Statistics

Let the 6^th number be x

Given that mean of 11 results = 30

∴ sum of 11 numbers = 11 × 30 = 330

Mean of the first 6 results = 28

Sum of first 6 numbers = 6 × 28 = 168

Mean of the last 6 results = 32

Sum of the last 6 results = 6 × 32 = 192

Therefore,

Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers

168 + 192 - x = 330

⇒ 360 – x = 330

⇒ x = 30

Let the 9^th observation be x

Given that mean of 17 observations = 20

∴ sum of 17 observations = 17 × 20 = 340

Mean of the first 9 observations= 23

Sum of first 9 observations = 9 × 23 = 207

Mean of the last 9 observations = 18

Sum of the last 9 observations = 9 × 18 = 162

Therefore,

Sum of first 9 observations + sum of last 9 observations – 9^th observation = sum of 17 observations

207 + 162 - x = 340

⇒ 369 – x = 340

⇒ x = 29

Let the weight of 11^th student be x

Given that mean weight of 21 students = 52kg

∴ sum of 21 students weight = 21 × 52 = 1092kg

Mean weight of the first 11 students = 50kg

Sum of first 11 students weight = 11 × 50 = 550kg

Mean weight of the last 11 students = 54kg

Sum of the last 11 students weight = 11 × 54 = 594kg

Therefore,

Sum of first 11 students weight + sum of last 11 students weight – weight of the 11^th student = sum of 21 students weight

550+594 - x = 1092

⇒ 1144 – x = 1092

⇒ x = 52

Hence, weight of 11^th student is 52kg

Let the weight of 13^th student be x

Given that mean weight of 25 students = 60kg

∴ sum of 25 students weight = 25 × 60 = 1500kg

Mean weight of the first 13 students = 57kg

Sum of first 13 students weight = 13 × 57 = 741kg

Mean weight of the last 13 students = 63kg

Sum of the last 13 students weight = 13 × 63 = 819kg

Therefore,

Sum of first 13 students weight + sum of last 13 students weight – weight of the 13^th student = sum of 25 students weight

741 + 819 - x = 1500

⇒ 1560 – x = 1500

⇒ x = 60

Hence, weight of 13^th student is 60kg

Let the 12^th observation be x

Given that mean of 23 observations = 34

∴ sum of 23 observations = 23 × 34 = 782

Mean of the first 12 observations = 32

Sum of first 12 observations = 12 × 32 = 384

Mean of the last 12 observations = 38

Sum of the last 12 observations = 12 × 38 = 456

Therefore,

Sum of first 12 observations + sum of last 12 observations – 12^th observation = sum of 23 observations

384 + 456 - x = 782

⇒ 840 – x = 782

⇒ x = 58

Let the 6^th number be x

Given that mean of 11 results = 35

∴ sum of 11 numbers = 11 × 35 = 385

Mean of the first 6 results = 32

Sum of first 6 numbers = 6 × 32 = 192

Mean of the last 6 results = 37

Sum of the last 6 results = 6 × 37 = 222

Therefore,

Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers

192 + 222 - x = 385

⇒ 414 – x = 385

⇒ x = 29

Let the 13^th observation be x

Given that mean of 25 observations = 36

∴ sum of 25 observations = 25 × 36 = 900

Mean of the first 13 observations = 32

Sum of first 13 observations = 13 × 32 = 416

Mean of the last 13 observations = 39

Sum of the last 13 observations = 13 × 39 = 507

Therefore,

Sum of first 13 observations + sum of last 13 observations – 13^th observation = sum of 25 observations

