Concept:
Relation Between Mean Median and Mode:
Mean – Mode = 3 (Mean – Median)
Calculation:
As we know,
Mean – Mode = 3 (Mean – Median)
⇒ Mean – Mode = 3Mean – 3Median
⇒ 3Median  Mode = 2Mean
⇒ \(\frac 1 2\)(3Median  Mode) = Mean
∴ k = \(\frac 1 2\)
Concept:
Mean: It is the sum of all observations upon a number of observations.
If x1, x2, x3, …., xn are the observations, then mean is given as
\(\rm \bar{X} = \frac{sum\ of\ observations}{number\ of\ observations}\)
\(\rm \bar{X}= \frac{x_{1}+x_{2}+...+x_{n}}{n}\)
Calculation:
We have X̅ = 44 and observations 5, 7, 6, 11, x, 13.
So, mean \(\bar{X} = \frac{5+7+6+11+x+13}{6}\)
⇒ 44 × 6 = 42 + x
⇒ x = 222
Concept:
Suppose there are ‘n’ observations {\({\rm{\;}}{{\rm{x}}_{\rm{1}}},{{\rm{x}}_{{\rm{2\;}}}},{{\rm{x}}_{{\rm{3\;}}}}, \ldots ,{{\rm{x}}_{{\rm{n\;}}}}\)}
Mean \(\left( {{\rm{\bar x}}} \right) = \frac{{{\rm{\;}}({{\rm{x}}_1} + {{\rm{x}}_{2{\rm{\;}}}} + {{\rm{x}}_{3{\rm{\;}}}} + \ldots + {{\rm{x}}_{{\rm{n\;}}}})}}{{\rm{n}}}\) \( = {\rm{\;}}\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)
Sum of the first n natural numbers = \(\rm \frac{n(n+1)}{2}\)
Calculation:
To find: Mean of first n natural numbers
As we know, Sum of first n natural numbers = \(\rm \frac{n(n+1)}{2}\)
Now, Mean = \(\rm \dfrac { \frac{n(n+1)}{2}}{n}\)
= \(\rm \frac{(n+1)}{2}\)
Concept:
Mean = \(\frac{{{\rm{sum\;of\;the\;terms\;}}}}{{{\rm{total\;number\;of\;terms\;}}}}\)
If number of terms is even, take average of two middle values
Calculation:
Given data: 3, 5, 1, 6, 5, 9, 5, 2, 8, 6
Mean = x, median = y, and mode = z
Number of terms = n = 10
\({\rm{mean}} = {\rm{x}} = \frac{{3 + {\rm{\;}}5 + 1 + 6 + 5 + 9 + 5 + 2 + 8 + 6}}{{10}}\)
\(= \frac{{50}}{{10}}\)
= 5
Arrange given data in ascending order: 1, 2, 3, 5, 5, 5, 6, 6, 8, 9
Here number of terms is even,
So, median will be average of two middle terms
i.e., \({\rm{y}} = {\rm{\;}}\frac{{5{\rm{th\;term}} + 6{\rm{th\;term\;}}}}{2}\)
\(= \frac{{5\; + \;5}}{2}\)
= 5
Now, mode is the most common number
In given set 5 is the most common number
∴ Mode = z = 5
∴ x = y = z
Hence, option (4) is correct.
A random sample of 20 people is classified in the following table according to their ages:
Age 
Frequency 
15 – 25 
2 
25 – 35 
4 
35 – 45 
6 
45 – 55 
5 
55  65 
3 
What is the mean age of this group of people?
Concept:
Calculation:
Age 
Frequency (f) 
x 
xf 
15 – 25 
2 
20 
40 
25 – 35 
4 
30 
120 
35 – 45 
6 
40 
240 
45 – 55 
5 
50 
250 
55  65 
3 
60 
180 

\(\sum {\rm{f}} = 20\) 

\(\sum {\rm{xf}} = 830\) 
We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)
\(\therefore {\rm{Mean}} = \frac{{830}}{{20}} = 41.5\)
Concept:
The ‘less than’ ogive curve and the ‘more than’ ogive curve intersect at the median.
Concept:
For the observations \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)
Mean (\(\rm\overline x\)) = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)
Calculation:
Given mean of the n observation is M
M = \(\rm x_1  k +x_2  k+ x_3  k +..........+x_nk\over n\)
M = \(\rm (x_1 +x_2 + x_3 +..........+x_n)nk\over n\)
M = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}{nk\over n}\)
M = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)  k
M + k = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)
∴ The required mean of x1, x2, x3, _ _ _, xn = M + k
Given:
Mode = 24
Mean = 60
Formula used:
Mean  Mode = 3(Mean  median)
Calculation:
Let Median be x.
(60  24) = 3(60  x)
⇒ 36/3 = (60  x)
⇒ 12 = (60  x)
⇒ 60  12 = x
⇒ x = 48
∴ The median is 48.
Concept:
Median: The median is the middle number in a sorted ascending or descending list of numbers.
Case 1: If the number of observations (n) is even
\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)
Case 2: If the number of observations (n) is odd
\({\rm{Median\;}} = {\rm{value\;of\;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{\;observation}}\)
Calculation:
Given values 4, 6, 3, 8, 5, 2, 7, 9
Arrange the observations in ascending order:
2, 3, 4, 5, 6, 7, 8, 9
Here, n = 8 = even
As we know, If n is even then,
\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)
= \(\rm \frac{4^{th} \;\text{observation}+5^{th} \;\text{observation}}{2} \)
= \(\frac{6+5}{2} = 5.5\)
Hence Median = 5.5
Concept:
Mean: It is the sum of all observations upon a number of observations.
