⛪Home⇐ Class X Statistics


Mind Maps
 
 
 


Class X - Maths: Statistics





Contents
  • Mean
  • Mean Of Grouped Data
  • Class Limits in Exclusive and Inclusive Form
  • construct a pie chart
  • Frequency Distribution of Un-grouped and Grouped Data
  • Mode
  • Real Life Statistics
  • Mean of the Tabulated Data
  • Median
  • Construction of Bar Graphs
  • Properties of Arithmetic Mean
  • Average
  • Word Problems on Average
Mean

Mean or average or arithmetic mean is one of the representative values of data. We can find the mean of observations by dividing the sum of all the observations by the total number of observations

Mean of raw data:

If x1, x2, x3, ......xn are n observations, then
Arithmetic Mean =` (x_1, x_2, x_3, ......x_n)/n`
= `[sum(x_i)]/n`
`sum` (Sigma) is a Greek letter showing summation

1. Weights of 6 boys in a group are 63, 57, 39, 41, 45, 45. Find the mean weight.
Solution:

Number of observations = 6
Sum of all the observations = 63 + 57 + 39 + 41 + 45 + 45
= 290
Therefore, arithmetic mean = `(290)/6` = 48.3

Mean of tabulated data:

If x1, x2, x3, ....xnare n observations, and f1, f2, f3, .....fn represent frequency of n observations.
Then mean of the tabulated data is given by
= `(f_1 x_1 + f_2 x_2 + f_3 x_3 + .... f_n x_n)/(f_1 + f_2 + f_3 + ....f_n)`
= `[sum(f_ix_i)]/[sum f_i] `


2. A die is thrown 20 times and the following scores were recorded 6, 3, 2, 4, 5, 5, 6, 1, 3, 3, 5, 6, 6, 1, 3, 3, 5, 6, 6, 2.
Prepare the frequency table of scores on the upper face of the die and find the mean score.

Number on the upper
face of die
Number of times it occurs
(frequency)
fixi
1 2 1 `xx`2 = 2
2 2 2 `xx` 2 = 4
3 5 3`xx` 5 = 15
4 1 4 `xx` 1 = 4
5 4 5 `xx` 4 = 20
6 6 6` xx` 6 = 36

Therefore, mean of the data = `[sum(f_ix_i)]/[sum f_i] `
= `(2 + 4 + 15 + 4 + 20 + 36)/(20)`
= `(81)/(20)`
= 4.05


Mean of the data=`[sum(f_ix_i)]/[sum f_i] `


3. If the mean of the following distribution is 9, find the value of p.

X 4 6 p + 7 10 15
f 5 10 10 7 8

Solution:
Calculation of mean

xi fi xifi
4 5 20
6 10 60
p + 7 10 10(p + 7)
10 7 70
15 8 120

∑fi = 5 + 10 + 10 + 7 + 8 = 40
∑ fixi = 270 + 10(p + 7)
Mean =`[sum(f_ix_i)]/[sum f_i] `
9 = `(270 + 10(p + 7))/(40)`
270 + 10p + 70 = 9 `xx` 40
340 +10p = 360
10p = 360 - 340
10p = 20
p = `(20)/(10)`
p = 2


Mean Of Grouped Data

While calculating the mean of the grouped data, the values x1, x2, x3, ....xn are taken as the mid-values or the class marks of various class intervals. If the frequency distribution is inclusive, then it should be first converted to exclusive distribution.


4. The following table shows the number of plants in 20 houses in a group.Find the mean number of plans per house

Number of Plants (0 - 2) (2 - 4) (4 - 6) (6 - 8) (8 - 10) (10 - 12) (12 - 14)
Number of Houses 1 2 2 4 6 2 3

Solution:
We have

Number of PlantNumber of Houses
(fi)
Class Mark
(xi)
fixi
(0 - 2) 1 1 `(1 xx 1)` = 1
(2 - 4) 2 3 `(2 xx 3)`= 6
(4 - 6) 2 5 `(2 xx 5)`= 10
(6 - 8) 4 7 `(4 xx 7)`= 28
(8 - 10) 6 9 `(6 xx 9)` = 54
(10 - 12)2 11 `(2 xx 11)` = 22
(12 -14) 3 13 `(3 xx 13)` = 39

