Statistics

Statistics is a science which deals with collection, presentation, analysis and finally interpretation of numerical data.

Fundamental characteristics of statistics are:-

i) Data given or collected for a definite purpose can’t be used for another purpose.

ii) Numerical facts always constitute a specific data.

iii) A single observation does not form data. Data should be aggregate of facts.

iv) Qualitative characteristics cannot be measured numerically. Hence, they do not form Data.

Primary Data: A Data collected by the investigator himself with a definite plan in mind is called primary data.

Secondary Data: A data collected by someone, other than investigator is called secondary data.

Primary data are highly reliable than secondary data because they are collected by investigator himself with a definite plan in mind so they are more relevant. While secondary data are not being fully relevant to the investigation.

(i) Variate:- Any character which consists of several values is called Variate.

(ii) Class interval:- A Group in which Raw data is condensed is called class-interval.

(iii) Class size:- The difference between true upper limit and true lower limit is termed as class-size.

(iv) Class mark:- it is given as

(v) Class limit:- each class is bounded by two figures which termed as class limits.

(vi) True class limits:- in exclusive form of distribution, true lower limit of a class is obtained by subtracting from lower limit and true upper limit is obtained by adding 0.5 to the upper limit.

(vii) Frequency of a class:- Number of data values that fall in the range specified by that class is called frequency of that class.

(viii) Cumulative frequency of a class:- The cumulative Frequency corresponding to that class is the sum of all frequencies up to and including that class.

Minimum observation = 0

Maximum observation = 6

The classes of equal size covering the given data are:

0-2, 2-4, 4-6, 6-8

So, frequency distribution table will be as given below,

Minimum observation = 1

Maximum observation = 24

The classes of equal size covering the given data are:

0-5, 5-10, 10-20, 20-25

So, frequency distribution table will be as given below,

Minimum observation = 6

Maximum observation = 23

The classes of equal size covering the given data are:

6-9, 9-12, 12-15, 15-18, 18-21, 21-24, 24-27

So, grouped frequency table will be as given below,

Minimum observation = 210

Maximum observation = 320

The classes of equal size covering the given data are:

210-230, 230-250, 250-270, 270-290, 290-310, and 310-330

So, frequency table will be as given below,

Minimum observation = 30

Maximum observation = 120

Frequency Table :-

Cumulative frequency table: -

Minimum observation = 800

Maximum observation = 900

∴ Range = 900 – 800 = 100

Class size = 10

Number of classes =

Minimum data = 52

Maximum data = 130

Range = 130 – 52 = 78

Let the class size = 10

∴ Number of classes = or 8 classes

The Cumulative frequency table can be drawn as given below:

Grouped frequency table can be drawn as shown below;

Frequency table can be represented as below;

The frequency table can be represented as below:

The given class interval is in the form of inclusive form.

To find out the true lower limit of the class we need to draw the table in continuous frequency distribution in exclusive form.

So, as we can see that true lower limit of class 21 – 30 is 20.5

As the class intervals are given in exclusive form so the true upper class limit of the class 10 – 20 is 20.

The given class interval is in the form of inclusive form.

To find out the class size we need to first find out the true class limits.

Now,

By converting the table in exclusive form;

We get,

Class size = True upper limit – True lower limit

Class size = 20.5 – 10.5 = 10

Class Interval = 21 – 30 (given)

Upper class limit = 30

Lower class limit = 21

Class marks = 1/2 [upper limit + lower limit]

Class marks = 1/2 [30 + 21]

∴ Class marks = 51/ 2 = 25.5

Observations are = x, x + 2, x + 4, x + 6, x + 8

No. of observations = 5

Given mean = 11

⇒ 11 =

⇒ x + x + 2 + x + 4 + x + 6 + x + 8 = 11 × 5

⇒ 5x + 20 = 55

⇒ 5x = 55 – 20

x = = 7

So, value of x = 7.

Arrange the scores in ascending order;

Scores are = 2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 28, 48

Total points = 16 (even)

Median = 1/2 [value of(n/2)th term + value of (n/2 + 1)th term]

⇒ Median = 1/2 [value of 8th term + value of 9th term]

⇒ Median = 1/2 [10 + 14]

∴ Median = 24/2 = 12

The median of the points scored by the team is 12.

To draw the bar graph,

On x-axis take the games and on y-axis take the no. of students.

Now draw the bar graph;

Heights of the players = 148cm, 154cm, 153cm, 140cm, 150cm

Number of players = 5

Mean height per player =

∴ Mean = 149cm

So, mean height per player is 149cm.

By arranging the marks in ascending order, we have;

5, 10, 19, 20, 23, 23, 23, 34, 36, 37, 38

In the given data 23 occurred the maximum number of times,

So, the model mark = 23

To find true class limit should know the class size;

Class size = 31 – 26 = 5

Mid value = 5/2 = 2.5

When class mark is 26;

Upper class limit = 26 + 2.5 = 28.5

Lower class limit = 26 – 2.5 = 23.5

When class mark is 31;

Upper class limit = 31 + 2.5 = 33.5

Lower class limit = 31 – 2.5 = 28.5

When class mark is 36;

Upper class limit = 36 + 2.5 = 38.5

Lower class limit = 36 – 2.5 = 33.5

When class mark is 41;

Upper class limit = 41 + 2.5 = 43.5

Lower class limit = 41 – 2.5 = 38.5

When class mark is 46;

Upper class limit = 46 + 2.5 = 48.5

Lower class limit = 46 – 2.5 = 43.5

When class mark is 51;

Upper class limit = 51 + 2.5 = 53.5

Lower class limit = 51 – 2.5 = 48.5

So, true class limits are;

23.5 - 28.5, 28.5 - 33.5, 33.5 - 38.5, 38.5 - 43.5, 43.5 - 48.5, 48.5 - 53.5

First draw the table as shown below;

