Statistics

Mean =

= = 7.025

Mean =

= = 25

Given,

Mean = 20.6

= 20.6

= 20.6

530 + 25p = 20.6 (50)

25p = 20.6 (50) – 530

p = = 20

Given,

Mean = 15

= 15

= 15

10p + 445 = 15 (p + 27)

10p + 445 = 15p + 405

15p – 10p = 445 – 405

5p = 40

p = = 8

Given,

Mean = 16.6

= 16.6

= 16.6

24p + 1228 = 1660

24p = 1660 – 1228

24p = 432

p = = 18

Given,

Mean = 12.58

= 12.58

= 12.58

524 + 7p = 12.58 (50)

7p = 629 – 524

p = = 15

Given,

Mean = 7.68

= 7.68

= 7.68

9p + 303 = 7.68 (p + 41)

9p – 7.68p = 314.88 – 303

1.32p = 11.88

p = = 9

Given,

Mean = 20

= 20

= 20

295 + 100p + 5p² = 100p + 300

295 + 5p² = 300

5p² = 300 – 295

5p² – 5 = 0

5 (p² – 1) = 0

p² – 1 = 0

(p + 1) (p – 1) = 0

p = � 1

If p + 1 = 0, p = - 1 (Reject)

Or p – 1 = 0, p = 1

Mean =

= = 17.45

Therefore, mean age is 17.45 years.

Let the number of candidates from school III be P.

Given,

Average score of all schools = 66

= 66

= 66

= 66

10340 + 55P = 66P + 9768

10340 – 9768 = 66P – 55P

P = = 52

Mean number of heads per toss =

= = 2.47

Therefore, mean = 2.47

Given, mean = 50

= 50

= 50

30f1 + 70f2 + 3480 = 50 * 120

30f1 + 70f2 + 3480 = 6000 (i)

Also, ∑y = 120

17 + f1 + 32 + f2 + 19 = 120

f1 + f2 = 52

f1 = 52 – f2

Substituting value of f1 in (i), we get

30 (52 – f2) + 70f2 + 3480 = 6000

40f2 = 960

f2 = 24

Hence, f1 = 52 – 24 = 28

Therefore, f1 = 28 and f2 = 24

Given,

Mean = 14

= 14

= 14

360 + 10k = 336 + 14k

24 = 4k

k = 6

Hence, the value of k is 6

Given,

Mean = 25

= 25

= 25

15k + 390 = 25 (14 + k)

15k + 390 = 350 + 25k

40 = 10k

k = 4

Given,

Mean = 18.75

= 18.75

= 18.75

7p + 460 = 18.75 (32)

7p + 460 = 600

7p = 140

p = 20

Hence, the value of p is 20

Let the assumed mean (A) = 3

Mean number of calls = A +

= 3 + =

= 3.54

Let the assumed mean (A) = 2

Mean number of heads per toss = A +

= 2 + = 2 + 0.47

= 2.47

Let the assumed mean (A) = 4

Average number of branches per plant = A +

= 4 + =

= 3.62 (approx.)

Let the assumed mean (A) = 2

Average number of children per family = A +

= 2 + =

= 2.35 (approx)

Let the assumed mean (A) = 25

Average number of marks = A +

= 25 + =

= = 26.08 (approx)

Let the assumed mean (A) = 3

Mean number of students absent per day = A +

= 3 + =

= 3.53 (approx)

To find : the average number of misprints per page.
Solution :
Use the shortcut method to find the mean of given data.
For that,
Let the assumed mean be (A) = 2,
The deviation of values x_i from assumed mean be d_i = x_i – A.
Now to find the mean:
First multiply the frequencies in column (ii) with the value of deviations in column (iii) as f_id_{i.
Construct the table using above information.
The table is as follows:}

Now add the sum of all entries in column (iii) to obtain
and the sum of all frequencies in the column (ii) to obtain
So,
Average number of misprints per day = A +
where, N = total number of observations

⇒ Mean = 2 +
Mean =

Mean =
Mean = 0.73

Let the assumed mean (A) = 2

Average number of accidents per day workers = A +

= 2 + =

= 0.83

Let the assumed mean (A) = 25

Mean = A +

= 25 + =

= 22.075

Let the assumed mean (A) = 275

We have, A = 275

h = 50

Mean = A + h *

= 275 + 50 *

= 275 – 8.75

= 266.25

Let us find class marks (x_i) for each interval by using the relation.

Class mark (x_i) =

Now we may compute x_i and f_ix_i as follows:

Fro, the table we may observe that,

∑f_i = 20

∑f_ix_i = 162

Mean, X̅ =

= = 8.1

So mean number of plants per house is 8.1

We have used for the direct method values x_i and f_i are very small.

Let the assumed mean (A) = 150

We have, A = 150

h = 20

Mean = A + h *

= 150 + 20 *

= 150 - = 150 – 4.8

= 145.2

We may find class mark of each interval (x_i) by using the relation.

x_i =

Class size of this data = 3

Now taking 75.5 as assumed mean (A), we may calculate d_i, u_i, f_iu_i as following:

Now we may observe from table that ∑f_i = 30, ∑f_iu_i = 4

Mean (X̅) = d_i + * h

= 75.5 + * 3

= 75.5 + 0.4 = 75.9

So mean of heart beat per minute of women are 75.9 beats per minute.

