What is a mean, median and mode

Mean

• Mean is the same as the average and is a useful property for understanding data sets.
• It is the value if all data values were made the same.
• Calculated by adding up all the values and then dividing the total by how many numbers were added.
• The equation of the mean is therefore, x = ∑/n
  ∑ representing the sum all date set values
  n, representing total number of the values.

Median

• Median is the middle number in a data set when the numbers are in order.
• In a data set with an odd number of members, the middle number is the median.
• In a data set with an even number of members, the mean of the two middle numbers is the median.

Mode

• Mode is the value that occurs most often in a data set.
• It is the number that is repeated the most, not how many times it was repeated.
• A data set can have no mode, one mode, or more than one mode.

Real-World Example

• Heights of ten male students in a class were measure in centimeters (cm) as
  175, 166, 181, 164, 181, 171, 167, 182, 172, 176.
  
  

• Find the mean height of the above data set.

  
  First step, we recommend arranging the data in an ascending order
  Hence, the data set will now look as follow: 164, 166, 167, 171, 172, 175, 176, 181, 181, 182.
  mean = ∑/n
  mean = (164 + 166 + 167 + 171 + 172 + 175 + 176 + 181 + 181 + 182)/10
  mean = 1735/10
  mean = 173.5
  Therefore, the average or mean height of these 10 students is 173.5

• Find the median height of the above data set.

  
  There is a very important factor you need to consider when working with the median, which is:
  Is the total number of data set even or odd. If the total is even, the median is the mean of the two middle numbers in the data set, e.g,:
  The median of the data set above is the sum of the middle values because the total number of data is 10, which is an even number divide by 2.
  Median = (∑ of the 5th and 6th value)/2
  Median = (172 + 175)/2
  Median = 187
  Additionally, if the data set total is odd, e.g., 164, 166, 167, 171, 172, 175, 176, 181, 181
  In this scenario, the median is simply 172, because 172 is the exact middle number of the data set
  

• Find the median height of the above data set.

  The mode of this data is 181. Because it a value found more than once in the data set above
  

Variance and Standard deviation

Finding the formula of a quadratic sequence:

• Tn = an² + bn + c
• -145; -122; -101; -82;...
• First, you find the first differences
  -122 - (-145) = -122 + 145 = 23
  -101 - (-122) = -101 + 122 = 21
  -82 - (-101) = -82 + 101 = 19.

Then, find the difference of the first differences:

• 21 - 23 = -2
  19 - 21 = -2
• Find a by 2a = the 2nd difference
  2a = -2
  2a/2 = -2/2
  Therefore, a = -1

Find b by 3a + b = the first value of the first differences

• 3a + b = 23
  3(-1) + b = 23
  -3 + b = 23
  b = 23 + 3
  b = 26

Last but not least, find c by a + b + c = the first value of the sequence

• a + b + c = -145
  -1 + 26 + c = -144
  25 + c = -145
  c = -170

Therefore, the quadratic equation for this sequence is:

• Tn=-1n²+26n-170
  

Cumulative frequency distribution table

• Cumulative frequency distribution table contains three colums. Class Interval, frequency and cumulative frequency, respectively.

Class Interval	Frequency	Cumulative Frequency
0 ≤ x <10	0	0
10 ≤ x <20	0	0
20 ≤ x <30	2	2
30 ≤x < 40	4	6
40 ≤ x <50	6	12
50 ≤ x <60	5	17
60 ≤ x <70	7	24
70 ≤ x <80	3	27
80 ≤ x <90	2	29
90 ≤ x <100	1	30

Frequency

• Is the number of values falling within a given class interval.
  e.g. The frequency of students that marks between 40 to 50, is 6 according to the graph.

Acumulative frequency

• Is the sum of all values falling under the overall lowest class interval to the current highest class interval.
  e.g. The cumulative frequency of students with marks between 40 to 50, is 12 because 0 + 0 + 2 + 4 + 6 = 12.

Finding the estimated mean

First step

• Find the mean of each class interval.
  20 ≤ x <30 = 20 + 30 / 2 = 25
  30 ≤ x <40 = 30 + 40 / 2 = 35
  40 ≤ x <50 = 40 + 50 / 2 = 45
  50 ≤ x <60 = 20 + 10 / 2 = 55
  60 ≤ x <70 = 20 + 10 / 2 = 65
  70 ≤ x <80 = 40 + 50 / 2 = 75
  80 ≤ x <90 = 20 + 10 / 2 = 85
  90 ≤ x <100 = 20 + 10 / 2 = 95
  

• Second step

• Multiply each mean by the frequency of its class interval.
  25 * 2 = 50
  35 * 4 = 140
  45 * 6 = 270
  55 * 5 = 275
  65 * 7 = 455
  75 * 3 = 225
  85 * 2 = 170
  95 * 1 = 95
  

Third step

• Find the sum of all those products.
  ∑ = 50 + 140 + 270 + 275 + 455 + 225 + 170 + 95
  ∑ = 1680

• Final step

  The sum is then divided by the last cumulative frequency values, in this example that is 30. Mean = ∑/n
  Mean = 1680/30
  Mean = 56.

Modal Class Determination

• The modal class, is the class interval with highest frequency.
• In the table above, that 60 ≤ x ≤70

Introducing additional properties of data sets

• Min
• Max
• Range
• Quartiles
• Interquartile range
• Outliers

Minimum (Min)

Minimum is the lowest value of the data set.

• Given the data below, the minimum is 2.
  2, 4, 6, 8, 10, 12, 14, 16.

Maximum (Max)

Maximum is the highest value of the data set.

• Given the data below, the maximum is 16.
  2, 4, 6, 8, 10, 12, 14, 16.

Range

Is the difference between the highest value and the lowest value of the data set.

• 2, 4, 6, 8, 10, 12, 14, 16.
  Range = max - min
  Range = 16 - 2
  Range = 14

Quartiles

quartiles are a type of quantiles which divide the number of data points into four parts, or quarters, of more-or-less equal size.

• There are three quartiles (Q1, Q2 and Q3)

• Second quartile (Q2)

  Q2 is the median of the data set.
  So what is the second quartile of the data set below?
  2, 4, 6, 8, 10, 12, 14, 16.
  Median or Q2 = 10 + 8/2
  Median or Q2 = 9
  Thus second quartile of the data set above is 9.

• First Quartile (Q1)

  is the median of the values in the data set lower than the median of the whole data set.
  Therefore Q1 of the data set below is the median of 2, 4, 6, 8.
  2, 4, 6, 8, 9, 10, 12, 14, 16.
  Q1 = 6 + 4/ 2
  Q1 = 10/2
  Q1 = 5

• Quartile 3 (Q3)

  Is the median of the values in the data set that are higher than the median of whole data set.
  Hence, Q3 of the data set below is the median 10, 12, 14, 16.
  2, 4, 6, 8, 9, 10, 12, 14, 16.
  Q3 = 12 + 14/ 2
  Q3 = 26/2
  Q3 = 13

Interquartile range (IR)

• IR is the difference between Q3 and Q1
  IQR = Q3 - Q1
  = 13 - 5
  = 8

Outliers

• Any values of the data set that are less than Q1 - 1.5 x IQR or greater than Q3 + 1.5 x IQR.
  So, does our above data set have outliers?
  Q1 - 1.5 x IQR
  5 - 1.5 x 8
  5 - 12
  -7
  
  Q3 + 1.5 x IQR
  13 + 1.5 x 8
  13 + 12
  25
  Therefore, the above data set have no outliers because all the values fall within -7 and 25.