Given mean = 8

8 =

⇒ 8(41 + p) = 303 + 9p

⇒ 328 + 8p = 30 + 9p

⇒ 9p – 8p = 328 – 303

⇒ p = 25

So, the value of p = 25

10 observations in ascending order = 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41

n = 10 (even)

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

Median = 1/2 [value of 5th term + value of 6th term]

Median = 1/2 [(x + 1) + (x + 3)] = (2x + 4)/2

Given median = 24

⇒

⇒ 2x + 4 = 24 × 2

⇒ 2x + 4 = 48

⇒ 2x = 48 – 4 = 44

⇒ x = 44/2 = 22

By arranging the given data in ascending form we get;

10, 10, 11, 11, 12, 12, 13, 14, 15, 17

Draw the table;

∑f = N = 10 (even)

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

= 1/2 [value of 5th term + value of 6th term]

= 1/2 [12 + 12] = 24/2 = 12

Now,

Mean = (∑fx)/(∑f) = 125/10 = 12.5

Mode = 3(median) – 2(mean)

= 3 × 12 – 2 × 12.5

= 36 – 25 = 11

Arrange the data in ascending form:

Draw the cumulative frequency table;

∑f = N = 40 (even)

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

= 1/2 [value of 20th term + value of 21st term]

= 1/2 [7 + 10] = 17/2 = 8.5

Let suppose the numbers are = n₁, n₂, n₃, n₄, n₅, n₆

Given mean = 23

So,

n₁ + n₂ + n₃ + n₄ + n₅ + n₆ = 23×6 = 136 ……….(i)

Let suppose excluded number be n₄

N = 5

Mean of remaining numbers = (given)

n₁ + n₂ + n₃ + n₅ + n₆ = 20 × 5 = 100 ……….(ii)

Subtract the (ii) equation from (i) equation

We get;

n₄ = 138 – 100 = 38

So, the excluded number is 38.

Cumulative frequency is the total of frequency.

Draw the histogram with the help of given data.

Calculated mean of marks of 50 students = 39

According to the given mean the sum of these marks will be = 39 ×50 = 1950

Correct sum will be = incorrect sum + (correct marks – incorrect marks)

= 1950 + 43 – 23

= 1993 – 23 = 1970

Correct mean = (correct sum)/50 = 1970/50 = 39.4

So, the correct mean is 39.4

Draw the table;

Mean =

∑f = 12, ∑fx = 771

The mean weight is 64.25kg

Draw the cumulative frequency table;

N = ∑f = 50 (even)

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

⇒ Median = 1/2 [value of 25th term + value of 26th term]

⇒ Median = 1/2 [154 + 155] = 309/2 = 154.5cm

(i) The bar graph shows the marks obtained by a student in various subjects in an examination.

(ii) The student scores very good in mathematics, as the height of the corresponding bar is the highest.

(iii) The student scores least marks in Hindi, as the height of the corresponding bar is the lowest.

(iv) Average marks =

The given frequency distribution is in exclusive form, we will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 40 rupees on X-axis,

1 big division = 2 workers on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

Thus, we get the following histogram.

The given frequency distribution is in exclusive form, we will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 50 rupees on X-axis,

1 big division = 1 store on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

Thus, we get the following histogram.

The given frequency distribution is in exclusive form, we will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 6cm on X-axis,

1 big division = 2 students on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

Thus, we get the following histogram.

The given frequency distribution is in exclusive form, we will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 5 units on X-axis,

1 big division = 50 units on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

Thus, we get the following histogram.

The given frequency distribution is in inclusive form. So, convert it in Exclusive form.

We will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 8 units on X-axis,

1 big division = 2 units on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

Now, we get following Histogram:

We will represent the class intervals along the X-axis and the corresponding frequency on the Y axis.

Now take the scale of,

1 big division = 7 years on X-axis,

1 big division = 50 person on Y axis

We will draw the rectangles with the class intervals as basis and the corresponding frequency as the height.

The given frequency distribution is in inclusive form. So, convert it in Exclusive form.

Histogram as shown below:

In the given frequency distribution, Class sizes are different.

So, we calculate adjusted frequency for each class, As, minimum class size = 4

Adjusted frequency of a class =

Histogram is as follows:

Let’s take two classes interval, first at beginning (0-10) and second at the end (70-80) each with frequency zero.

Now we can draw the frequency table with the help of these two classes,

Now plot the following points on the graph,

A (5,0)

B (15,2)

C (25, 5)

D (35, 12)

E (45, 19)

F (55, 9)

G (65, 4)

H (75,0)

Join the points with line segments

AB, BC, CD, DE, EF, FG, GH, to obtain required frequency polygon. As shown in the figure.

We take imagined class 0 – 10 and 70 – 80, each with frequency zero. The class marks of above classes are 5 and 75 respectively.

So, we plot the points A (5,0) and B (75,0). We join A with the midpoint of the top of the first rectangle and B with the mid-point of the last rectangle.

Thus we obtain a complete frequency polygon,

We take imagined class 15-20and 0-55, each with frequency 0. The class marks of above classes are 17.5 and 52.5 respectively.

So, we plot the points A (17.5, 0) and B (52.5, 0). We join A with the midpoint of the top of the first rectangle and B with the mid-point of the last rectangle.

Thus we obtain a complete frequency polygon,

We take imagined class 560-600 and 840-880, each with frequency 0. The class marks of above classes are 5 and 75 respectively.

As we can see in the figure the x-axis starts at 560, a break is indicated near the origin to show that the graph is drawn with a scale beginning at 560, not at origin.

So, we plot the points A (580, 0) and B (860, 0). We join A with the midpoint of the top of the first rectangle and B with the mid-point of the last rectangle.