416 + 507 - x = 900

⇒ 923 – x = 900

⇒ x = 23

Now,

⇒25(14+k) = 390+ 15k

⇒350 + 25k = 390 + 15k

⇒ 25k – 15k = 390 – 350

⇒ 10k = 40

⇒ k = 4

Now,

Now,

⇒62.8 (40 + x) = 2640 + 50x

⇒2512 + 62.8x = 2640 + 50x

⇒ 62.8x – 50x = 2640 – 2512

⇒ 12.8x = 128

⇒ x = 10

Now,

⇒14 (24+p) = 360 + 10p

⇒336 + 14p = 360 + 10p

⇒ 14p – 10p = 360 – 336

⇒ 4p = 24

⇒ p = 6

Now,

⇒7.5(41+p) = 303 + 9p

⇒307.5 + 7.5p = 303 + 9p

⇒ 7.5p – 9p = 303 – 307.5

⇒ -1.5p = -4.5

⇒ p = 3

Now,

Now,

Now,

⇒53 (72+p) = 3340 + 70p

⇒3816 + 53p = 3340 + 70p

⇒ 53p – 70p = 3340 – 3816

⇒ -17p = -476

⇒ p = 28

Now,

⇒ 5(96+ f₁) = 4320 + 70f₁

⇒480 + 5f₁ = 4320 + 70f₁

⇒ 5f₁ – 70f₁ = 4320 - 480

⇒ -65f₁ = +3840

⇒ f₁ = - 59.07

This is not possible as frequency can not be negative.

Now,

Now,

[given: ∑f_i = 50]

⇒ 62.8(50) = 2060 + 30f₁ +70f₂

⇒ 3140 = 2060 + 30f₁ +70f₂

⇒ 3140 - 2060 = 30f₁ +70f₂

⇒ 1080 = 30f₁ +70f₂

⇒ 108 = 3f₁ +7f₂ …(i)

and 30 + f₁ +f₂ = 50

⇒ f₁ +f₂ = 50 – 30

⇒ f₁ +f₂ = 20

⇒ f₁ = 20 – f₂ …(ii)

Now, putting the value of f₁ in eq. (i), we get

3(20 –f₂) + 7f₂ = 108

⇒ 60 – 3f₂ + 7f₂ = 108

⇒ 4f₂ = 108 – 60

⇒ 4f₂ = 48

⇒ f₂ = 12

Now, substitute the value of f₂ in eq. (ii), we get

f₁ = 20 – 12

⇒ f₁ = 8

Now,

[given: ∑f_i = 50]

⇒ 57.6(50) = 1940 + 30f₁ +70f₂

⇒ 2880 = 1940 + 30f₁ +70f₂

⇒ 2880 – 1940 = 30f₁ +70f₂

⇒ 940 = 30f₁ +70f₂

⇒ 94 = 3f₁ +7f₂ …(i)

and 32 + f₁ +f₂ = 50

⇒ f₁ +f₂ = 50 – 32

⇒ f₁ +f₂ = 18

⇒ f₁ = 18 – f₂ …(ii)

Now, putting the value of f₁ in eq. (i), we get

3(18 –f₂) + 7f₂ = 94

⇒ 54 – 3f₂ + 7f₂ = 94

⇒ 4f₂ = 94 – 54

⇒ 4f₂ = 40

⇒ f₂ = 10

Now, substitute the value of f₂ in eq. (ii), we get

f₁ = 18 – 10

⇒ f₁ = 8

Now,

Now,

Here, the class size varies, and x_i’s are large. Now, we apply the step deviation method with a = 120 and h = 20

Now,

⇒ = 112.2

Here, we can see that the class interval is not continuous. So, we make it continuous.

Now,

⇒ x̄ = 36.36

Here, the x_i’s are large. Now, we apply the step deviation method with a = 350 and h = 100

Now,

⇒x̄ = 350 - 46

⇒x̄ = 304

Hence, the mean marks scored by a student in the class-test is 304

Now,

Here, the x_i’s are large. Now, we apply the step deviation method with a = 250 and h = 100

Now,

⇒x̄ = 250 + 14

⇒x̄ = 264

Here, the x_i’s are large. Now, we apply the step deviation method with a = 175 and h = 50

Now,

⇒x̄ = 175 - 30

⇒x̄ = 145

Hence, the mean marks scored by a student in the class-test is 145

Here, the x_i’s are large. Now, we apply the step deviation method with a = 3250 and h = 500

Now,

⇒x̄ = 3250 – 587.5

⇒x̄ = 2662.5

Hence, the mean expenditure on education of a family is Rs 2662.5

Here, we can see that 64 observation has the maximum frequency.