Let A assumed mean be 15.

A = 15, h = 6

Mean = A + h *

= 15 + 6 *

= 15 + 0.45

= 15.45

Let the assumed mean (A) be 100.

A = 100, h = 20

Mean = A + h *

= 100 + 20 *

= 100 + 12.2

= 112.2

Let the assumed mean (A) = 20

We have, A = 20

h = 8

Mean = A + h *

= 20 + 8 *

= 20 + 1.4 = 21.4

Let the assumed mean (A) = 15

We have, A = 15

h = 6

Mean = A + h *

= 15 + 6 *

= 15 + 0.75 = 15.75

Let the assumed mean (A) = 25

We have, A = 25

h = 10

Mean = A + h *

= 25 + 10 *

= 25 +

= 25 + = 26.333

Let the assumed mean (A) = 20

We have, A = 20

h = 8

Mean = A + h *

= 20 + 8 *

= 20 + 1 = 21

Let the assumed (A) = 20

We have, A = 20

h = 8

Mean = A + h *

= 20 + 8 *

= 20 - = 20 – 3.6

= 16.4

Let the assumed mean (A) = 60

We have, A = 60

h = 20

Mean = A + h *

= 60 + 20 *

= 60 + 5.6

= 65.6

Let the assumed mean (A) = 50

We have, A = 50

h = 10

Mean = A + h *

= 50 + 10 *

= 50 – 0.5

= 49.5

Let the assumed mean (A) = 42

We have, A = 42

h = 5

Mean = A + h *

= 42 + 5 *

= 42 -

=

= 36.357

By direct method

Mean =

= = 13.25

By assumed mean method

Let, the assumed mean (A) = 6.5

Mean = A +

= 6.5 +

= 6.5 + 6.75

= 13.25

Let the assumed mean (A) = 1650

We have, A = 1650

h = 100

Mean = A + h *

= 1650 + 100 *

= 1650 +

= =

= 1663.46

From Direct method:

Mean =

= = 25.857

Assumed mean method: let assumed mean (A) = 25

Mean = A +

Mean = A +

= 25 + = 25 + 0.857

= 25.857

Step deviation method: Let the assumed mean (A) = 25

Mean = A + * h

= 25 + * 10

= 25 + 0.857 = 25.857

Given, Sum of frequency = 50

5+ f1 + 10 + f2 + 7 + 8 = 50

f1 + f2 = 50 – 5 – 10 – 7 – 8

f1 + f2 = 20

3f1 + 3f2 = 60 (i) [Multiply by 3]

And mean = 62.8

= 62.8

= 62.8

30f1 + 70f2 = 3140 – 2060

30f1 + 70f2 = 1080

3f1 + 7f2 = 108 (ii) [Divide by 10]

Subtract (i) from (ii), we get

3f1 + 7f2 – 3f1 – 3f2 = 108 – 60

4f2 = 48

f2 = 12

Put value of f2 in (i), we get

3f1 + 3 * 12 = 60

3f1 + 36 = 60

3f1 = 24

f1 = 8

So, f1 = 8 and f2 = 12

Given, Mean = 18

Let missing frequency be V

Mean =

18 =

792 + 18V = 752 + 20V

792 – 752 = 20V – 18V

40 = 2V

V = 20

Given, mean =

27 =

1161 + 27p = 1245 + 15p

27p – 15p = 1245 – 1161

12p = 84

p = 7

We may observe that the class intervals are not continuous. There is a gap between two class intervals so we have to add from lower class limit of each interval.

Class size (h) of this data = 3

Now taking 57 as assumed mean, we can calculate as follows:

Mean = A + * h

= 57 + * 3

= 57 +

= 57 + 0.1875 = 57.1875

= 57.19

Number of mangoes kept in packing box is 57.19

We may calculate class marks (x_i) for each interval by using the relation

x_i =

Class size = 50

Now taking 225 as assumed mean we can calculate as follows:

Mean (x̅) = A + * h

= 225 + * 50

= 225 – 14 = 211

So mean expenditure on food is 211

We may calculate class marks (x_i) for each interval by using the relation

x_i =

Class size = 0.04

Now taking 0.14 as assumed mean (A), we can calculate as follows:

Mean (x̅) = A + * h

= 0.14 + * (0.04)

= 0.14 – 0.04133

= 0.099ppm

SO, mean concentration of SO₂ in the air is 0.099ppm

We may calculate class marks (x_i) for each interval by using the relation

x_i =

Now taking assumed mean (A) = 16

Mean (x̅) = A +

= 16 +

= 16 – 3.62 = 12.38

So, mean number of days is 12.38 days for which students were absent.