∑fi = 1 + 2 + 2 + 4 + 6 + 2 + 3 = 20
∑fixi =1 + 6 + 10 + 28 + 54 + 22 + 39 = 160
Therefore, mean =`[sum(f_ix_i)]/[sum f_i] `
= `(160)/(20) = 8` plants

Class Limits in Exclusive and Inclusive Form

In class boundaries and class limits in exclusive and inclusive form we will mainly discuss about;

  • Class limits and class boundaries in exclusive and inclusive forms
  • Class interval, class mark, range of the data

Class limits in exclusive and inclusive form:

  • In exclusive form, the lower and upper limits are known as true lower limit and true upper limit of the class interval.
    Thus, class limits of 10 - 20 class intervals in the exclusive form are 10 and 20.
  • In inclusive form, class limits are obtained by subtracting 0.5 from lower limit and adding 0.5 to the upper limit.
    Thus, class limits of 10 -20 class interval in the inclusive form are 9.5 - 20.5.
  • Class size: Difference between the true upper limit and true lower limit of a class interval is called the class size. Class size remains the same for all class intervals. For the class interval 10 - 20 Class size is 10, i.e., (20 -10 = 10)
  • Class mark: Mid-value of each class interval is called its class mark. Class mark = `1/2` (Upper limit + Lower limit) For the class interval 10 - 20 Class mark = `(10 + 20)/2` =`(30)/2` = 15
  • Range: The difference between the maximum value and the minimum value of the observation is called the range.
    In the above data, the maximum value is 24 and the minimum value is 0.
    Therefore, range = 24 - 0 = 24

Frequency Distribution of Ungrouped and Grouped Data

Frequency distribution of ungrouped data:

Given below are marks obtained-by 20 students in Maths out of 25. 21, 23, 19, 17, 12, 15, 15, 17, 17, 19, 23, 23, 21, 23, 25, 25, 21, 19, 19, 19

Statistics

Frequency distribution of grouped data:

  • The presentation of the above data can be expressed into groups. These groups are called classes or the class interval.
  • Each class interval is bounded by two figures called the class limits.
MarksNumber of Students
(Frequency)
(0 - 10)0
(10 - 20)11
(20 - 30)9

Note: The lower value of a class interval is called lower limit and upper value of that class interval is called the upper limit. Thus, each class interval has lower and upper limits.

Exclusive form of data

  • This above table is expressed in the exclusive form.
  • In this, the class intervals are 0 - 10, 10 - 20, 20 - 30. In this, we include lower limit but exclude upper limit.
  • So, 10 - 20 means values from 10 and more but less than 20.
  • 20 -30 would mean values from 20 and more but less than 30.
  • Data in the inclusive form:
    Marks obtained by 20 students of class VIII in Maths text are given below.
  • 23, 0, 14, 10, 15, 3, 8, 16, 18, 20, 1, 3, 20, 23, 24, 15, 24, 22, 14, 13

Let us represent this data in the inclusive form.

Marks Number of Students
(Frequency)
(0 - 10) 6
(11 - 20) 9
(21 - 30) 5
  • Here, also we arrange the data into different groups called class intervals, i.e., 0 - 10, 11 - 20, 21 -30.
  • 0 to 10 means between 0 and 10 including 0 and 10.
  • Here, 0 is the lower limit and 10 is the upper limit. 11 to 20 means between 11 and 20 including 11 and 20.
  • Here, 11 is the lower limit and 20 is the upper limit.
  • When the data is expressed in the inclusive form, it is converted to exclusive form by subtracting 0.5 from lower limit and adding it to upper limit of each class interval.
  • 11 - 20 is expressed in the inclusive form which can be changed and taken as 10.5 - 20.5 which is the exclusive form of the data.
  • Similarly, 21 -30 can be taken as 20.5 - 30.5.
Mode

For a set of given observation, mode is that observation which occurs maximum number of times.
1. Find the mode of the given set of number 2, 2, 3, 5, 4, 3, 2, 3, 3, 5
Solution:

Arranging the number with same values together, we get
2, 2, 2, 3, 3, 3, 3, 4, 5, 5
We observe that 3 occurs maximum number of times, i.e., four.
Therefore, mode of this data is 3.

Note:A data may not have a mode.


2. The data 3, 4, 1, 5, 4, 2 has no mode because no number occurs more number of times then any other number.

A data may have more than one mode.