Hence, Mode = 64

Here, the marks 8 has the maximum frequency ‘16’.

∴ Mode = 8

Here, the class interval 12 has the maximum frequency ‘9’.

∴ Mode = 12

Here, the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5.

So, the modal class is 3 – 5.

Now, modal class = 3 – 5, lower limit (l) of modal class = 3, class size(h) = 2

frequency (f₁) of the modal class = 8

frequency (f₀) of class preceding the modal class = 7

frequency (f₂) of class succeeding the modal class = 2

Now, let us substitute these values in the formula

= 3.286

Here, the maximum number of students i.e. 71 have got marks in the interval 60 – 70.

So, the modal class is 60 – 70.

Now, modal class = 60 – 70, lower limit (l) of modal class = 60, class size(h) = 10

frequency (f₁) of the modal class = 71

frequency (f₀) of class preceding the modal class = 31

frequency (f₂) of class succeeding the modal class = 52

Now, let us substitute these values in the formula

= 60 + 6.78

= 66.78

Here, the maximum class frequency is 28, and the class corresponding to this frequency is 30 – 40.

So, the modal class is 30 – 40.

Now, modal class = 30 – 40, lower limit (l) of modal class = 30, class size(h) = 10

frequency (f₁) of the modal class = 28

frequency (f₀) of class preceding the modal class = 25

frequency (f₂) of class succeeding the modal class = 25

Now, let us substitute these values in the formula

= 30 + 5

= 35

Here, the maximum number of batsman i.e. 18 have scored the runs in the interval 4000 – 5000.

So, the modal class is 4000 – 5000 .

Now, modal class = 4000 – 5000,

lower limit (l) of modal class = 4000,

class size(h) = 1000

frequency (f₁) of the modal class = 18

frequency (f₀) of class preceding the modal class = 4

frequency (f₂) of class succeeding the modal class = 9

Now, let us substitute these values in the formula

= 4000 + 608.7

= 4608.7

Here, the maximum class frequency is 20, and the class corresponding to this frequency is 40 – 50.

So, the modal class is 40 – 50.

Now, modal class = 40 – 50, lower limit (l) of modal class = 40, class size(h) = 10

frequency (f₁) of the modal class = 20

frequency (f₀) of class preceding the modal class = 12

frequency (f₂) of class succeeding the modal class = 11

Now, let us substitute these values in the formula

= 40 + 4.7

= 44.7

Here, the maximum class frequency is 50, and the class corresponding to this frequency is 16 – 20.

So, the modal class is 16 – 20.

Now, modal class = 16 – 20, lower limit (l) of modal class = 16, class size(h) = 4

frequency (f₁) of the modal class = 50

frequency (f₀) of class preceding the modal class = 30

frequency (f₂) of class succeeding the modal class = 40

Now, let us substitute these values in the formula

= 16 + 2.67

= 18.67

Here, the maximum number of persons i.e. 30 have income in the interval 300 – 400.

So, the modal class is 300 – 400.

lower limit (l) of modal class = 300,

class size(h) = 100

frequency (f₁) of the modal class = 30

frequency (f₀) of class preceding the modal class = 18

frequency (f₂) of class succeeding the modal class = 20

Now, let us substitute these values in the formula

= 300 + 54.54

= Rs. 354.54

Here, the maximum no. of students is 9, and the class corresponding to the frequency is 20 – 30.

So, the modal class is 20 – 30.

lower limit (l) of modal class = 20,

class size(h) = 10

frequency (f₁) of the modal class = 9

frequency (f₀) of class preceding the modal class = 5

frequency (f₂) of class succeeding the modal class = 3

Now, let us substitute these values in the formula

= 20 + 4

= 24

Here, the maximum no. of students is 20, and the class corresponding to this frequency is 20 – 30

So, the modal class is 20 – 30.

lower limit (l) of modal class = 20,

class size(h) = 10

frequency (f₁) of the modal class = 20

frequency (f₀) of class preceding the modal class = 6

frequency (f₂) of class succeeding the modal class = 10

Now, let us substitute these values in the formula

= 20 + 5.83

= 25.83

Here, the maximum frequency is 25, and the class corresponding to this frequency is 40 - 50

So, the modal class is 40 – 50.

lower limit (l) of modal class = 40,

class size(h) = 10

frequency (f₁) of the modal class = 25

frequency (f₀) of class preceding the modal class = 12

frequency (f₂) of class succeeding the modal class = 10

Now, let us substitute these values in the formula

= 40 + 4.64

Mode = 44.64(approx.)