We may calculate class marks (x_i) for each interval by using the relation

x_i =

Class size (h) for this data = 10

Now taking 70 as assumed mean (A) we can calculate as follows:

Mean (x̅) = A + * h

= 70 + * 10

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.437

Lives in hours of 15 pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Arrange the above in ascending order:

694, 696, 699, 705, 710, 712, 716, 719, 724, 725, 728, 729, 734, 745

N = 15 (Odd)

Median = Term

= Term

= 16^th Term = 716

We have, N = 420

= = 210

The cumulative frequency just greater than is 275 then 165.5-168.5 is the median class such that,

l = 165.5, f = 142, F = 133 and h = 168.5-165.5 = 3

Median = l + * h

= 165.5 + * 3

= 165.5 + * 3

= 165.5 + 1.63

= 167.13

We have, N = 100

= = 50

The cumulative frequency just greater than is 67 then the median class 94.5-104.5 such that,

l = 94.5, f = 33, F = 34, h = 104.5 – 94.5 = 10

Median = l + * h

= 94.5 + * 10

= 94.5 + * 10

= 94.5 + 4.85 = 99.35

We have, N = 140

= = 70

The cumulative frequency is just greater than 98 then median class is 55-65 such that

l = 55, f = 40, F = 58, h = 65 – 55 = 10

Median = l + * h

= 55 + * 10

= 55 + * 10

= 55 + 3 = 58

Therefore, Median = 58

We have, N = 250

= = 125

The cumulative frequency is just greater than is 127 then median class is 50-60 such that:

l = 50, f = 31, F = 96, h = 60-50 = 10

Median = l +

= 50 + * 10

= 50 +

= = 59.35

Given, Median = 35

The median class = 30-40

l = 30, h = 10, f = 40 and F = 30 + f₁

Median = l +

35 = 30 + * 10

5 =

f₁ = 55 – 20 = 35

Given, Sum of frequencies = 170

= 10 + 20 + f₁ + 40 + f₂ + 25 + 15 = 170

= 10 + 20 + 35 + 40 + f₂ + 25 + 15 = 170

= f₂ = 170 – 145 = 25

Therefore, f₁ = 35 and f₂ = 25

Given, Median = 24

Then median class = 20-30

l = 20, h = 10, f = x and F = 30

Median = l +

24 = 20 + * 10

4x = 275 + 5x – 300

4x – 5x = -25

-x = -25

x = 25

Therefore, missing frequency = 25

Given, N = 200

= 46 + x + y + 25 + 10 + 5 = 200

= x + y = 200 – 46 – 25 – 10 – 5

= x + y = 114 (i)

And Mean = 1.46

= 1.46

= = 1.46

= x + 2y + 140 = 292

= x + 2y = 292 – 140

= x + 2y = 152 (ii)

Subtract (i) from (ii), we get

X + 2y – x – y = 152 – 114

y = 38

Put the value of y in (i), we get

x = 114 – 38 = 76

We have, N = 200

= = 100

The cumulative frequency just more than is 122 so the median is 1

(i)

Given, Median = 46

Then, median class = 40-50

Therefore, l = 40, h = 10, f = 65, F = 42 + x

Median = l +

46 = 40 + * 10

= 73 – x

39 = 73 – x

x = 73 – 39

x = 34

Given, N = 230

= 12 + 30 + 34 + 65 + y + 25 + 18 = 230

= 184 + y = 230

= y = 230 – 184 = 46

(ii)

Mean =

= = 45.87

N = 357

= = 178.5

The cumulative frequency just greater than is 193 then the median class is 19.5-24.5 such that:

l = 19.5, f = 140, F = 53, h = 5

Median = l + * h

= 19.5 + * 5 = 23.98

Nearly half the women were married between the age 15 and 25

Given, Median = 28.5

So, median class is 20-30

l, lower class = 20,
frequency of median class, f = 20,
Cumulative frequency of class preceding the median class, F = 5 + f₁,
height of class, h = 10
N is 60,

28.5 = 20 + * 10

28.5 – 20 = * 10

8.5 =
17 = 25 - f₁

f₁ = 25 – 17

f₁ = 8

Given, sum of frequencies = 60

i.e. 5 + f₁ + 20 + 15 + f₂ + 5 = 60
Put the value of f₁

therefore, 5 + 8 + 20 + 15 + f₂ + 5 = 60

hence, f₂ = 7

Therefore, f₁ = 8 and f₂ = 7

Given, Median = 525

Then median class = 500-600

l = 500, f = 20, F = 36 + f₁, h = 100

Median = l +

525 = 500 + * 100

525 – 500 = * 100

25 = (14 – f₁) 5

5f₁ = 45

f₁ = 9

Given, sum of frequencies = 100

= 2 + 5 + f₁ + 12 + 17 + 20 + f₂ + 9 + 7 + 4 = 100

= 2 + 5 + 9 + 12 + 17 + 20 + f₂ + 9 + 7 + 4 = 100

=85 + f₂ = 100

f₂ = 15

Therefore, f₁ = 9 and f₂ = 15

Given, Median = 32.5

Then median class = 30-40

l = 30, h = 10, f = 12, F = 14 + f₁

Median = l +

32.5 = 30 + * 10

2.5 = * 5

15 = (6 – f₁) 5

3 = 6 – f₁

f₁ = 3

Given, sum of frequencies = 40

= 3 + 5 + 9 + 12 + f₂ + 3 + 2 = 40

= 34 + f₂ = 40

= f₂ = 6

Therefore, f₁ = 3 and f₂ = 6

(i)

We have, N = 100

= = 50

The cumulative frequency just greater than is 65 then median Class 70-90, such that:

l = 70, f = 22, F = 43, h = 20

Median = l + × h

= 70 + × 20

= 70 +

= 70 + 6.36 = 76.36

(ii)

We have, N = 150

= = 7

The cumulative frequency is just more than is 90 then, the Median Class is 110-120 such that:

l = 110, f = 45, F = 45, h = 10

Median = l + × h

= 110 + × 10

= 110 +

= 110 + 6.67

= 116.67

To calculate the median height we need to find the class interval and their corresponding frequencies.