Thus we obtain a complete frequency polygon,

We take the imagined classes (-9, 0) at the beginning and (61-70) at the end, each of the frequency 0.

Thus we have,

We plot the following points on the graph,

A (-4.5, 0)

B (5.5, 8)

C (15.5, 3)

D (25.5, 6)

E (35.5, 12)

F (45.5, 2)

G (55.5, 7)

H (65.5, 0)

Now we draw the line segment AB, BC, CD, DE, EF, FG, GH, to obtain the frequency polygon.

(i) First eight natural no’s are = 1,2,3,4,5,6,7,8

Sum of these numbers = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Arithmetic Mean =

(ii) First ten odd numbers are = 1,3,5,7,9,11,13,15,17,19

Sum of these numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

Arithmetic Mean =

(iii) First five prime numbers are = 2,3,5,7,11

Sum of these numbers = 2 + 3 + 5 + 7 + 11 = 28

Arithmetic Mean =

(iv) First six even numbers are = 2,4,6,8,10,12

Sum of these numbers = 2 + 4 + 6 + 8 + 10 + 12 = 42

Arithmetic Mean = 7.

(v) First seven multiples of 5 are = 5,10,15,20,25,30,35

Sum of these numbers = 5 + 10 + 15 + 20 + 25 + 30 + 35 = 140

Arithmetic Mean =

(vi) All factors of 20 are = 1,2,4,5,10,20

Sum of numbers = 1 + 2 + 4 + 5 + 10 + 20 = 42

Arithmetic Mean =

Sum of numbers of children in families = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30

Total number of family = 10

Mean number of children per family =

Sum of numbers of books issued in a week = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1632

Total number of days = 7

No. of books issued per day =

Sum of temperature during whole week = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 =

Total number of days = 6

Mean temperature =

Sum of percentage of marks = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474

Number of students = 12

Mean percentage of marks =

Sum of numbers = 7 + 9 + 11 + 13 + x + 21 = 61 + x

Number of observation = 6

Arithmetic Mean =

= 13 =

= 61 + x = 13 × 6 = 78

= x = 78 – 61 = 17.

Given,

Mean of 24 numbers = 35

Sum of these numbers = 24 × 35 = 840.

Every number is increased by 3,

Total increment = 24 × 3 = 72

New sum = 840 + 72 = 912

New Mean =

Given,

Mean of 20 numbers = 43

Sum of these numbers = 20 × 43 = 860

When, every number is decreased by 6,

Total decrease in sum = 20 × 6 = 120

New sum will be = 860 – 120 = 740

New Mean =

Given,

Mean of 15 numbers = 27

Sum of these numbers = 15 × 27 = 405

When, every number is multiplied by 4,

The new sum becomes = 405 × 4 = 1620

New Mean =

Given,

Mean of 12 numbers = 40

Sum of these numbers = 12 × 40 = 480

When each number is divided by 8

The new sum becomes =

New mean =

Given,

Mean of 20 numbers = 18.

Sum of these numbers = 20 × 18 = 360.

When 3 is added to each of the first ten numbers,

Total increment in sum = 10 × 3 = 30.

New sum becomes = 360 + 30 = 390

Mean of New set of numbers =

Given,

Mean weight of 6 boys = 48 kg

Sum of their weights = 6 × 48 = 288 kg

Sum of weight of 5 boys = 51 + 45 + 49 + 46 + 44 = 235 kg

Hence,

Weight of sixth boy = total weight – weight of 5 boys = 288 – 235 = 53 kg.

Given,

Mean of marks of 50 students = 39

Sum of marks = 50 × 39 = 1950

It is given that he misread 43 as 23, so,

Correct sum = Sum obtained + (43 – 23) = 1950 + 20 = 1970.

Correct Mean = =

Given,

Mean of 100 items = 64

Sum of items = 100 × 64 = 6400.

It is given that 26 and 9 were misread as 36 and 90, so,

Correct sum = 6400 + (36 – 26) + (90 – 9) = 6400 + 91 = 6491

Correct Mean =

Given,

Mean of 6 numbers = 23

Sum of numbers = 23 × 6 = 138

Mean of 5 numbers = 20 (Given)

Sum of 5 numbers = 20 × 5 = 100

So,

The excluded number = Sum of 6 numbers – Sum of 5 numbers = 138 – 100 = 38.

Given,

Mean marks of 7 students = 226

Total marks of 7 students = 7×226 = 1582

Total marks of 6 students = (340 + 180 + 260 + 56 + 275 + 307)

Marks of the 7^th students = (total marks of 7 students) - (total marks of 6 students)

= 1582 – 1418 = 164

Hence the marks of the 7^th student are 164.

Mean weight of 34 students = 46.5kg

Total weight of 34 students = 34×46.5 = 1581 kg

If the weight of a teacher is included then mean arises = 500gm = 0.5 kg

So new mean = 46.5 + 0.5 = 47kg

Total weight = 47×35 = 1645 kg

Weight of a teacher = 1645 – 1581 = 84kg

Hence the weight of a teacher is 64 kg.

Mean weight of 36 students = 41kg

Total weight of 36 students = 36×41 = 1476 kg

If one student leaves the class then mean decreases = 200gm = 0.2 kg

So new mean = 41 - 0.2 = 40.8kg

Total weight of 35 students = 40.8×35 = 1428 kg

Weight of the student = 1476 – 1428 = 48kg

Hence the weight of the student who left the class is 48 kg.

Mean weight of 39 students = 40kg

Total weight of 39 students = 39×40 = 1560 kg

If a new student is admitted to the class, the average decrease by = 200g = 0.2 kg

So new average weight = 40 - 0.2 = 39.8kg

Total weight of 39 students and 1 new student = 39.8×40 = 1592 kg

Weight of new student = 1592 – 1560 = 32kg

Hence the weight of the new student is 32 kg.