Given: Mode = Rs. 24

Frequency (f_i) = 100

Class size = 10

Let frequency of the class 10 – 20 be f₀. Then, the frequency of the class 30 – 40 will be equal to 100 – 14 – f₀ – 27 – 15 = 44 – f₀.

It is given that mode = 24, therefore, modal class is 20 – 30.

Thus, lower limit of the modal class (l) = 20

frequency (f₀) of class preceding the modal class = f₀

frequency (f₂) of class succeeding the modal class = 44 – f₀

⇒ 4 = 27 - f₀

⇒ -23 = -f₀

⇒ f₀ = 23

∴ f₂ = 44 – 23 = 21

Hence, the missing frequencies are 23 and 21.

Now,

⇒x̄ = 37.5 – 8.28

⇒x̄ = 29.22

∴ Mean = 29.22

Here, the maximum no. of students per teacher i.e. 10, and the class corresponding to this frequency is 30 – 35

So, the modal class is 30 – 35 .

lower limit (l) of modal class = 30,

class size(h) = 5

frequency (f₁) of the modal class = 10

frequency (f₀) of class preceding the modal class = 9

frequency (f₂) of class succeeding the modal class = 3

Now, let us substitute these values in the formula

= 30 + 0.625

Mode = 30.625(approx.)

Now, items from 23 to 32 have value of variate 14 as shown by cumulative frequency.

∴ Median = 14

Now, persons from 100 to 110 have daily wages 29 as shown by cumulative frequency.

∴ Median = 29

We have n = 190

So,

The cumulative Frequency just greater than is 120 then the median class is 15 – 20 such that

the lower limit (l) = 15

cumulative frequency of the class preceding 15 – 20 (cf) = 92

the frequency of the median class 15 – 20 =28,

class size (h) = 5

Using the formula, , we have

= 15 + 0.53

= 15.53

We have n = 30

So,

The cumulative Frequency just greater than is 19 then the median class is 55 – 60 such that

the lower limit (l) = 55

cumulative frequency of the class preceding 55 – 60 (cf) = 13

frequency of the median class 55 – 60 =6,

class size (h) = 5

Using the formula,,we have

= 55 + 1.66

= 56.66

So, the median weight of the students is 56.66kg

We have n = 50

So,

The cumulative Frequency just greater than is 36 then the median class is 10.75 – 11.25 such that

the lower limit (l) = 10.75

cumulative frequency of the class preceding 10.75 – 11.25 (cf) = 19

frequency of the median class 10.75 – 11.25 = 17,

class size (h) = 0.5

Using the formula,, we have

= 10.75 + 0.176

= 10.93

We have n = 80

So,

The cumulative Frequency just greater than is 56 then the median class is 40 – 50 such that

the lower limit (l) = 40

cumulative frequency of the class preceding 40 – 50 (cf) = 37

the frequency of the median class 40 – 50 =19,

class size (h) = 10

Using the formula, , we have

= 40 + 1.58

=41.58

We have n = 100

So,

The cumulative Frequency just greater than is 78 then the median class is 35 – 40 such that

the lower limit (l) = 35

cumulative frequency of the class preceding 35 – 40 (cf) = 45

frequency of the median class 35 – 40 = 33,

class size (h) = 5

Using the formula,, we have

= 35 + 0.757

= 35.76

So, the median age of the policy holders is 35.76years

We have n = 51

So,

The cumulative Frequency just greater than is 29 then the median class is 145-150 such that

the lower limit (l) = 145

cumulative frequency of the class preceding 145-150 (cf) = 11

frequency of the median class 145-150 = 18,

class size (h) = 5

Using the formula,, we have

= 145 + 4.027

= 149.03

So, the median height of the girls is 149.03cm

We have n = 400

So,

The cumulative Frequency just greater than is 216 then the median class is 3000-3500 such that