The given distribution being of the less than type 140, 145, 150,….165 give the upper limit of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150,….160-165. Observe that from the given distribution, we find that there are 4 girls with height less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140-145 is 11 – 4 = 7

Similarly, the frequency of 145-150 is 29 – 11 = 19, for 150-155 it is 40 – 29 = 11 and so on so our frequency distribution table with the given cumulative frequency becomes:

Now N = 51

So, = = 25.5

This observation lies in the class 145 – 150

Then, l (lower limit) = 145

f = 11 and h = 5

Median = 145 + * 5

= 145 + 4.03

= 149.03

So, the median height of the girls is 149.03 cm. This means that the height of the about 50% of the girls is less than this height and 50% are taller than this height.

: Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can define class intervals with their respective cumulative frequencies as below:

Now from the table we may observe that N = 100

Cumulative frequency just greater than (N = 50) is 78belonging to interval 35-40.

So, Median Class = 35 – 40

Lower limit (l) = 35

Class size (h) = 5

Frequency (f) = 33 and F = 45

Median = l + * h

= 35 + () * 5

= 35 + * 5

= 35 + 0.76

= 35.76

So, Median age is 35.76 years.

: The given data is not having continuous class intervals. So, we have to add and subtract 0.5 to upper class limit and lower class limit.

Median class = 144.5-153.5

l = 144.5, h = 9, f = 12 and F = 17

Median = l + * h

= 144.5 + ( * 9

= 144.5 +

= 146.75

So, median length is 146.75 mm.

We can find cumulative frequencies with their respective class intervals as below:

Median class = 3000-3500

l = 3000, f = 86, F = 130, h = 500

Median = l + * h

= 3000 +

= 3406.98 hours

So, median life time is 3406.98 hours.

We may find cumulative frequencies with their respective class intervals as below:

Median class = 55-60

l = 55, f = 6, F = 13 and h = 5

Median = l + * h

= 55 + () * 5

= 55 +

= 56.666

So, median weight is 56.67 kg.

Mode is the value which occurs maximum number of times in a data.
(i)

Mode = 5 (Since, its frequency is 5 which is maximum)

(ii)

Mode = 3 (Since, its frequency is 5 which is maximum)

(iii)

Mode = 15 (Since, its frequency is 4 which is maximum)

Model shirt size= 40 (Since, it occurs maximum number of times)

Mode = l +
Where
l = lower limit of the modal class
h=width of the modal class
f₁ = frequency of the class preceding the modal class
f₂ = frequency of the class following the modal class

(i)

Here, the maximum frequency is 28 then the corresponding class 40-50 is the model class

l= 40, h= 50-40= 10, f= 28, f₁=12, f₂ =20

Mode = l +

Mode = 40 +

Mode =40 + 6.67

Mode =46.67

(ii)

Here, the maximum frequency is 75, then the corresponding interval 20- 25 is modal class

l= 20, h=5, f=75, f₁=45, f₂ =35

Mode = l +

Mode = 20 +

Mode =20 + 2.14

Mode =22.14

(iii)

Here, the maximum frequency is 50, then the corresponding interval 35-40 is modal class

l= 35, h=5, f=50, f₁=34, f₂ =42

Mode = l +

Mode = 35 +

Mode =35 + 3.33

Mode =38.33

For Group A:

Here, the maximum frequency is 78, the corresponding class interval 18 -20 is modal class

l=18, h=2, f=78, f₁=50, f₂ =46

Mode = l +

= 18 +

=18+

=18 + 0.93 =18.93 years

For Group B:

Here, the maximum frequency is 89, the corresponding class interval 18 -20 is modal class

l=18, h=2, f=89, f₁=54, f₂ =40

Mode = l +

= 18 +

=18+ =18 + 0.83 =18.33

Hence, the modal age of group A is higher than that of group B.

Here, the maximum frequency is 20, the corresponding class interval 50-60 is modal class

l=50, h=10, f=20, f₁=13, f₂ =5

Mode = l +

Mode = 50 +

Mode =50+ =50 + 3.18

Mode =53.18

Here, the maximum frequency is 142, the corresponding class interval 165.5-168.5 is modal class

l=165.5, h=3, f=142, f₁=118, f₂ =127

Mode = l +

= 165.5 +

=18+ =165.5 +1.85

=167.35 cm

We may compute class marks (x_i) as per the relation

X_i=

Now, let assumed mean (A) = 30

∑f_i=80, ∑f_id_i=430

Mean=

= 30 + = 30+ 5.375

=35.38

It represents that on an average the age of patients admitted was 35.38 years. As we can observe that the maximum class frequency 23 belonging to class interval 35-45.