3. The data 2, 5, 1, 3, 5, 7, 6, 3, 8 have two modes 3 and 5.

Therefore, each is repeated two times which is maximum


4. The height of 50 plants in a garden are given below. Find the mode of the data.

Height (cm)10 25 30 40 45
Number Plants 13 15 12 8 2

Solution:
The frequency of 25 is maximum.
So, the mode of this data is 15.


Real Life Statistics

  • In real life statistics, we come across numerical data in the newspapers, magazines and television regarding different aspects like increase or decrease in population, profit made by a company in different years, weather report, etc.
  • These numerical facts are also represented by graphs which are easy to understand.

Definition of statistics:

  • The term statistics is derived from a Latin word status meaning condition. The branch of mathematics which deals with the collection, presentation, analysis and interpretation of the numerical data is called statistics.
  • We have already learned how to draw and read the pie graph, column graph and line graph in the statistics sixth grade. Now, we will learn the terms related to statistics to handle the raw data, make frequency distribution table and calculate mean.

Line Graph

In line graph:

  • Points are plotted on the graph related to two variables
  • Points are joined by the line segments.

How to Construct a Line Graph?
Steps of construction of line graph:

  • On a graph, draw two lines perpendicular to each other intersecting at O. The horizontal line is x-axis and vertical line is y-axis.
  • Mark points at equal intervals along x-axis and write the names of the data items whose values are to be marked.
  • Along the y-axis, choose an appropriate scale considering the given values.
  • Now, make the points Join each point with the successive point using a ruler. Thus, a line graph is obtained

Solve Examples on Line Graph

1. The following table gives the information of the sum scored by McKay in 10 matches. Represent this information using line graph.

Match 1 2 3 4 5 6 7 8 9 10
Runs Scored 80 30 50 90 40 60 75 20 15 90

Statistics

2. Mobile phones sold by a shop in a certain week are as follows:

DaysMonTuesWedThursFriSat
No. of Mobile Phones Sold201612244 10

Represent the above data by a line graph.


Statistics

Mean of the Tabulated Data

In the mean of the tabulated data, if the frequencies of n observations x1, x2, x3, .... xn are f1, f2, f3 ...... fn, then Mean of the tabulated data = `((f_1x_1 + f_2x_2 + f_3x_3 ..... f_nx_n))/(( f_1 + f_2 + f_3 .... f_n)) = (sum(f_ix_i))/(sum f_i) `


Worked-out examples on mean of the tabulated data

1. Find the mean weight of 50 girls from the following table

Weight in kg 40 42 34 36 46
No.of girls 6 6 15 14 7

Solution:
Mean = `(f_1x_1 + f_2x_2 + f_3x_3+ f_4x_4 + f_5x_5)/(f_1 + f_2 + f_3 + f_4 + f_5)`

= `((40 xx 6) + (42 xx 6 )+ (34xx 15 )+ (36 xx 14) + (46 xx 7))/(8 + 6 + 15 + 14 + 7)`

= `(240 + 252 + 510 + 504 + 322)/(50)`
= `(1828)/(50)`
= 36.56


2. If the mean of the following frequency distributions is 9, find the value of 'a'. Write the tally marks also.

Variable (xi) 4 6 8 10 12 15
Frequency (fi) 8 9 17 a 8 4

Solution:
Frequency distribution table

Statistics
Mean = `(sum(f_ix_i))/(sum f_i) `
But given mean = 9
So, we have `(378 + 10a)/(46 + a)` = 9
378 + 10a = 9(46 + a)
378 + 10a = 414 + 9a
10a - 9a = 414 - 378
a = 36


Median

  • Median of a group of observation is the value which lies in the middle of the data (when arranged in an ascending or descending order) with half of the observations above it and the other half below it.
  • When the number of observations (n) is odd. Then, median is `[(n + 1)/2]^ (th)` observation.
  • When the number of observations (n) is even. Then median is the mean of `(n/2)^(th)` and `(n/2+ 1)^(th) `observation.

i.e., Median =`{(n/2)^(th) observation +(n/2 + 1)^(th) observation}/2`

Let us observe the following solved problems using step-by-step explanation.


Worked-out examples on median

1. Find the median of the data 25, 37, 47, 18, 19, 26, 36.
Solution:

Arranging the data in ascending order, we get 18, 19, 25, 26, 36, 37, 47
Here, the number of observations is odd, i.e., 7.
Therefore, median = `((n + 1)/2)^(th)` observation.
= `((7 + 1)/2)^(th)`observation.
= `(8/2)^(th)` observation
= 4th observation.
4th observation is 26.
Therefore, median of the data is 26.