the lower limit (l) = 3000

cumulative frequency of the class preceding 3000-3500 (cf) = 130

frequency of the median class 3000-3500 = 86,

class size (h) = 500

Using the formula,, we have

= 3000 + 406.98

= 3406.98

So, median life time of lamps is 3406.98hours

We have n = 100

So,

The cumulative Frequency just greater than is 62 then the median class is 15.5 – 20.5 such that

the lower limit (l) = 15.5

cumulative frequency of the class preceding 15.5-20.5 (cf) = 30

frequency of the median class 15.5 – 20.5 = 32,

class size (h) = 5

Using the formula,, we have

= 15.5 + 3.125

= 18.625 = 18.63

So, the median numbers of letters in the names is 18.63

We have n = 40

So,

The cumulative Frequency just greater than is 29 then the median class is 144.5 – 153.5 such that

the lower limit (l) = 144.5

cumulative frequency of the class preceding 144.5 – 153.5 (cf) = 17

frequency of the median class 144.5 – 153.5 = 12,

class size (h) = 9

Using the formula,, we have

= 144.5 + 2.25

= 146.75

So, the median length of the leaves is 146.75mm

Here, the class mark is given.

∴Class size = 45 – 35 = 10

If a is a class mark and h is the size of the class interval, then the lower limit and upper limit of the class interval are and repectively.

∴, we have h = 10

∴Lower Limit of first class interval

The upper limit of first class interval

∴ The first class interval is 30 – 40

Hence, the class intervals are 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80, 80 – 90.

Now, We find the median

We have n = 90

So,

The cumulative Frequency just greater than is 46 then the median class is 40 – 50 such that

the lower limit (l) = 40

cumulative frequency of the class preceding 40 – 50 (cf) = 20

frequency of the median class 40 – 50 = 26,

class size (h) = 10

Using the formula,, we have

= 40 + 9.61

= 49.61

Here, we can see that the intervals are unequal.

Firstly, we convert the unequal class intervals into equal class intervals.

We have n = 80

So,

The cumulative Frequency just greater than is 45 then the median class is 12-18 such that

the lower limit (l) = 12

cumulative frequency of the class preceding 12 – 18 (cf) = 17

frequency of the median class 12 – 18 = 28,

class size (h) = 6

Using the formula,,we have

= 12 + 4.928

= 14.93

Given Median =46

Then, median Class = 40 – 50

the lower limit (l) = 40

cumulative frequency of the class preceding 40 – 50 (cf) = 42 + x

frequency of the median class 40 – 50 = 65,

class size (h) = 10

Total frequencies (n) = 229

So, 150 + x + y = 229

⇒ x + y = 229 – 150

⇒ x + y = 79 …(i)

Using the formula,,we have

⇒39 = 72.5 – x

⇒ x = 33.5

Putting the value of x in eq. (i), we get

⇒ 33.5 + y = 79

⇒ y = 79 – 33.5

⇒ y = 45.5

Given Median =28.5

Then, median Class = 20 – 30

the lower limit (l) = 20

cumulative frequency of the class preceding 20 – 30 (cf) = 5 + x

frequency of the median class 20 – 30 = 20,

class size (h) = 10

Total frequencies (n) = 60

So, 45 + x + y = 60

⇒ x + y = 60 – 45

⇒ x + y = 15 …(i)

Using the formula,, we have

⇒8.5 × 2 = 25-x

⇒17 = 25 – x

⇒ x = 8

Putting the value of x in eq. (i), we get

⇒ 8 + y = 15

⇒ y = 15 – 8

⇒ y = 7

Given Median =525

Then, median Class = 500-600

the lower limit (l) = 500

cumulative frequency of the class preceding 500-600(cf) = 36 + x

frequency of the median class 500-600 = 20,

class size (h) = 100

Total frequencies (n) = 100

So, 76 + x + y = 100

⇒ x + y = 100 – 76

⇒ x + y = 24 …(i)