So, modal class= 35-45

Lower limit (l) of modal class =35

Frequency (f₁) of the modal class=23

h=10,

Frequency (f₀)of class preceding the modal class=21

Frequency (f₂) of class succeeding the modal class =14

Now,

=

=35 + 1.81 =36.8years

From the data as given above we may observe that maximum class frequency 61 belonging to the class interval 60 -80

So, modal class = 60-80

l=60, f₁=61, f₀=52, f₂=38, h=20

=

=60 +

=65.625hours

We may observe that the given data be maximum class frequency is 40 belonging to 1500- 2000 intervals

So, modal class = 1500- 2000

l=1500, f=40, f₀=24, f₂=33, h=50

=

=1500 + 347.826

=1847.826

So, modal class monthly expenditure was Rs. 1847.83

We may compute class marks (x_i) as per the relation:

X_i=

h= 500, A=2750

∑x_i=200,

(x̅) mean =

(x̅) =

=2750-87.5

=2662.5

So, mean monthly expenditure was Rs. 2662.50

To find: The mean and mode of the given table.
Solution:

Mode is calculated as:

Where

l=lower limit of modal class

h=width of the modal class

f=frequency of the modal class

f₁=frequency of the class preceding the modal class

f₂= frequency of the class following the modal class

Since, the maximum class frequency is 10

Hence, modal class interval= 30-35

h= 5, l=30, f= 10,f₁=9 and f₂=3



⇒ Mode = 30 + 0.625
⇒ Mode = 30.625
Now to find mean,
Use the formula,

Where

A = assumed mean

d_i=x_i-A

h=length of class intervals

u_i= (x_i-A)/h

N=sum of all frequencies

Here A = 32.5
h = 5
N = 35
So,


⇒ Mean = 32.5-3.28
⇒ Mean = 29.21

From the given data we may observe that maximum class frequency is 18 belonging to the class interval 4000-5000

So, modal class= 4000-5000

Lower limit, l= 4000

f₀=4, f₂=9, f=18,h = 1000

=

=4000 + =4608.7 runs

From the given data we may observe that maximum class frequency is 20 belonging to the class interval 40-50

So, modal class= 40-50

Lower limit, l= 40

f₀=12, f₂=11, f=20,h = 10

=

=40 +

=40 + 4.7 =44.7

Mean= =

We have, N= 68,

N/2 = 34

Hence, medium class =125- 145, such that

l=125, f’=20, f=22, h=20

Median = l +

Here, we may observe that maximum class frequency is 20 belonging to the class interval 125-145

So, modal class= 125-145

Lower limit, l= 125

f₀=13, f₂=14, f=20,h = 20

=

Mean= =

We have, N= 100,

N/2 = 50

Hence, median class =7-10, such that

l=7, f’=40, f=36, h=3

Median = l +

Here, we may observe that maximum class frequency is 40 belonging to the class interval 7-10

So, modal class= 7-10

Lower limit, l= 7

f₀=30, f₂=16, f=40,h = 3

=

Mean= =

We have, N= 50,

N/2 = 25

Hence, median class =60-80, such that

l=60, f’=12, f=24, h=20

Median = l +

Here, we may observe that maximum class frequency is 12 belonging to the class interval 60-80

So, modal class= 60-80

Lower limit, l= 60

f₀=10, f₂=6, f=12,h = 20

=

Mean= =

We have, N= 25,

N/2 = 12.5

Hence, median class =150-200, such that

l=150, f’=6, f=10, h=50

Median = l +

Here, we may observe that maximum class frequency is 6 belonging to the class interval 150-200

So, modal class= 150-200

Lower limit, l= 150

f₀=5, f₂=5, f=6, h = 50

=

Mean= =

We have, N= 50,

N/2 = 25

Hence, medium class =120-140, such that

l=120, f’=14, f=12, h=20

Median = l +

Here, we may observe that maximum class frequency is 14 belonging to the class interval 120-140

So, modal class= 120-140

Lower limit, l= 120

f₀=12, f₂=8, f=14,h = 20

=

We first prepare the cumulative frequency distribution by less than method as given below:

Now we mark the upper class limits along x-axis and cumulative frequency along y-axis. Thus, we plot the points (1,4); (2,13); (3,35); (4,63); (5,87); (6,99); (7,107); (8,113); (9,118); (10,120)

We first prepare the cumulative frequency distribution by less than method as given below:

Now we mark the upper class limits along x-axis and cumulative frequency along y-axis. Thus, we plot the points: (640,16); (680,61); (720,217); (760,501); (800,673); (840,732); (880,750)

The given frequency of distribution is not continuous. So, we first make it continuous and prepare cumulative frequency distribution as under:

Now we mark the upper class limits along x-axis and cumulative frequency along y-axis. Thus, we plot the points: (4,5,2); (9,5,8); (14,5,18); (19,5,23); (24,5,26)

We have,

Less than method, it is given that:

Now we mark the upper class limits along x-axis and cumulative frequency along y-axis. Thus, we plot the points: (7,26); (14,57); (21,92); (28,134); (35,216); (42,287); (49,341); (56,360)

More than method: We will prepare the cumulative frequency table by more than method as given below:

Now we mark,

On x-axis lower class limit and on y-axis Cumulative frequency

Thus, we plot graph as (0,360); (7,334); (14,303); (21,263); (28,226); (35,144); (42,73); (49,19)

More than method:

Now we mark,

On x-axis lower class limit and on y-axis Cumulative frequency

Thus, we plot graph as: (5,30); (10,28); (15,16); (20,14); (25,10); (30,7); (35,3)

Less than method:

Now we mark the upper class limit on x-axis and the cumulative frequency on y-axis. Thus, we plot the points: (10,2); (15,14); (20,16); (25,20); (30,23); (35,27); (40,30)

We first prepare cumulative frequency table by less than method as given below:

Now we mark on x-axis upper class limit and on y-axis cumulative frequency. Thus, we plot the points: (120, 12); (140,26); (160,34); (180,40); (200,50)

Less than method:

Now on x-axis upper class limits and on y-axis cumulative frequency, we plot the points: (55,2); (60,10) ;(65,22); (70,46); (75,84); (80,100)

More than method:

Now, Mark on x-axis lower class limit and on y-axis cumulative frequency. We plot the points: (50,100); (55,98); (60,90); (65,78); (70,54); (75,16)

Less than method:

It is given that on x-axis upper class limit and on y-axis cumulative frequency. We plot the points: (38,0); (40,3); (42,5); (49,9); (46,14); (48,28); (50,32); (52,35)

More than method:

X -axis lower class limit and y-axis cumulative frequency, we plot the points: (38,35); (40,32); (42,30); (44,26); (46,21); (48,7); (50,3)

We find the two types of cumulative frequency curves intersect at point P.

The value of M is 46.5 kg

Verification,

We have

Now, N = 35

Therefore, = = 17.5

The cumulative frequency is just greater than is 28 and the corresponding classes 46-48

Thus, 46-48 is the median class such that,

l = 46, f = 14, C₁ = 14 and h = 2

Median = l + * h

= 46 + * 2

= 46 + = 46 + 0.5

= 46.5 kg

Hence, verified.

The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

If x₁, x₂, … , x_n are observations with frequencies f₁, f₂, … , f_n i.e. x₁ occurs f₁ times and x₂ occurs f₂ times and so on, then we have

Sum of the values of the observations = f₁x₁ + f₂x₂ + … + f_nx_n

and Number of observations = f₁ + f₂ + … + f_n

So, mean() of observations is given by

Or,

In summation form, it can be shorted to

Which also can also be written as,

And it is understood i varies from 1 to n.

There are three measures of central tendency

1) Mean 2) Median 3) Mode

Suppose x₁, x₂, … , x_n are n observations with mean as x.

By definition of mean, [i.e. The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations]

We have,

and

nx = x₁ + x₂ + … + x_n …[1]

So, in this case we have assumed mean(a) is equal to mean of the observations(x)

And we know that

d_i = x_i - a

where, d_i is deviation of a (i.e. assumed mean) from each of x_i i.e. observations.

So, In the above case we have

d₁ = x₁ - x

d₂ = x₂ - x

.

.

.

d_n = x_n - x

and sum of deviations

d₁ + d₂ + … + d_n = x₁ - x + x₂ - x + … + x_n - x

= x₁ + x₂ + … + x_n - (x + x + … {upto n times})

= nx - nx [Using 1]

= 0

Hence, sum of deviations is zero.

Suppose x₁, x₂, … , x_n are n observations with mean as x.

By definition of mean, [i.e. The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations]

We have,

and

nx = x₁ + x₂ + … + x_n …[1]

So, in this case we have assumed mean(a) is equal to mean of the observations(x)

And we know that

d_i = x_i - a

where, d_i is deviation of a (i.e. assumed mean) from each of x_i i.e. observations.

So, In the above case we have

d₁ = x₁ - x

d₂ = x₂ - x

.

.

.

d_n = x_n - x

and sum of deviations

d₁ + d₂ + … + d_n = x₁ - x + x₂ - x + … + x_n - x

= x₁ + x₂ + … + x_n - (x + x + … {upto n times})

= nx - nx [Using 1]

= 0

Hence, sum of deviations is zero.

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have given series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

…[1]

As,

Mean

⇒ Mean

Median

As we know that, the x-coordinate of the point of intersection of the more than ogive and less than ogive give us median of the data.

4

As we know that, the x-coordinate of the point of intersection of the more than ogive and less than ogive give us median of the data.

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

Median can be find graphically by drawing any of the ogive or both ogives.

And Mode can be find graphically by drawing histogram of the given data.

But mean can't be determined graphically.

We know that,

Mode = 3 Median – 2 Mean

As we know that, the x-coordinate of the point of intersection of the more than ogive and less than ogive give us median of the data.

So, median can be determined graphically.

There are two ways in which median can be determined graphically.

(1) By drawing any of the ogive

In this case, we first compute , where N is the sum of frequencies and then we locate the point M corresponding to Nth cumulative frequency on curve, and the x-coordinate of M gives the median.

(2) By drawing both of the ogives

We draw both ogive curves [i.e. less than ogive and greater than ogive] and intersection of both ogives gives the value of median.

The following steps must be followed to find the mode graphically.

1. Represent the given data in the form of a Histogram. The frequency determines the height of each bar. Identify the highest rectangle. This corresponds to the modal class of the series.

2. Join the top corners of the modal bar with the immediately next corners of the adjacent bars. The two lines must be cutting each other.

3. Let the point where the joining lines cut each other be 'A'. Draw a perpendicular line from point A onto the x-axis. The point 'P' where the perpendicular will meet the x-axis will give the mode.

As class of maximum frequency is called modal class.

Modal class in above case is 20-25 as 75 is maximum frequency.

We know that, For finding median from a less than ogive or more than ogive curve, we follow below steps.

