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

We form the table as below:

Therefore, mean  

We form the table as below:

Therefore, mean

We form the table as below:

Therefore, mean number of days

We form the table as below:

Therefore, mean daily expenses

We form the table as below:

Therefore, mean heartbeats per minute

Here, class size, h = 10.

Let the assumed mean be A = 35.

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

Therefore,

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

Arithmetic mean

Arithmetic mean

We have

Now,

Also, arithmetic mean = 188

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

Hence, mean marks per student = 28

Let the assumed mean be 150, h = 20

Hence, Mean = 146.25

Let A = 50 be the assumed mean, we have

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

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

Hence, the mean of the frequency distribution is 544.

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

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

Hence, Mean = 39.36 years

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

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

Hence, Mean = 34.87 years

We prepare the frequency table, given below

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.

We prepare the frequency table given below:

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.

We prepare the frequency table, given below:

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

We prepare the cumulative frequency table as given below:

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

Frequency table is given below:

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

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

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

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

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

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

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

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

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

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

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

Hence, mode = 28.5

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

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

Hence, mode = Rs. 7727.27

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

Hence mode = 17.3 years

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

Hence, mode = 85.71

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

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

Hence, mode = 23.28

Mean:

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

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

(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

     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

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

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

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

'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:

'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:

'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:

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

'More than type' distribution is as follows:

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'.

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

(i) Less than series:

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

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

We may prepare less than series and more than series

(i)Less than series

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

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.

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.

Correct option: (d)

Range is not a measure of central tendency.

Correct option: (a)

Mean cannot be determined graphically.

Correct option: (a)

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

Correct option: (c)

Mode can be obtained graphically from a histogram.

Correct option: (d)

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

Correct option: (b)

The cumulative frequency table is useful in determining the median.

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.

Correct option: (c)

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

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.

Correct option: (b)

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

Correct option: (c)

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

Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 - 40.

Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median - 2Mean

= 3 x 9 - 2 x 8.9

= 27 - 17.8

= 9.2

Correct option: (c)

For a symmetrical distribution, we have

Mean = mode = median

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

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 20^th number is (-S), then their sum is zero.

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

(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.

Correct option: (b)

The cumulative frequency table is useful in determining the median.

Correct option: (c)

Mean = 27

Median = 33

Mode = 3Median - 2Mean

= 3 x 33 - 2 x 27

= 99 - 54

= 45

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

= 2 + 4 + 15 + 54

= 75

The frequency table is as follows:

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

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

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 :

Less Than Series:

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:

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 

Less Than Series:

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:

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 :