Average monthly salary of 20 workers = 7650 Rs

Total salary of 20 workers per month = 20×7650 = 153000Rs

The wage of one member of the group = 8100 Rs

New average salary of 20 workers and manager = 8200 per month

Let suppose salary of manager is x

The new average (20 workers + manager) = (given)

= 153000 + x = 8200 × 21

153000 + x = 172200

x = 172200 – 153000

x = 19200 Rs

Hence salary of manager is 19200Rs.

Average monthly wage of 10 persons = 9,000Rs

Total wage of 10 person = 10×9000 = 90,000 Rs

The wage of one member of the group = 8100 Rs

When one member leaves the group then total wage of 9 person = 90000 -81900 Rs

The wage of new member = 7200 Rs

When new member join to the group then total wage of 10 person = 81900 + 7200 = 89100Rs

New monthly average =

Hence new monthly average is 8910 Rs.

Average monthly consumption of petrol for 7 months = 330 liters

Total consumption for 7 months = 7×330 = 2310 liters

Average consumption for next 5 months = 270 liters

Total consumption for next 5 months = 5×270 = 1350 liters

Total consumption during whole year = 2310 + 1350 = 3660 liters

Now,

Average consumption per month during whole year =

= 305 liter/per month

Mean of 15 numbers = 18

Sum of 15 numbers = 18×15 = 270

Mean of remaining 10 numbers = 13

Sum of remaining 10 numbers = 10×13 = 130

Mean of 25 numbers =

=

Hence mean of 25 numbers is 16.

Mean weight of 60 students = 52.75 kg

Sum of weight of 60 students = 60×52.75 = 3165 kg

Mean weight of 25 students of them = 51kg

Sum of weight of 25 students = 51×25 = 1275 kg

Sum of weight of remaining 35 students = Sum of weight of 60 students - Sum of weight of 25 students

= 3165 – 1275 = 1890 kg

Mean of remaining 35 students =

Hence mean of remaining students is 54kg.

Let the average weight of 10 oarsmen be x kg

Sum of weight of 10 oarsmen = 10x kg

When new man is added to the crew in place of another man then the average increases by = 1.5 kg

So, new average weight = (x + 1.5)kg

New average weight =

10x – 58 + weight of new mean = 10(x + 1.5)

10x – 58 + weight of new mean = 10x + 15

Weight of new mean = 15 + 10x – 10x + 58

Weight of new mean = 73 kg

Hence Weight of the new man in the crew is 73kg.

Mean of 8 numbers = 35

Sum of 8 numbers = 8×35 = 280

New mean = 35-3 = 32

Sum of 7 numbers = 32×7 = 224

Excluded number = sum of 8 numbers – sum of 7 numbers

= 280 – 224 = 56

So, the excluded number is 56.

Calculated mean of 150 items = 60

Incorrect sum of 150 items = 60×150 = 9000

Correct sum of 150 items = (incorrect sum) - (incorrect items) + (correct items)

= 9000 – (52 + 8) + (152 + 88)

= 9000 – 60 + 240

= 9240 – 60 = 9180

Correct mean =

So, correct mean is 61.2.

Mean of 31 results = 60

Sum of 31 results = 31×60 = 1860

Mean of first 16 results = 58

Sum of first 16 results = 58×16 = 928

Mean of last 16 results = 62

Sum of last 16 results = 62×16 = 992

16^th result = sum of first 16^th results + sum of last 16 results – sum of 31 results

= 992 + 928 – 1860 = 60

So, 16^th result is 60.

Given,

Mean of 11 number = 42

So, Sum of numbers = 11 × 42 = 462.

Mean of first 6 numbers = 37.

Hence, sum of first six numbers = 6 × 37 = 222.

Mean of last 6 numbers = 46

Sum of last 6 numbers = 6 × 46 = 276.

So,

The 6^th number = (sum of first 6 numbers and last 6 numbers) – (Sum of 11 numbers)

= (222 + 276) – 462 = 498 – 462 = 36.

The 6^th number is 36.

Given,

Mean weight of 25 students = 52 kg

Total weight of 25 students = 25 × 52 = 1300 kg

Mean weight of first 13 students = 48 kg

Total weight of First 13 students = 13 × 48 = 624 kg

Mean weight of last 13 students = 55 kg

Total weight of last 13 students = 13 × 55 = 715 kg

So,

Weight of 13^th student = (sum of weight of first 13 and last 13 students) – (wt. of 25 students)

= (624 + 715) – 1300 = 1339 – 1300 = 39 kg

Hence, Weight of 13^th student = 39 kg.

Given,

Mean score of 25 observations = 80

Total score of 25 observations = 25 × 80 = 2000.

Mean score of another 55 observations = 60

Total score of 55 observations = 55 × 60 = 3300.

Total observations = 55 × 25 = 80

Mean score of whole set of observations =

Hence, Mean score of whole set is 66.25.

Given,

Marks obtained in English = 36

Marks obtained in Hindi = 44

Marks obtained in Maths = 75

Marks obtained in Science = x

Average marks of all subjects = 50

We know that,

Average marks =

= 50 =

= 50 × 4 = 155 + x

= x = 200 – 155 = 45.

Hence, Marks in science = 45.

Given,

Mean salary of 75 workers = Rs.5680

Total salary of 75 workers = 75 × 5680 = Rs.426000

Mean salary of 25 workers of them = Rs.5400

Total salary of 25 workers = 25 × 5400 = Rs.135000

Mean salary of 30 workers in them = Rs.5700

Total salary of 30 workers = 30 × 5700 = Rs.171000

Let Mean salary of remaining 20 workers = Rs. X

Total salary of 20 workers will be = Rs.20x

So, we have,

Salary of 75 workers = salary of (25 workers + 30 workers + 20 workers)

426000 = 135000 + 171000 + 20x

= 20x = 426000 – 306000 = 120000

= x =

Hence, Salary of 20 workers = Rs.6000.