2. Find the median of the data 24, 33, 30, 22, 21, 25, 34, 27.
Solution:

Here, the number of observations is even, i.e., 8.
Arranging the data in ascending order, we get 21, 22, 24, 25, 27, 30, 33, 34
Therefore, median = `{(n/2)^(th) observation + (n/2 + 1)^(th) observation}/2`
= `(8/2)^(th)` observation + `(8/2 + 1)^(th)` observation
= 4th observation + (4 + 1)th observation
= `(25 + 27)/2`
= `(52)/2`
= 26
Therefore, the median of the given data is 26.


Construction of Bar Graphs

Bar graph is the simplest way to represent a data.

  • In consists of rectangular bars of equal width.
  • The space between the two consecutive bars must be the same.
  • Bars can be marked both vertically and horizontally but normally we use vertical bars.
  • The height of bar represents the frequency of the corresponding observation.

For example, let us observe the following data of the bar graph.
The following data gives the information of the number of children involved in different activities.

Activities Dance Music Art Cricket Football
No. of Children 30 40 25 20 53

How to Construct a Bar Graph?

Steps in construction of bar graphs/column graph:

  • On a graph, draw two lines perpendicular to each other, intersecting at 0.
  • The horizontal line is x-axis and vertical line is y-axis.
  • Along the horizontal axis, choose the uniform width of bars and uniform gap between the bars and write the names of the data items whose values are to be marked.
  • Along the vertical axis, choose a suitable scale in order to determine the heights of the bars for the given values. (Frequency is taken along y-axis).
  • Calculate the heights of the bars according to the scale chosen and draw the bars.
  • Bar graph gives the information of the number of children involved in different activities.
Statistics


Solved examples on construction of bar graphs:

1. The percentage of total income spent under various heads by a family is given below.

Different needs FoodClothing Health Education House Rent Miscellaneous
% Age of Total
Number
40% 10% 10% 15% 20% 5%

Represent the above data in the form of bar graph.

Statistics

2. 150 students of class VI have popular school subjects as given below:

Subject French English Maths Geography Science
Number of Students 30 20 26 38 34

Draw the column graph/bar graph representing the above data.
Solution:
Take the subjects along x-axis, and the number of students along y-axis


Statistics

Bar graph gives the information of favourite subjects of 150 students.


3. The vehicular traffic at a busy road crossing in a particular place was recorded on a particular day from 6am to 2 pm and the data was rounded off to the nearest tens.

Time in Hours (6 - 7) (7 - 8) (8 - 9) (9 - 10) (10 - 11) (11 - 12) (12 - 1) (1 - 2)
Number of Vehicles 100 450 1250 1050 750 600 550 200

Statistics

Bar graph gives the information of number of vehicles passing through the crossing during different intervals of time.

Use of Tally Marks
  • Here we will learn how to use of tall When the observations are large, it may not be easy to find the frequencies by simply counting, so we make the use of bass (|, \) called the tally marks.
  • Tallies are usually marked in bunches of five.
  • The first four tallies are marked vertically. The fifth tally in a bunch is marked diagonally across the earlier 4.
Statistics

The illustration of the example will help us to learn the use of tally marks:

1. The scores obtained in 20 throws of a dice are: 5, 4, 3, 2, 1, 1, 2, 5, 4, 6, 6, 6, 3, 2, 1, 4, 3, 2, 2, 4 Prepare the frequency table for the above scores.
Solution:

Frequency table of scores in 20 throws.

Statistics

Here
1 occurs 3 times
2 occurs 5 times
3 occurs 3 times
4 occurs 4 times
5 occurs 2 times
6 occurs 3 times
y marks.


Properties of Arithmetic Mean

  • To solve different types of problems on average we need to follow the properties of arithmetic mean.
  • Here we will learn about all the properties and proof the arithmetic mean showing the step-by-step explanation

What are the properties of arithmetic mean?
The properties are explained below with suitable illustration.