Using the formula,,we have

⇒25 = (14 – x) × 5

⇒5 = 14 – x

⇒ x = 9

Putting the value of x in eq. (i), we get

⇒ 9 + y = 24

⇒ y = 24 – 9

⇒ y = 15

Now,

⇒x̄ = 45 + 5.44

⇒x̄ = 50.44

We have n = 250

So,

The cumulative Frequency just greater than is 127 then the median class is 50-60 such that

the lower limit (l) = 50

cumulative frequency of the class preceding 50 – 60 (cf) = 96

frequency of the median class 50 – 60 = 31,

class size (h) = 10

Using the formula,,we have

= 50 + 9.35

= 59.35

Now,

⇒x̄ = 24.5 + 1.99

⇒x̄ = 26.5

Now, we calculate the median

We have n = 274

So,

The cumulative Frequency just greater than is 152 then the median class is 21 – 28 such that

the lower limit (l) = 21

cumulative frequency of the class preceding 21 – 28 (cf) = 80

frequency of the median class 21-28 = 72,

class size (h) = 7

Using the formula,,we have

= 21 + 5.57

= 26.57

Now, we have to find the mode

Here, the maximum class frequency is 72, and the class corresponding to this frequency is 21 – 28.

So, the modal class is 21 – 28.

Now, modal class = 21 – 28, lower limit (l) of modal class = 21, class size(h) = 7

frequency (f₁) of the modal class = 72

frequency (f₀) of class preceding the modal class = 36

frequency (f₂) of class succeeding the modal class = 51

Now, let us substitute these values in the formula

= 21 + 4.42

= 25.42

We have n = 100

So,

The cumulative Frequency just greater than is 36 then the median class is 7 – 10 such that

the lower limit (l) = 7

cumulative frequency of the class preceding 7 – 10 (cf) = 36

frequency of the median class 7 – 10 = 40,

class size (h) = 3

Using the formula,,we have

= 7 + 1.05

= 8.05

Now, we calculate the Mean

Now,

⇒ = 11.5 – 3.18

⇒ = 8.32

Now, we have to find the mode

Here, the maximum class frequency is 40, and the class corresponding to this frequency is 7 – 10.

So, the modal class is 7 – 10.

Now, modal class = 7 – 10, lower limit (l) of modal class = 7, class size(h) = 3

frequency (f₁) of the modal class = 40

frequency (f₀) of class preceding the modal class = 30

frequency (f₂) of class succeeding the modal class = 16

Now, let us substitute these values in the formula

= 7 + 0.88

= 7.88

Now, taking upper class limits on x- axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, taking lower class limits on x- axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, taking upper class limits on x- axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, we have n = 250

The cumulative Frequency just greater than is 140 then the median class is 300-400 such that

the lower limit (l) = 300

cumulative frequency of the class preceding 300-400 (cf) = 75

frequency of the median class 300-400 = 65,

class size (h) = 100

Using the formula,,we have

= 300 +76.9

= 376.9

=377 (approx.)

Now, taking lower class limits on x-axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, we have n = 201

The cumulative Frequency just greater than is 156 then the median class is 40 – 60 such that

the lower limit (l) = 40

cumulative frequency of the class preceding 40 - 60 (cf) = 92

frequency of the median class 40-60 = 64,

class size (h) = 20

Using the formula, , we have

= 40 +2.65

= 42.65

= 42.7

Now, taking upper-class limits on x-axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, we have n = 30

The cumulative Frequency just greater than is 16 then the median class is 15 – 20 such that

the lower limit (l) = 15

cumulative frequency of the class preceding 15 – 20 (cf) = 14

the frequency of the median class 15 – 20 = 2,

class size (h) = 5

Using the formula, , we have

= 15 + 2.5

= 17.5

So, the median profit is Rs. 17.5

Now, taking monthly income on x-axis and their respective frequencies on y-axis we can draw its ogive as follows:

Now, we have n = 600

The cumulative Frequency just greater than is 430 then the median class is 150 – 225 such that

the lower limit (l) = 150

cumulative frequency of the class preceding 150 – 225 (cf) = 230

frequency of the median class 150 – 225 = 200,

class size (h) = 75

Using the formula,,we have

= 150 +26.25

= 176.25

= 176(approx.)