Given,

Speed of ship in sailing direction (x) = 15 km/h

Speed of ship in sailing back (y) = 10 km/h

So, By using direct formula,

Average speed of ship in whole journey =

Hence, Average speed of ship = 12 km/h

Given,

Average weight of 50 students in class = 44 kg

Total weight of 50 students = 50 × 44 = 2200 kg

Number of girls = 50 – number of boys = 50 – 40 = 10

Average weight of 10 girls = 40 kg

Total weight of 10 girls = 10 × 40 = 400 kg

Hence,

Total weight of 40 boys in class = total weight of class – weight of girls

= 2200 – 400 = 1800 kg

Average weight of boys =

Hence, Average weight of 40 boys = 45 kg.

Let’s draw the table and calculate the total frequency.

To calculate mean we need the relative value of variables which is x_i×f_i. To attain x_i×f_{i .}we have to multiply the value of variables (x_i)to the frequency of the value (f₁).

By putting the formula of;

Mean =

= 118.50

Let’s draw the table and calculate the relative value of variables.

By putting the formula of mean we get;

Mean =

= 64.25 kg

So the mean weight of the workers will be 64.25kg.

Let’s draw the table and calculate the relative value of variables∑xi×fi

By putting the formula of mean we get;

Mean =

= 17.45 years

So the mean age of the students will be 17.45years.

Let’s draw the table and calculate the relative value of variables.

By putting the formula of mean we get;

Mean =

= 55

So the mean of the given distribution will be 55.

Let’s draw the table and calculate the relative value of variables ∑xi×fi

By putting the formula of mean we get;

Mean = (given)

So we have,

303 + 9P = 8 (41 + P)

303 + 9P = 328 + 8P

9P-8P = 328-303

P = 25

Hence the value of the P is 25.

Let’s draw the table and calculate the relative value of variables ∑xi×fi

By putting the formula of mean we get;

Mean = = 28.25 (given)

28.25(50 + P) = 1445 + 25P

28.25×50 + 28.25×P = 1445 + 25P

1412.50 + 28.25P = 1445 + 25P =

28.25P – 25P = 1445 – 1412.50

3.25P = 32.5

= 10

Hence the value of P is 10.

By putting the formula of mean we get;

Mean = = 16.6 (given)

16.6×100 = 1228 + 24P

1660 = 1228 + 24P

24P = 1660-1228

24P = 432

So, the value of P is 18.

Let’s draw the table and calculate the relative value of variables ∑xi×fi

We have,

∑f_i = 120 (given)

∑f_i = 68 + f₁ + f₂ = 120

f₁ + f₂ = 120 - 68

f₁ + f₂ = 52…. Equation (i)

Now we have,

Mean = 50 (given)

Mean = = 50

50 =

50×68 + f₁ + f₂ = 3480 + 30f₁ + 70f₂

3400 + 50 f₁ + 50f₂ = 3480 + 30f₁ + 70f₂

50f₁ - 30f₁ + 50 f₂ -70f₂ = 3480 – 3400

20f₁-20f₂ = 80

20(f₁-f₂) = 80

f₁-f₂ = ……. Equation (ii)

by adding equation (i) and (ii) we get,

∑f₁ = 56

From equation (ii)

f₁-f₂ = 4

28 – f₂ = 4

f₂ = 28-4 = 24

Hence missing frequencies are f₁ = 28 and f₂ = 24.

Let assume mean be A = 900

Now we can arrange data in below format

We have,

Mean =

= 891.21

Hence the mean is 891.21

Let the assumed mean be A = 67

So, we arrange the given data as under :

We know,

Mean =

Hence, The mean height of the plants = 67.45 cm.

Let the assumed mean be A = 21

= h = x₂ – x₁ = 19 – 18 = 1

Thus, we prepare the table given below:

We Know that,

Mean =

Hence, The Mean value of given distribution = 20.62.

Let the assumed Mean be A = 1400

= h = x₂ – x₁ = 600 – 200 = 400

Thus, we prepare the table as below:

We know that,

Mean =

=

Hence, The Mean height is 984.51.

To find median first we must arrange the data in either ascending order or descending order.
(i) By arranging the data in ascending order

We have;

2, 2, 3, 5, 7, 9, 9, 10, 11

Total number of observations, N = 9 (Odd number)

Median = value of the term

= value of 5^th term

Median = 7

(ii) By arranging the data in ascending order

We have;

6, 8, 9, 15, 16, 18, 21, 22, 25

N = 9 (odd number)

Median = value of the term

= value of 5^th term

Median = 16

(iii) By arranging the data in ascending order

We have;

6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

N = 11 (Odd number)

Median = value of the term

= value of 6^th term

Median = 16

(iv) By arranging the data in ascending order

We have;

0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10

N = 13 (Odd number)

Median = value of the term

= value of 7^th term

Median = 4

(i) By arranging the data in ascending order

9, 10, 17, 19, 21, 22, 32, 35

We have;

N = 8 (even number)

Median =

Median = 20

(ii) By arranging the data in ascending order

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

We have;

N = 10 (even number)

Median =

Median = 61.5

(iii) By arranging the data in ascending order

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81

We have;

N = 12 (even number)

Median =

Median = 16

By arranging the data in ascending order

We have;

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

N = 15 (odd number)

Median = value of the term

=

= value of 8^th term

= 23

Hence the Median marks are 23.