If x_{1}, x_{2}, x_{3}, …., x_{n} are the observations, then mean is given as:
\(\bar{X'}=\frac{x_1+x_2+...+x_n}{n}\)
Calculation:
Given: Mean X̅ = 30
New mean be \(\bar{X'}=28\)
So, \(\bar{X'}=\frac{x_1+x_2+...+x_4}{4}\)
\(\Rightarrow 28=\frac{x_1+x_2+...+x_4}{4}\)
⇒112 = x_{1} + x_{2} + ⋯ + x_{4}
Now, \(\bar{X}=\frac{x_1+x_2+...+x_5}{5}\)
\(\Rightarrow 30=\frac{x_1+x_2+...+x_5}{5}\)
⇒ 150 = x_{1} + x_{2} + ⋯ + x_{5}
⇒ 150 = 112 + x_{5}
⇒ 38 = x_{5}
The excluded number is 38.
Concept:
Median: Let a data of n observations is given, then for finding median first arrange the observation either in increasing order or decreasing order.
Calculation:
Arranging the given observation in increasing order,
4.8, 0, 2.6, 3.5, 3.9, 4.6, 5.2, 6.1, 7.6, 8.2, 9.3, 12.7
Number of observation, n = 12
So, we know that if n is even for median then,
Median \(= \frac{{{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{term\;}} + {{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)}^{{\rm{th}}}}{\rm{term}}}}{2}\)
\( = \frac{{4.6\; +\; 5.2}}{2}\)
= 9.8/2
= 4.9
What is the median of the following scores?
3, 5, 4, 9, 8
The median is also a measure of central tendency. Unlike arithmetic mean, this median is based on the position of a given observation in a series arranged in an ascending or I descending order. Therefore, it is called a positional average.
Median: 3, 5, 4, 9, 8
Hence, we can conclude that the median of the above scores is 5.
Calculation:
Let the average initially be x
Then the sum of the observations = 15x
As the tens digit is recorded 5 more, therefore the sum is 50 more then the correct sum.
Correct sum of the observations = 15x  50
The correct average = \(\rm 15x50\over15\) = x  \(10\over3\)
∴ The average is reduced by \(10\over3\)
Concept:
Mean = Sum of total observations/Total number of observation
Calculation:
Given:
The mean of 10 observations x1, x2, x3. ... x10 is 20.
Mean of x1 + 2, x2 + 4, x3 + 6, ... x10 + 20 = Mean of (x1, x2, x3. ... x10) + Mean of (2, 4, 6, ..... 20) (1)
Now Mean of (2, 4, 6, ..... 20) is:
(2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10
\(=\frac{2\times55}{10}=11\)
From equation 1)
mean of x1 + 2, x2 + 4, x3 + 6, ... x10 + 20 = 20 + 11 = 31
Consider the following statements in respect of class intervals of grouped frequency distribution:
1. Class intervals need not be mutually exclusive.
2. Class intervals should be exhaustive.
3. Class intervals need not be of equal width. Which of the above statements are correct?Concept:
Properties of class intervals:
So, statement 2 and 3 are correct.
Concept:
\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\), where xi is observations and n is the total number of observation.
Calculation:
Given: Mean of seven numbers is 8
Here n = 10,
Let seven numbers be x_{1}, x_{2}, x_{3},..., x_{7}
\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{7}}}}{{7}} = 8\)
⇒ x_{1} + x_{2} + x_{3} +...+ x_{7} = 56 …. (1)
Now, A new number 16 is added.
Mean of eight numbers = \( {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{7}} + 16}}{{8}} \)
From equation (1) we get
Mean of eight numbers = \(\rm \frac{56+16}{8} = \frac{72}{8} =9\)
Concept:
Mean: It is the average of given observation. Let x_{1}, x_{2}, …, x_{n} be n observations, then
Mean \(= {\rm{\bar X}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)
Calculation:
Given: Mean (X̅) = 20
Mean is given as, \(\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + \ldots + {{\rm{x}}_{\rm{n}}}}}{{\rm{n}}} = {\rm{\bar X}}\)
Sum of incorrect observations = n X̅
= 100 × 20
= 2000
Removing incorrect observations, sum of observations = 2000 – 21 – 21 – 18 – 20
= 1920
So, new mean \(= \frac{{1920}}{{96}}\)
= 20
Concept:
Calculation:
Given: Mean = 32, Median = 33
It was recorded as 40 instead of 35.
So, sum will decrease and so mean.
Median remains same because it depend on sequence of data.
The empirical relationship between Mean, Median, and Mode is :
Mode = 3 Median – 2 Mean
Hence, Option 4 is correct.
Important Points
1. The mode is not always unique. A data set can have more than on mode, or the mode may not exist for a data set.
2. The median is the middle point in a data set—half of the data points are smaller than the median and half of the data points are larger.
What is the median value of the following data?
18, 26, 15, 20, 23, 15, 9, 21, 12, 24
Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it. The Median is the “middle” element when the data set is arranged in order of magnitude. Since the median is determined by the position of different values, it remains unaffected if, say, the size of the largest value increases. The median can be easily computed by sorting the data from smallest to largest and finding out the middle value.
In the above situation, the median can be calculated as:
Concept: Since the number of terms is 10 (which is an even number), we will use the below formula:
\(nth\ term + (n+1)th\ term \over2\), where n = \(N\over 2\)= \(10\over2\) = 5
Calculation:
Given: 18, 26, 15, 20, 23, 15, 9, 21, 12, 24
Arranging these terms in ascending order, we get,
9, 12, 15, 15, 18, 20, 21, 23, 24, 26
So, the 5th and the 6th terms are 18 and 20,
Therefore,
Median = \(18+20\over 2\) = 19