8 6 10 110 130 150= A 170 190 -2 -1 0 1 2 -24 -14 0 6 20 12 26 34 40 50 N = 50

Let assumed mean A = 150 and h = 20

(i)Mean

(ii)

Cumulative frequency just after 25 is 26

Corresponding frequency median class is 120 - 140

So, l = 120, f = 14, h = 20, c = 12

(iii)Mode = 3Median - 2Mode

= 3(138.6) - 2(145.2)

= 415.8 - 190.4

= 125.4

Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4

### Solution 7

 Class Frequency Mid-value C.F. 100-150 150-200 200-250 250-300 300-350 6 7 12 3 2 125 175 225 275 325 -2 -1 0 1 2 -12 -7 0 3 4 6 13 25 28 30 N = 30

Let assumed mean = 225 and h = 50

(i)Mean =

(ii)

Cumulative frequency just after 15 is 25

corresponding class interval is 200 - 250

Median class is 200 - 250

Cumulative frequency c just before this class = 13

Hence, Mean = 205 and Median = 208.33

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18E

### Solution 1

We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the 'less than type' ogive as follows:

At y = 26.5, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 68 units

Hence, median marks = 68

### Solution 2

 Number of wickets Less than 15 Less than 30 Less than 45 Less than 60 Less than 75 Less than 90 Less than 105 Less than 120 Number of bowlers 2 5 9 17 39 54 70 80

We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the 'less than type' ogive as follows:

At y = 40, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 78 units

Hence, median number of wickets = 78

### Solution 3

'More than type' distribution is as follows:

We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the 'more than type' ogive as follows:

### Solution 4

'More than type' distribution is as follows:

We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the 'more than type' ogive as follows:

### Solution 5

'More than type' distribution is as follows:

We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the 'more than type' ogive as follows:

### Solution 6

We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the 'more than type' ogive as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 70.5 units

Hence, median production yield = 70.5 kg/ha

### Solution 7

'More than type' distribution is as follows:

 Production yield Number of farms More than 65 24 More than 60 54 More than 55 74 More than 50 90 More than 45 96 More than 40 100

On a graph paper, we plot the points A(40, 100), B(45, 96), C(50, 90), D(55, 74), E(60, 54) and F(65, 24).

Join these points to get a 'More than Ogive'.

### Solution 8

More than series

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)

Hence,

Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M

Then,OM = 590

Hence median = 590

### Solution 9

(i) Less than series:

 Marks No. of students Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 2 7 13 21 31 56 76 94 98 100

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)

Join these points free hand to get the curve representing "less than" cumulative curve.

(ii)From the given table we may prepare the 'more than' series as shown below

 Marks No. of students More than 45 More than 40 More than 35 More than 30 More than 25 More than 20 More than 15 More than 10 More than 5 More than 0 2 6 24 44 69 79 87 93 98 100

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)

Join these points free hand to get the required curve.

Here

Two curves intersect at point P(28, 50)

Hence, the median = 28

### Solution 10

We may prepare less than series and more than series

(i)Less than series

 Height in (cm) Frequency Less than 140 Less than 144 Less than 148 Less than 152 Less than 156 Less than 160 Less than 164 Less than 168 Less than 172 Less than 176 Less than 180 0 3 12 36 67 109 173 248 330 416 450

Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)

(ii)More than series

 Height in cm C.F. More than 140 More than 144 More than 148 More than 152 More than 156 More than 160 More than 164 More than 168 More than 172 More than 176 More than 180 450 447 438 414 383 341 277 202 120 34 0

Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)

The curves intersect at (167, 225).

Hence, 167 is the median.

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18F

### Solution 2

Class having maximum frequency is the modal class.

Here, maximum frequency = 27

Hence, the modal class is 40 - 50.

Thus, the lower limit of the modal class is 40.

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise MCQ

### Solution 1

Correct option: (d)

Range is not a measure of central tendency.

### Solution 2

Correct option: (a)

Mean cannot be determined graphically.

### Solution 3

Correct option: (a)

Since mean is the average of all observations, it is influenced by extreme values.

### Solution 4

Correct option: (c)

Mode can be obtained graphically from a histogram.

### Solution 5

Correct option: (d)

Ogives are used to determine the median of a frequency distribution.

### Solution 6

Correct option: (b)

The cumulative frequency table is useful in determining the median.

### Solution 7

Correct option: (b)

Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.

### Solution 10

Correct option: (c)

di's are the deviations from A of midpoints of the classes.

### Solution 11

Correct option: (b)

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

### Solution 12

Correct option: (b)

Mode = (3 x median) - (2 x mean)

### Solution 13

Correct option: (c)

Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5

### Solution 15

Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 - 40.

### Solution 18

Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median - 2Mean

= 3 x 9 - 2 x 8.9

= 27 - 17.8

= 9.2

### Solution 23

Correct option: (c)

For a symmetrical distribution, we have

Mean = mode = median

### Solution 24

Correct option: (c)

Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having income range 20000 to 25000 = 50 - 37 = 13

### Solution 26

Correct option: (d)

Mean of 20 numbers = 0

Hence, sum of 20 numbers = 0 x 20 = 0

Now, the mean can be zero if

sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),

sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),

…….

sum of 19 numbers is (S) and the 20th number is (-S), then their sum is zero.

So, at the most, 19 numbers can be greater than zero.

### Solution 29

(a) - (s)

The most frequent value in a data is known as mode.

(b) - (r)

Mean cannot be determined graphically.

(c) - (q)

An ogive is used to determine median.

(d) - (p)

Standard deviation is not a measure of central tendency.

## Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Test Yourself

### Solution 1

Correct option: (b)

The cumulative frequency table is useful in determining the median.

### Solution 2

Correct option: (c)

Mean = 27

Median = 33

Mode = 3Median - 2Mean

= 3 x 33 - 2 x 27

= 99 - 54

= 45

### Solution 6

Number of athletes who completed the race in less than 14.6 seconds

= 2 + 4 + 15 + 54

= 75

### Solution 8

The frequency table is as follows:

 Classes Profit (in lakhs Rs.) Frequency Number of shops 5 - 10 2 10 - 15 12 15 - 20 2 20 - 25 4 25 - 30 3 30 - 35 4 35 - 40 3

The frequency corresponding to the class 20 - 25 is 4.

### Solution 13

 Marks Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Number of students 3 11 28 48 70

We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:

### Solution 15

 Marks obtained Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 Number of students 2 7 17 40 60 82 85 90 100

We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

Through P, a vertical line is drawn which meets OX at M.

OM = 56.

Hence, median = 56

Mode :

### Solution 20

Less Than Series:

 Class interval Frequency Less than 10 2 Less than 15 14 Less than 20 16 Less than 25 20 Less than 30 23 Less than 35 27 Less than 40 30

We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.

More Than Series:

 Class interval Frequency More than 5 30 More than 10 28 More than 15 16 More than 20 14 More than 25 10 More than 30 7 More than 35 3

We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.

The two curves intersect at L. Draw LM OX.

Thus, median = OM = 16

### Solution 21

Less Than Series:

 Class interval Frequency Less than 45 1 Less than 50 10 Less than 55 25 Less than 60 43 Less than 65 83 Less than 70 109 Less than 75 125 Less than 80 139 Less than 85 149

We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.

More Than Series:

 Class interval Frequency More than 40 149 More than 45 148 More than 50 139 More than 55 124 More than 60 106 More than 65 66 More than 70 40 More than 75 24 More than 80 10

We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.

Mode :