By arranging the data in ascending order

We have;

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

N = 9 (0dd number)

Median = value of the term

=

= value of 5^th term

Median height = 147.3 cm

By arranging the weight in ascending order,

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

We have;

N = 8 (even number)

Median =

Median weight = 13.85 kg

By arranging the ages of teachers in ascending order

32, 34, 36, 37, 40, 44, 47, 50, 53, 54

We have;

N = 10 (even number)

Median =

Median age= 42 years

By arranging the data in ascending order,

10, 13, 15, x + 1, x + 3, 30, 32, 35, 41

We have;

N = 10 (even number)

Median = 24 (given)

Median =

24

24

48 = 2x + 4

2x = 44

x = 22

Hence the value of x = 22

By arranging the data in ascending order,

N = 41 (odd number)

Median = value of the term

=

Median is the 21^st term which is 50.

As we can see student from 20^th to 28^th comes under the weight of 50 so the median weight of the student is 50 kg.

By arranging the data in the following format,

N = 37 (odd number)

Median = value of the term

=

Median is the 19^th observation, which comes under the observations from 18^th to 21^st.

So the median value of observations is 22.

By arranging the data in table form,

N = 43 (odd number)

Median = value of the term

=

Median is the marks of 22^nd student as we know students from 11^th to 26^th got 25 marks.

So the median value of observations is 25.

By arranging the data in ascending order,

We have,

N = 50 (even number)

Median =

As we can see height of 25^th student is 154 and 26^th student has the height of 155 cm

So the median height is 154.5cm

By arranging data in ascending order,

We have,

N = 60 (even number)

Median =

As we can see variate from 18^th to 30^th has the size of 20 and from 31^st to 34^th has the size of 23.

Hence median is 21.5.

By assigning the given data in ascending order

0, 0, 1, 2, 3, 4, 5, 6, 6, 6, 6

As we can clearly see that 6 has occurred maximum times, and we know that mode is the most appeared value in the set of data values, so 6 is the mode of the given data.

By arranging the given data in ascending order,

15, 20, 22, 23, 25, 25, 27, 40

As we can clearly see that 25 has occurred maximum times, so 25 is the mode of the given data.

By arranging the given data in ascending order,

1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9

As we can clearly see that 9 has occurred maximum times, so 9 is the mode of the given data.

By arranging the given data in ascending order,

9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60

As we can clearly see that the player has scored 50 maximum times, so 50 is the modal of the given data.

To calculate mode by empirical method we have to arrange the data in the form of frequency table,

As we can see 50 has occurred the maximum times so 50 is the mode of the given data.

By arranging the given data in ascending order we will get,

10, 10, 11, 11, 12, 12, 13, 14, 15, 17

We have;

N = 10 (even number)

Median =

Now,

∑fx = 125 and ∑f = N = 10

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×12 - 2×12.5

= 36 – 25

= 11

We can draw the table in the given format

We have;

N = 25 (odd number)

Median =

= value of 13^th term

= 13

Now,

∑fx = 332 and ∑f = N = 25

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×13 - 2×13.28

= 39 – 26.56

= 12.44

Draw the table as below,

We have;

N = 34 (even number)

Median =

Now,

∑fx = 407 and ∑f = N = 34

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×12 - 2×11.97

= 36 – 23.94

= 12.06

Draw the table as below,

We have;

N = 40 (even number)

Median =

Now,

∑fx = 1161 and ∑f = N = 40

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×30 - 2×29

= 90 - 68

= 32

Draw the table as below,

We have;

N = 50 (even number)

Median =

Now,

∑fx = 2935 and ∑f = N = 50

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×59.7 - 2×58.7

= 179.1 – 117.4

= 61.1 kg

We can draw the table as below,

We have;

N = 80 (even number)

Median =

Now,

∑fx = 1992 and ∑f = N = 80

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×28 - 2×24.9

= 84 –49.8

= 34.2

Draw the tables as given below,

We have;

N = 106 (even number)

Median =

Now,

∑fx = 2650 and ∑f = N = 106

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×25 - 2×25

= 75 - 50

= 25

Draw the table as below,

We have;

N = 15(0dd number)

Median = value of the term

= value of 8^th term

= 53

Now,

∑fx = 796 and ∑f = N = 15

Mean =

Hence,

Mode = 3(median) – 2 (mean)

= 3×53 - 2×53.06

= 159 – 106.12

= 52.88

So, Mean, median and mode are 53, 53.06 and 52.88

Range = maximum value – minimum value

= 32 – 12 = 20

؞ Range = 20

Class interval = 100 – 120 (given)

We know that,

Class marks = 1/2 (upper class limit + lower class limit)

= 1/2 (100 + 120) = 1/2 ×220 = 110

So, Class marks = 110.

The number 20 is included in class-interval 20 – 30,

Because,

Class interval 20 – 30 contains values, which are either equal to 20 or less than 30.

Class marks = 20 (given)

Class size = 20 – 15 = 5

Lower class limit = (20 – 5/2 ) = 20 – 2.5 = 17.5

Upper class limit = (20 + 5/2) = 20 + 2.5 = 22.5.

Hence,

Class mark 20 will lie in class 17.5 – 22.5

Let the upper limit of class = u

Let lower limit of class = l

Mild value = 10 (given)

We know that,

Mild value = 1/2 (u + l)

10 = 1/2 (u + l)

= u + l = 20..............(i)

Width of class = 6 (given)

= u – l = 6 ................(ii)

Subtracting equation (ii) from equation (i), we get,

2l = 14

= l = 14/2 = 7.

lower limit = 7.

We know that,

Mid-point = 1/2 (upper class limit + lower class limit)

m = 1/2 (upper class limit + lower class limit)

upper class limit = u (given)

u + lower class limit = 2m

lower class limit = 2m – u .

Width of each class = 5 (given)

Total no. Of classes = 5 (given)

Lower class limit of lowest class = 10 (given)

Total width of class = 5 × 5 = 25

upper class limit of highest class = 10 + 25 = 35.