Property 1:

If x is the arithmetic mean of n observations x1, x2, x3, .... xn; then
(x1 - x) + (x2 - x) + (x3 - x) + .... + (xn - x) = 0.
proof:

We know that `x = (x_1 + x_2 + x_3 + ... + x_n)/n`
(x1 + x2 + x3 + ..... + xn) = nx........ (A)
Therefore, (x1 - x) + (x2 - x) + (x3 - x) + ..... + (xn - x)
= (x1 + x2 + x3 + ..... + xn) - nx
= (nx - nx), [using (A)].
= 0.
Hence, (x1 - x) + (x2 - x) + (x3 - x) + ..... + (xn - x) = 0.


Property 2:

The mean of n observations x1, x2, x3, .... xn is x. If each observation is increased by p, the mean of the new observations is (x + p).
proof :

`x = (x_1 + x_2 + x_3 + ... + x_n)/n`
x1 + x2 + x3 + ... + xn = nx ...... (A)
Mean of (x1 + p), (x2 + p), ...,(xn + p)
= `((x_1 + p) + (x_2 + p) + ... + (x_n + p))/n `
= `((x_1+ x_2 + .... + xn) + np)/n`
= `(nx + np)/n`, [using (A)].
= `(n(x + p))/n`
= (x + p).
Hence, the mean of the new observations is (x + p).


Property 3:

The mean of n observations x1, x2, x3, .... xn is x. If each observation is decreased by p, the mean of the new observations is (x - p).
proof:

`x = (x_1 + x_2 + x_3 + ... + x_n)/n`
(x1 + x2 + x3 + .... + xn) = nx ..... (A)
Mean of (x1 - p), (x2 - p), ...., (xn - p)
= `((x_1 - p) + (x_2 - p) + ... + (x_1 - p))/n`
= `((x_1 + x_2 + ... + x_n) - np)/n`
= `(nx - np)/n`, [using (A)].
= `(n(x - p))/n`
= (x - p).
Hence, the mean of the new observations is (x - p).


Property 4:

The mean of n observations x1, x2, x3, .... xn is x. If each observation is multiplied by a non-zero number p, the mean of the new observations is px.
proof:

`x = (x_1 + x_2 + x_3 + ... + x_n)/n`
x1 + x2 + ....+ xn = nx ..... (A)
Mean of px1, px2, ...,pxn,
=` (px_1 + px_2+ .... + px_n)/n `
= `(p(x_1 + x_2+ .... + x_n))/n`
= `(p(nx))/n`, [using (A)].
= px.
Hence, the mean of the new observations is px.


Property 5:

The mean of n observations x1, x2, x3, .... xn is x. If each observation is divided by a non-zero number p, the mean of the new observations is (x/p).
proof:

`x = (x_1 + x_2 + x_3 + ... + x_n)/n`
x1 + x2 + ... + xn = nx ..... (A)
Mean of `(x_1)/p, (x_2)/p,......, (x_n)/p`
= `1/n xx (x_1)/p + (x_2)/p +... (x_n)/p`
= `(x_1 + x_2 + ... + x_n)/(np)`
= `(nx)/(np)`, [using (A)].
= `x/p`.


Properties Questions on Arithmetic Mean

Worked out examples on properties questions on arithmetic mean

1. The mean of eight numbers is 25. If five is subtracted from each number, what will be the new mean?
Solution:
Let the given numbers be x1, x2, x3, .... x8.
Then, the mean of these numbers = `(x_1+ x_2+ x_3+ ....+ x_8)/8`.
Therefore, `(x_1+ x_2+ x_3+ ....+ x_8)/8` = 25
=> `(x_1+ x_2+ x_3+ ....+ x_8)/8` = 200 .....(A)
The new numbers are (x1 - 5), (x2 - 5), .... ,(x8 - 5)
Mean of the new numbers =` ((x_1 - 5) + (x_2 - 5) + .... + (x_8 - 5))/8`
=` ((x_1+ x_2+ x_3+ ....+ x_8) - 40)/8`
= `(200 - 40)/8`, [using (A)]
= `(160)/8`
= 20
Hence, the new mean is 20.