1. we find the sum of all frequencies or the last cumulative frequency in our given data, let that be N

2. Then we calculate and locate the point corresponding to on the curve.

3. The X coordinate of the point located i.e. the class corresponding to cumulative frequency is the median of data.

From the graph, we locate last cumulative frequency as 40 i.e. sum of all the frequencies is 40.

i.e. N = 40 and

Median is the marks corresponding to student.

In order to find the median, we first locate the point corresponding to 20^th student on Y axis.

And from graph, that point is (50, 20)

So, marks corresponding to 20^th student is 50.

So, the median of above data is 50

By Definition of mode, mode is most frequent value.

Given, mean is ,

Let x₁, x₂, …, x_n are n observations.

And we know

The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

i.e.

…[1]

Given as the first term is increased by 1 and 2^nd term is increased by 2 and so on. Then the terms will be

x₁ + 1, x₂ + 2, …,x_n + n

Let the new mean be x

…[2]

Now, we have series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

Using this in equation [2] and using equation [1] we have

First, we prepare the cumulative frequency table for above data

We know that median class of a data is the class-interval corresponding to cumulative frequency just greater than

Where,

N = sum of all frequencies

As N = 100 therefore

And

Cumulative frequency just greater than 50 is 60 which lies corresponding to class 40-50

Hence, 40-50 is median class.

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

Distance between mode and mean = mode - mean

Distance between median and mean = median - mean

Given that,

(mode - mean) = k(median - mean)

⇒ mode - mean = k median - k mean

⇒ mode = k median - k mean + mean

⇒ mode = k median - (k-1) mean

Comparing it with empirical relation, i.e.

mode = 3 Median - 2 mode

We get,

k = 3

Let the draw the frequency distribution table for the above data

As we know the mean(

In this case,

So we have

⇒ 31.2 + 2.6y = 28 + 3y

⇒ 3.2 = 0.4y

⇒ y = 8

We know, class marks of a class interval is

For 10-25

Lower limit = 10

Upper limit = 25

Class mark

For 35-55

Lower limit = 35

Upper limit = 55

Class mark

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

First, we prepare the cumulative frequency table for above data

We know that median class of a data is the class-interval corresponding to cumulative frequency just greater than

Where,

N = sum of all frequencies

As N = 25 therefore

And

Cumulative frequency just greater than 25 is 26 which lies corresponding to class 30-40

Hence, 30-40 is median class.

If x₁, x₂, … , x_n are observations with frequencies f₁, f₂, … , f_n i.e. x₁ occurs f₁ times and x₂ occurs f₂ times and so on, then we have

Sum of the values of the observations = f₁x₁ + f₂x₂ + … + f_nx_n

and Number of observations = f₁ + f₂ + … + f_n

So, mean() of observations is given by

Or,

In summation form, it can be shorted to

Which also can also be written as,

And it is understood i varies from 1 to n.

Terms are x, x + 3, x + 6, x + 9, x + 12

No of terms = 5

We know that

Mean

⇒ 50 = 5x + 30

⇒ 5x = 20

⇒ x = 4

Terms are 24, 25, 26, x + 2, x + 3, 30, 31, 34

No of terms = 8

We know that, if even no of terms or observations are given, then the median of data is mean of the values of term and term. Where n is no of terms.

In this case, n = 8

i.e. median of above data is mean of 4^th and 5^th term

⇒ median

⇒ 55 = 2x + 5

⇒ 2x = 50

⇒ x = 25

Terms are 6, 7, x - 2, x, 17, 20

No of terms = 6

We know that, if even no of terms or observations are given, then the median of data is mean of the values of term and term. Where n is no of terms.

In this case, n = 6

i.e. median of above data is mean of 3^rd and 4^th term

⇒ median

⇒ 16 = x - 1 [As median is 16]

⇒ x = 17

The first ten prime no's are :

2, 3, 5, 7, 11, 13, 17, 23, 29, 31

Clearly, the data is in ascending order. and

No of terms, n = 10

We know that, if even no of terms or observations are given, then the median of data is mean of the values of term and term. Where n is no of terms.

In this case, n = 10

i.e. median of above data is mean of 5^th and 6^th term

⇒ median

⇒ median = 12

As we know, mode of any data is the observation which occurs most.

In this case, 48 occurs two times and 43 is the mode of data

Therefore, 43 should occur more than two times.

And this is possible if and only if

x = 43

⇒ x + 3 = 43 + 3 = 46

Hence, correct option is (C).

Given: The mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15.

To find: The value of x.

Solution: As we know, mode of any data is the observation which occurs most.

In this case, 17 occurs two times which implies 17 is the mode.

But it is given that 15 is the mode of data.

In order for 15 to be mode it has to occur more than 2 times.

As 15 is already occurring 2 times, the possibility of it occurring more than 2 times is that x should be 15.

⇒ x = 15

Hence, correct option is (B).