Given,

Mid-point of class = m

We know that,

Mid-point = upper class limit + lower class limit)

m = upper class limit + lower class limit)

lower class limit = l (given)

upper class limit + l = 2m

upper class limit = 2m – l .

Given,

Mid value of class interval = 42

Class size = 10

Let the upper limit of class = u

Let lower limit of class = l

We know that,

Mild value = 1/2 (u + l)

42 = 1/2 (u + l)

= u + l = 84..............(i)

Class size = u – l

= u – l = 10 (given)

Adding equation (ii) & equation (i), we get,

2u = 94

⇒ u = 94/2 = 47.

From equation (ii)

u – l = 10

l = 47 – 10 = 37.

Mean of 5 observations =

Mean = 11 (given)

⇒

⇒ 5x + 20 = 55

⇒ 5x = 55 – 20 = 35

⇒ x = 7.

Mean of 5 observations =

⇒ Mean of 5 observations

⇒ Mean = 9 (given)

⇒

⇒ 5x + 25 = 45

⇒ 5x = 45 – 25 = 20

⇒ x = 4.

Now,

Mean of three observations =

.

Mean of x₁, x₂, x_3.........x_n =

..............(i)

Now,

)

⇒ _i - =

⇒ = 0.

Let the observations be x₁, x₂, x_3.........x_n

Mean = .....(i)

If each observation is increased by 5, we get,

x₁ + 5, x₂ + 5, x₃ + 5_.........x_n + 5

New mean

⇒ Mean =

Thus, the mean is also increased by 5.

(given)..............(i)

(given)..............(ii)

Now, their combined mean is,

From equation (i) and eq.(ii)..

= )

It is given that ... (i)

Required mean =

From equation (i)

Sum of all terms =

Total number of factors =

Required mean

Given:

Mean weight of 6 boys = 48 kg

Let the weight of the 6^th boy = x kg

Mean weight =

⇒

⇒ 235 + x = 48 × 6 = 288

⇒ x = 288 – 235 = 53 kg

Hence, the weight of the 6^th boy = 53 kg.

Given,

No. of students = 50

Mean marks secured by them = 39

Incorrect sum of marks secured = 39 × 50 = 1950.

Correct sum = incorrect sum – (incorrect marks) + correct marks

= 1950 – 23 + 43 = 1970

Correct mean = 39.4.

Given,

No. of items = 100

Mean of them = 64

Incorrect sum of 100 items = 64 × 100 = 6400.

Correct sum = incorrect sum – (incorrect marks) + correct marks

= 6400 – (26 + 9) + (36 + 90)

6400 – 35 + 126 = 6491.

Correct mean = 64.91.

Given:

No. of observations = 100

Mean of them = 50

Sum of observations = 100 × 50 = 5000

It is given that one of observation 50, is replaced by 150.

New sum = 5000 – 50 + 150 = 5100.

Resulting mean =

Mean of 25 observations 36.

Sum of 25 observations = 25 × 36 = 900.

Mean of first 13 observations = 32

Sum of first 13 observations = 13 × 32 = 416.

Means of last 13 observations = 40

Sum of last 13 observations = 13 × 40 = 520.

Hence,

13^th observation = (sum of first 13 observation + sum of last 13observations) – sum of 25 observations.

13^th observation = 416 + 520 – 900 = 936 – 900 = 36

Hence, 13^th observation = 36.

Let’s take 50 numbers as = n₁, n₂, …………, n₅₀

And mean = x

Then sum = 50x

Now each number is subtracted from 53,

We have,

53 - n₁, 53 - n₂, …………, 53 - n₅₀

Sum becomes = 53 × 50 – (n₁ + n₂ + …………, + n₅₀)

Given Mean = - 3.5

So,

= 2650 – 50x = 50 × (- 3.5) = -175

50x = 2875

So, mean of the given numbers will be 56.5

Given: Mean = 8

Mean =

⇒ 8 =

On cross multiplying both the sides we get,

⇒ 303 + 9p = 8(41 + p)

⇒ 303 + 9p = 328 + 8p

⇒ 9p – 8p = 328 – 303

⇒ p = 25

Arrange the scored runs in ascending order, we get,

0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56

Here, N = 11 (odd)

Hence,

Mean value = value of term = ^th term = 29

Arrange the weights in ascending order, we get,

31, 35, 36, 38, 40, 44, 50, 52, 55, 60

Here, N = 10 (even)

Hence,

Mean weight = value of

=

⇒ Mean weight

∴ Mean weight = 42 kg

Arrange the numbers in ascending order, we get,

3, 4, 4, 5, 6, 7, 7, 7, 12.

Here, N = 9 (odd)

Hence,

Median = value of term =

⇒ 5^th term which is 6.

Arrange the given numbers in ascending order,

We get,

22, 34, 39, 45, 54, 54, 56, 68, 78, 84

Here, N = 10 (even)

Hence,

Median =

= = 54

Let’s prepare a table,

Mode is the number which appeared maximum numbers of times.

In this given series 15 has the highest frequency,

So,

Mode = 15

When we draw a frequency polygon of a continuous frequency distribution, we need to plot the class marks of the given classes on the x-axis.