2. The mean of 14 numbers is 6. If 3 is added to every number, what will be the new mean?
Solution:
Let the given numbers be x1, x2, x3, .... x14.
Then, the mean of these numbers = `(x_1+ x_2+ x_3+ ....+ x_14)/(14)`
Therefore, `(x_1+ x_2+ x_3+ ....+ x_14)/(14) = 6`
` (x_1+ x_2+ x_3+ ....+ x_14) = 84`..... (A)
The new numbers are (x1 + 3), (x2 + 3), (x3 + 3), ....,(x14 + 3)
Mean of the new numbers
=` ((x_1 + 3)+ (x_2 + 3)+(x_3 + 3)+.... +(x_14 + 3))/(14)`
= `[(x_1+ x_2+ x_3+ ....+ x_14) + 42]/[14]`
= `(84 + 42)/(14)`, [Using (A)]
= `(126)/(14)`
= 9
Hence, the new mean is 9.


Average

Average means a number which is between the largest and the smallest number.
For example:

Teacher: Children, come one by one and pick up some toffees from the box.
Ron: I have got 8 toffees.
Shelly: I have got 2 toffees.
Mary: I have got 1 toffee.
Teacher: Don't worry! let me distribute these 24 toffees equally among you.
We divide 24 by 4;
`(24) / 4` = 6
Thus we observe that on an average each child gets 6 toffees
Or We can say that the average number of toffees each child gets is 6.


To find the average of a group of numbers, divide the sum of numbers by the total number of events.

  • Average = Sum of events by total number of events.
  • Average can be calculated only for similar quantities and not for dissimilar quantities.
  • Average of height and weight cannot be calculated. It should either be average height of all students or average weight of all students.

The formulas of solving average,

Average           =    Sum of all Items/Number of Items
Sum of  Items     =    Average x  Number of Items
Number of Items   =    Total /Average

Solved examples on finding the average of different set of of numbers

1. The maximum temperature of Miami during 8th September to 14th September is as follows:

8Monday20.6°
9Tuesday21.8°
10Wednesday23.8°
11Thursday27.7°
12Friday29°
13Saturday22.5°
14Sunday24°

Solution:
We know, Average = (Sum of all items)/(Number of items)
The average temperature of this week in Miami
= `(20.6 + 21.8 + 23.8 + 27.7 + 29 + 22.5 + 24)/7`
= `(169.4)/7`
= 24.2°


2. Find the average of 5, 7, 6, 8, 4, 9
Solution:

Sum of the given number = 5 + 7 + 6 + 8 + 4 + 9 = 39
Number of events = 6
Therefore, average of numbers =`( 39)/6 =( 13)/2` = 6.5


3. Find the average of 3.6, 2.7, 4.1, 1.5, and 5.3.
Solution:

Sum of the numbers = 3.6 + 2.7 + 4.1 + 1.5 + 5.3 = 17.2
Number of events = 5.
Average of numbers = `(17.2)/5 = 3.44`


4. The marks obtained by Sara in the first three Unit Tests in Maths are 85, 89 and 98.
Solution:

Sara's average marks in Maths Unit Tests are =` (85 + 89 + 98)/3`
= `(272)/3`
= 90.6
Thus, Sara's average marks in Maths Unit Tests = 90.6 %.


5. The average consumption of wheat by a family is 33 kgs in three months. If there are 15 members in the family, find the total consumption for three months.
Solution:

Average = 33 kgs.
No. of members = 15
Total = Average `xx` No. of members.
= 33 `xx` 15
= 495 kg.
Therefore, the total consumption of wheat for 3 months is 495 kg.


6. Total height of a class is 1300 cm. If the average height of a class is 65 cm, find the number of students in the class.
Solution:

Total height of a class = 1300 cm.
Average = 65 cm.
No. of students = Total / Average
= `(1300) / (65)`
= 20
Therefore, number of students in the class = 20


Word Problems on Average

1. The goal scored by a team is 6 matches are 1, 2, 5, 3, 4, 0. Find the average score of the team.
Solution:
Total goal scored by the team = 1 + 2 + 5 + 3 + 4 + 0 = 15
Total number of matches = 6
Therefore, average score of the team = `(15)/6 `= 2.5


2. The weights of 4 children in a medical text are given below:

Aaron34.2kg
Natalie36.8kg
Paraker41.3kg
Sophie40.5kg

(i) Find the average weight?
(ii) How many children have weight less than average?
(iii) Name the children whose weights are above average.
Solution:

(i) Total weight of 4 children =      34.2 kg
                                      36.8 kg
                                      41.3 kg
                                      40.5 kg
                                     __________
                                     152.8 kg
                                     __________

Average weight of one child = 152.8 kg / 4 = 38.2 kg
(ii) 2 children are having their weights less than the average weight.
(iii) The weight of Parker and Sophie is above average.