First data is:

1, 3, 4, 5, 7, 4

Given, mean = m

And we know,

Mean …[1]

No of observations = 6

Sum of all observations = 1 + 3 + 4 + 5 + 7 + 4 = 24

Hence,

Mean

⇒ m = 4 …[2]

Second data is :

3, 2, 2, 4, 3, 3, p

No of observations = 7

Sum of observations = 3 + 2 + 2 + 4 + 3 + 3 + p = 17 + p

Given,

Mean = m - 1

Using [1]

⇒ 21 = 17 + p

⇒ p = 4 …[3]

Hence, series is 3, 2, 2, 4, 3, 3, 4

For median, let us write our data in increasing order

2, 2, 3, 3, 3, 4, 4,

Also, as the no of terms in this data is odd

We know that if there are odd number of terms in a data, then the median of data is term. Where n is no of terms

n = 7

median = 4^th term = 3

⇒ q = 3 …[4]

From [3] and [4]

p + q = 4 + 3 = 7

If x₁, x₂, … , x_n are observations with frequencies f₁, f₂, … , f_n i.e. x₁ occurs f₁ times and x₂ occurs f₂ times and so on, then we have

mean() of observations is given by

Given, mean = 8.1

Putting this values in formula:

⇒ 162 = 132 + 5k

⇒ 5k = 30

⇒ k = 6

Terms are 6, 7, x, 8, y, 14

No of terms = 6

We know that

Mean

⇒ 54 = x + y + 35

⇒ x + y = 19

Hence, correct option is (B)

Given, mean is ,

Let x₁, x₂, …, x_n are n observations.

And we know

The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

i.e.

…[1]

Given as the first term is increased by 1 and 2^nd term is increased by 2 and so on. Then the terms will be

x₁ + 1, x₂ + 2, …,x_n + n

Let the new mean be x

…[2]

Now, we have series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

Using this in equation [2] and using equation [1] we have

First n natural numbers are

1, 2, 3, 4, …, n

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have given series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

….[1]

As,

Mean

⇒ Mean

Given, mean

⇒ 9n + 9 = 10n

⇒ n = 9

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

Given,

Mean = 24

Mode = 12

Putting values in the formula,

12 = 3 Median - 2(24)

⇒ 12 = 3 Median - 48

⇒ 3 Median = 60

⇒ Median = 20

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have first n odd natural numbers as

1, 3, …, 2n - 1

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 2

And no of terms is clearly n.

And last term is (2n - 1)

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

…[1]

As,

Mean

⇒ Mean

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have first n odd natural numbers as

1, 3, …, 2n - 1

Clearly the above series is an AP(Arithmetic progression) with first term, a = 1 and common difference, d = 2

And no of terms is clearly n.

And last term is (2n - 1)

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

…[1]

As,

Mean

⇒ Mean

Now, given mean

⇒ n = 81

Difference of mode and median, mode - median = 24 …[1]

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

⇒ 3 Mode - 2 Mode = 3 Median - 2 Mean

⇒ 3 Mode - 3 Median = 2 Mode - 2 Mean

⇒ 3(Mode - Median) = 2(Mode - Mean)

From [1] we have

3(24) = 2(Mode - Mean)

⇒ Mode - Mean = 36 …[2]

on substracting [1] from [2]

Mode - Mean - (Mode - Median) = 36 - 24

Mode - Mean - Mode + Median = 8

Median - Mode = 8

Hence, difference between median and mode is 8.

Terms are 7, 8, x, 11, 14

No of terms = 5

We know that

Mean

⇒ 5x = x + 40

⇒ 4x = 40

⇒ x = 10

Hence, correct option is (C)

Given, mode exceeds mean by 12 i.e. mode - mean = 12 …[1]

We know that empirical relation between mean, median and mode is

Mode = 3 Median - 2 Mean

⇒ 3 Mode - 2 Mode = 3 Median - 2 Mean

⇒ 3 Mode - 3 Median = 2 Mode - 2 Mean

⇒ 3(Mode - Median) = 2(Mode - Mean)

From [1] we have

3(Mode - Median) = 2(12)

⇒ Mode - Median = 8

i.e. Mode exceeds Median by 8.

First n natural numbers are

1, 2, 3, 4, …, n

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have given series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

Putting the values in above equation we have sum of series i.e.

…[1]

As,

Mean

⇒ Mean

Given, mean = 15

⇒ n + 1 = 30

⇒ n = 29

Given, mean is ,

and x₁, x₂, …, x_n are n observations.

And we know

The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

i.e.

…[1]

And we have another series

x₁ + a, x₂ + a, …,x_n + a

Let the new mean be x

Now, From [1] we have

Given, mean is ,

Let x₁, x₂, …, x_k are k observations.

And we know

The mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

i.e.

….[1]

Given, the terms are divided by m and increased by n. Then the terms will be

Let the new mean be x

Now, From [1] we have

Hence correct option is (A)

We have given,

Also,

Putting this in above equation

Now, given …[1]

using this

We have,

…[2]

We know,

If x₁, x₂, … , x_n are observations with frequencies f₁, f₂, … , f_n i.e. x₁ occurs f₁ times and x₂ occurs f₂ times and so on, then mean() of observations is given by

From [1] and [2]

Given series is,

30, 34, 35, 36, 37, 38, 39, 40

We know that, if even no of terms or observations are given, then the median of data is mean of the values of term and term. Where n is no of terms.

In this case, no of terms,n = 8

i.e. median of above data is mean of 4^th and 5^th term

In this case,

4^th term = 36

5^th term = 37

⇒ median

⇒ median = 37.5

If 35 is removed, the series will be

30, 34, 36, 37, 38, 39, 40

No of terms = 7

We know that if there are odd number of terms in a data, then the median of data is term. Where n is no of terms

n = 7

median = 4^th term = 37

Difference in both medians = 37.5 - 37 = 0.5

Hence, median increases by 0.5.