Formula of range = maximum value – minimum value

So, we have,

Maximum marks = 100

Minimum marks = 46

Range = 100 – 46 = 54

Formula to calculate the class mark, = 1/2 [ Upper class limit + lower class limit]

Upper class = 150

Lower class = 130

So, we get,

= 140

Given,

The mean of five numbers = 30

Now to calculate the sum of these five numbers we have to multiply the mean by 5,

So, we get,

Sum of five numbers = 5 × 30 = 150

Change in the mean when one number is excluded, it become = 28

Sum of remaining four numbers = 4 × 28 = 112

Now to find out the excluded number we have to subtract the sum of four numbers from the sum of five numbers,

We get,

The excluded number = sum of five numbers - sum of four numbers

= 150 – 112 = 38

Given data = 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39

We have n = 10 (even)

Median = 1/2 [Value of (n/2)th term + (n/2 + 1)th term]

Now, Median = 24 (given)

24 = 1/2 [ Value of 5^th term + Value of 6^th term]

24 × 2 = (x + 2) + (x + 4)

48 = 2x + 6

48 – 6 = 2x

42 = 2x

So,

x = 42/2 = 21

Let suppose 15 numbers are = n₁, n₂, …….. , n₁₅

Given mean = 25

25 =

⇒ n₁ + n₂ + …… + n₁₅ = 25 × 15

⇒ n₁ + n₂ + …… + n₁₅ = 375 … [equation (i)]

After subtracting 6 from each number the mean = 19

So, we have,

(n₁ – 6), (n₂ – 6), …….. (n₁₅ – 6)

From equation (i)

⇒ Mean = 25 – 6 = 19

It means assertion (A) is true,

Reason (R) ⇒ by empirical formula which is,

Mode = 3(median) – 2(mean) is true.

But Reason (R) is not correct explanation of assertion (A).

By arranging the data in ascending form, we get,

51, 55, 58, 60, 62, 64, 65, 68, 70, 75, 80, 82, 85, 90, 95

Here, n = 15 (odd)

Median = value of th term

= Value of 8^th term

= 68

Means assertion is correct

Reason (R) is true and it is also the correct explanation of assertion (A).

Arrange the data in ascending form, we get,

2, 3, 3, 9, 9, 9, 16

In this given data 9 appears maximum numbers of time.

Mode = 9

Given mode = 16

So, the given statement is false.

By arranging the data in ascending form, we get,

3, 5, 14, 18, 20

Median = value[(n + 1)/2 the term]

Median = value of 3^rd term

Median = 14

Given median = 18

So, the given statement is false.

By arranging the data in ascending form, we get,

1, 1, 2, 3, 4, 5, 6, 7, 8, 9

n = 10 (even)

Median = 1/2 [Value of (n/2)th term + (n/2 + 1)th term]

= 1/2 [value of 5th term + value of 6th term]

= 1/2 (4 + 5)

= 9/2 = 4.5

Given median = 7

So, the given statement is false.

(A) → (q),

(B) → (r),

(C) → (s),

(D) → (p)

Explanation:

(A) First 10 odd numbers are = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Sum of these numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

n = 10

Mean = 100/10 = 10

(B) First 10 even numbers are = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Sum of these numbers = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110

n = 10

Mean = 110/10 = 11.0

(C) First 10 prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Sum of these numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129

n = 10

Mean = 129/10 = 12.9

(D) First 10 composite numbers are = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Sum of these numbers = 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 112

n = 10

Mean = 112/10 = 11.2

Class marks of the given frequency distribution are = 47, 52, 57, 62, 67, 72, 77

(i) Class size = 52 – 47 = 5

(ii) Class size = 5

Mid value = = 2.5

Class marks = 52

Upper class limit = 52 + 2.5 = 54.5

Lower class limit = 52 – 2.5 = 49.5

(iii) Here classes are in the exclusive form,

So true class limit for class mark 52 is 49.5 – 54.5

(A) When n is odd then formula for median is = Value of th term

So, the given statement is true.

(B) When n is even then formula for median become;

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

So, the given statement is true.

(C) Mode is the number or item which occur maximum numbers of time or which have the highest frequency.

So, the given statement is true.

(D) The correct formula is;

Mode = 3(median) – 2(mean)

So, the given statement is false.

(A) Given,

…..(i)

Now take L.H.S

So, given statement is true.

(B) Given,

Observations are = (x₁ + a), (x₂ + a), ……. (x_n + a)

⇒

From equation (i) we get,

So, given statement is true.

(C) Mean of x₁, x₂, …… x_n =

Given mean =

Observations are = ax₁, ax₂, ……..ax_n

⇒

∴ Mean = a ….. [From equation (i)]

So, the given statement is true.

(D) Let suppose the observations are =

……….( Given)

Sum = = nM…………..(i)

And

ax₁ + ax₂ + …………. + ax_n = aM

Then,

[from equation (i)]

a + M = aM (Which is not true)

(A) Given observations are = 4, 6, x, 8, 10, 13

Mean =

Now, given mean = 8

Sum of the observations = 4 + 6 + x + 8 + 10 + 13 = 8 × 6

⇒ 41 + x = 48

∴ x = 48 – 41 = 7

So, x = 7

The given statement is true.

(B) Given array = 59, 62, 65, x, x + 2, 72, 85, 99

n = 8 (even)

Median = 1/2 [value of (n/2)th term + value of (n/2 + 1)th term]

= 1/2 [value of 4th term + value of 5th term]

Median =

Given median = 67

67 = 2x + 2 = 67 × 2

2x + 2 = 134

2x = 134 – 2

X = = 66

So, x = 66

The given statement is true

(C) Given,

Observations are = 1, 3, 5, 7, 5, 2, 7, 5, 9, 3, p, 11

Mode = 5

Value of p = 7

We know that mode is the item which has highest frequency and in given statement 5 is the mode. If p = 7 then it will become the number with highest frequency. It means p can’t be equals to 7 as mode is 5.

So, the given statement is false.

(D) Given;

Mean of 10 observations = 15

Mean of 15 observations = 18

Mean of all 25 observation = 16.8

Now the sum of the observations;

Sum of 10 observations = 10 × 15 = 150

Sum of 15 observation = 15 × 18 = 270

Sum of 25 observation = Sum of 10 observations + Sum of 15 observations

Mean of 25 observations =

⇒ Mean =

⇒ Mean = 420/25

∴Mean = 16.8

So, the given statement is true.