3. Duration of sunshine in Aligarh city for one week of August 2002 are given as 12.20 hours, 12.18 hours, 12.17 hours, 12.15 hours, 12.13 hours, 12.12 hours, 12.10 hours. Find the average duration of sun shine.
Solution:

Total duration of sun shine = 12.20 + 12.18 + 12.17 + 12.15 + 12.13 + 12.12 + 12.10 = 85.05 hours.
Number of days = 7
Therefore, average duration of sunshine = `(85.05)/7` = 12.15 hours


Follow the explanation to solve the word problems on arithmetic mean (average)

1. The heights of five runners are 160 cm, 137 cm, 149 cm, 153 cm and 161 cm respectively. Find the mean height per runner.
Solution:

Mean height = Sum of the heights of the runners/number of runners
= `(160 + 137 + 149 + 153 + 161)/5` cm
= `(760)/5` cm
= 152 cm.
Hence, the mean height is 152 cm.


2. Find the mean of the first five prime numbers.
Solution:

The first five prime numbers are 2, 3, 5, 7 and 11.
Mean = Sum of the first five prime numbers/number of prime numbers
= `(2 + 3 + 5 + 7 + 11)/5`
= `(28)/5`
= 5.6
Hence, their mean is 5.6


3. Find the mean of the first six multiples of 4.
Solution:

The first six multiples of 4 are 4, 8, 12, 16, 20 and 24.
Mean = Sum of the first six multiples of 4/number of multiples
= `(4 + 8 + 12 + 16 + 20 + 24)/6`
= `(84)/6`
= 14.
Hence, their mean is 14.


4. Find the arithmetic mean of the first 7 natural numbers.
Solution:

The first 7 natural numbers are 1, 2, 3, 4, 5, 6 and 7.
Let x denote their arithmetic mean.
Then mean = Sum of the first 7 natural numbers/number of natural numbers
x = `(1 + 2 + 3 + 4 + 5 + 6 + 7)/7`
= `(28)/7`
= 4
Hence, their mean is 4.


5. If the mean of 9, 8, 10, x, 12 is 15, find the value of x.
Solution:

Mean of the given numbers = `(9 + 8 + 10 + x + 12)/5 = (39 + x)/5`
According to the problem, mean = 15 (given).
Therefore, `(39 + x)/5 = 15`
`39 + x = 15 xx 5`
`39 + x = 75`
`39 - 39 + x = 75 - 39`
` x = 36`
Hence, `x = 36`.


6. If the mean of five observations `x, x + 4, x + 6, x + 8` and `x + 12` is 16, find the value of `x`.
Solution:

Mean of the given observations
= `(x + (x + 4) + (x + 6) + (x + 8) + (x + 12))/5`
= `(5x + 30)/5`
According to the problem, mean = 16 (given).
Therefore, `(5x + 30)/5 = 16`
`5x + 30 = 16 xx 5`
5x + 30 = 80
5x + 30 - 30 = 80 - 30
5x = 50
x = `(50)/5`
x = 10
Hence, x = 10.
148 + 153 + 146 + 147 + 154


7. The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of given numbers.
Solution:

Calculated mean of 40 numbers = 38.
Therefore, calculated sum of these numbers = (38 * 40) = 1520.
Correct sum of these numbers
= [1520 - (wrong item) + (correct item)]
= (1520 - 36 + 56)
= 1540.
Therefore, the correct mean = `(1540)/(40 )= 38.5`.


8. The mean of the heights of 6 boys is 152 cm. If the individual heights of five of them are 151 cm, 153 cm, 155 cm, 149 cm and 154 cm, find the height of the sixth boy.
Solution:

Mean height of 6 boys = 152 cm.
Sum of the heights of 6 boys = (152 `xx` 6) = 912 cm
Sum of the heights of 5 boys = (151 + 153 + 155 + 149 + 154) cm
= 762 cm.
Height of the sixth boy
= (sum of the heights of 6 boys) - (sum of the heights of 5 boys)
= (912 - 762) cm = 150 cm.
Hence, the height of the sixth boy is 150 cm.

Worksheet on word Problems on Average

1. Sam scored 98, 25, 105, 62 and 65 run in 5 matches. What was the average score per match?
2. The temperature of a city during a week was 35° C, 36°C, 34°C, 38°C, 40°C, 39°C and 44°C. What was the average daily temperature of the town for the week?
3. Jack earns Rs 100 on the first day, Rs 60 on the second day, Rs 150 on the third day, Rs 80 on the fourth day and Rs 90 on the fifth day. What was his average earnings?
4. Rex scored the following marks in 6 subjects:
English - 88
Hindi - 64
Sanskrit - 89
Maths - 96
Science - 87
Social Studies - 80
Find his average score in 6 subjects.
5. A train runs for 3 hours at a speed of 55 km per hour and for the next 2 hours at a speed of 65 km per hour. Find the average speed of the train for the 5 hours journey.
6. Which is more, the average of the 4 even natural numbers from 6 to 13 or the average on the 4 odd natural numbers from 6 to 13?
7. In a series of 4 cricket matches, the runs scored by India and Australia are given below:

MatchIndiaAustralia
First102208
Second202192
Third360241
Fourth276203

(i) What is the average score of India in one match?
(ii) What is the average score of Australia in one match?
(iii) Which team performed better?

Answers for the worksheet on word problems on average are given below.
Answers:
1. 71 runs.
2. 38°C
3. Rs 96
4. 84 marks
5. 59 km
6. Average of 4 odd natural numbers from 6 to 13 is more
7. (i) 235 runs
(ii) 211 runs
(iii) India

formulae

  • Mean or average or arithmetic mean is one of the representative values of data.
  • Mean of raw data:

    If x1, x2, x3, ......xn are n observations, then
    Arithmetic Mean = `(x_1, x_2, x_3, ......x_n)/n` = `[sum(x_i)]/n`
    `sum` (Sigma) is a Greek letter showing summation

Mean of the data=`[sum(f_ix_i)]/(sum f_i)`


  • Frequency distribution of grouped data:
  • The presentation of the above data can be expressed into groups. These groups are called classes or the class interval.
  • Each class interval is bounded by two figures called the class limits.

  • Class size:
    Difference between the true upper limit and true lower limit of a class interval is called the class size.
    Class size remains the same for all class intervals.
  • Class mark:
    Mid-value of each class interval is called its class mark. Class mark = `1/2` (Upper limit + Lower limit)
  • Range:
    The difference between the maximum value and the minimum value of the observation is called the range.

  • Mean of the Tabulated Data
  • In the mean of the tabulated data, if the frequencies of n observations
    x1, x2, x3, .... xn are f1, f2, f3 ...... fn, then

  • Mean of the tabulated data
    = `(f_1x_1 + f_2x_2 + f_3x_3 ..... f_nx_n)/( f_1 + f_2 + f_3 .... f_n)`
    = `[sum(f_ix_i)]/(sum f_i)`

Median

When the number of observations (n) is odd.
Median =

Statistics

When the number of observations (n) is even.Then median is the mean of `(n/2)^(th)` and `(n/2+ 1)^(th) `observation.
i.e., Median =`{(n/2)^(th) observation +(n/2 + 1)^(th) observation}/2`


What are the properties of arithmetic mean?

  • Property 1:
    If x is the arithmetic mean of n observations x1, x2, x3, .... xn; then
    (x1 - x) + (x2 - x) + (x3 - x) + .... + (xn - x) = 0.
  • Property 2:
    The mean of n observations x1, x2, x3, .... xn is x. If each observation is increased by p, the mean of the new observations is (x + p).
  • Property 3:
    The mean of n observations x1, x2, x3, .... xn is x. If each observation is decreased by p, the mean of the new observations is (x - p).
  • Property 4:
    The mean of n observations x1, x2, x3, .... xn is x. If each observation is multiplied by a non-zero number p, the mean of the new observations is px.
  • Property 5:
    The mean of n observations x1, x2, x3, .... xn is x. If each observation is divided by a non-zero number p, the mean of the new observations is `x/p`.

The formulas of solving average,

Average           =    Sum of all Items/Number of Items
Sum of  Items     =    Average x  Number of Items
Number of Items   =   Total /Average
- Facebook | - Twitter | - Google+ | - YouTube | - Mail
Meritpath...Way to Success!

Meritpath provides well organized smart e-learning study material with balanced passive and participatory teaching methodology. Meritpath is on-line e-learning education portal with dynamic interactive hands on sessions and worksheets.