STAM101 :: Lecture 04 ::Measures of averages
                  
				
Mean – median – mode – geometric mean – harmonic mean – computation of the above statistics for raw and grouped data - merits and demerits - measures of location – percentiles – quartiles - computation of the above statistics for raw and grouped data
In the study  of a population with respect to one in which we are interested we may get a  large number of observations. It is not possible to grasp any idea about the  characteristic when we look at all the observations. So it is better to get one  number for one group. That number must be a good representative one for all the  observations to give a clear picture of that characteristic. Such  representative number can be a central value for all these observations. This  central value is called a measure of central tendency or an average or a  measure of locations. There are five averages. Among them mean, median and mode  are called simple averages and the other two averages geometric mean and  harmonic mean are called special averages.
                  Arithmetic mean or mean
Arithmetic  mean or simply the mean of a variable is defined as the sum of the observations  divided by the number of observations. It is denoted by the symbol 
 If the variable x  assumes n values x1, x2 … xn then the mean is given by
![]()
![]()
This formula is for the ungrouped or raw  data.
 
Calculate the mean for pH levels of soil 6.8, 6.6, 5.2, 5.6, 5.8
Solution
Grouped Data
The mean for grouped data is obtained from the following formula:
Where x = the mid-point of individual class
f = the frequency of individual class
n = the sum of the frequencies or total frequencies in a sample.
Short-cut method
Where
A = any value in x
n = total frequency
c = width of the class interval
Example 2
Given the following frequency distribution, calculate the arithmetic mean
Marks : 64 63 62 61 60 59
Students : 8 18 12 9 7 6
Solution
X  | 
                      F  | 
                      Fx  | 
                      D=x-A  | 
                      Fd  | 
                    
64  | 
                      8  | 
                      512  | 
                      2  | 
                      16  | 
                    
63  | 
                      18  | 
                      1134  | 
                      1  | 
                      18  | 
                    
62  | 
                      12  | 
                      744  | 
                      0  | 
                      0  | 
                    
61  | 
                      9  | 
                      549  | 
                      -1  | 
                      -9  | 
                    
60  | 
                      7  | 
                      420  | 
                      -2  | 
                      -14  | 
                    
59  | 
                      6  | 
                      354  | 
                      -3  | 
                      -18  | 
                    
  | 
                      60  | 
                      3713  | 
                      
  | 
                      -7  | 
                    
Direct method
                    ![]()
                    ![]()
Short-cut method
                    ![]()
                  Here A  = 62 
  
 
  Example 3
                  For the frequency distribution of seed  yield of seasamum given in table, calculate the mean yield per plot.
Yield per plot in(in g)  | 
                      64.5-84.5  | 
                      84.5-104.5  | 
                      104.5-124.5  | 
                      124.5-144.5  | 
                    
No of plots  | 
                      3  | 
                      5  | 
                      7  | 
                      20  | 
                    
Solution
Yield ( in g)  | 
                      No of Plots (f)  | 
                      Mid X  | 
                      
  | 
                      Fd  | 
                    
64.5-84.5  | 
                      3  | 
                      74.5  | 
                      -1  | 
                      -3  | 
                    
84.5-104.5  | 
                      5  | 
                      94.5  | 
                      0  | 
                      0  | 
                    
104.5-124.5  | 
                      7  | 
                      114.5  | 
                      1  | 
                      7  | 
                    
124.5-144.5  | 
                      20  | 
                      134.5  | 
                      2  | 
                      40  | 
                    
Total  | 
                      35  | 
                      
  | 
                      
  | 
                      44  | 
                    
A=94.5
                  The  mean yield per plot is 
                  Direct  method:
  ![]()
                  = 
=119.64 gms
  Shortcut method
  ![]()
  ![]()
  Merits and demerits of Arithmetic mean
  Merits
                  1. It is rigidly defined.
                  2. It is easy to understand and easy to  calculate.
                  3. If the number of items is sufficiently  large, it is more accurate and more reliable.
                  4. It is a calculated value and is not  based on its position in the series.
                  5. It is possible to calculate even if  some of the details of the data are lacking.
                  6. Of all averages, it is affected least  by fluctuations of sampling.
                  7. It provides a good basis for  comparison.
  Demerits
                  1. It cannot be obtained by inspection  nor located through a frequency graph.
                  2. It cannot be in the study of  qualitative phenomena not capable of numerical measurement i.e.   
                  Intelligence, beauty, honesty etc.,
                  3. It can ignore any single item only at  the risk of losing its accuracy.
                  4. It is affected very much by extreme  values.
                  5. It cannot be calculated for open-end  classes.
                  6. It may lead to fallacious conclusions,  if the details of the data from which it is computed are 
                  not given.
  Median 
                  The median is the middle most item  that divides the group into two equal parts, one part comprising all values  greater, and the other, all values less than that item.
  Ungrouped or Raw data 
                  Arrange the given values in the ascending  order. If the number of values are odd, median is the middle value
                  If the number of values are even, median  is the mean of middle two values.
                  By formula
                  When n is odd, Median = Md = 
 
                  When n is even, Average of ![]()
Example 4
                  If the weights of  sorghum ear heads are 45, 60,48,100,65 gms, calculate the median
  Solution
                  Here n = 5
                  First arrange it  in ascending order
                  45, 48, 60, 65,  100
                  Median =![]()
                  =
=60
  Example 5 
                  If the sorghum  ear- heads are 5,48, 60, 65, 65, 100 gms, calculate the median.
  Solution 
                  Here n = 6
  ![]()
  
 ![]()
  ![]()
  Grouped data
                  In a grouped  distribution, values are associated with frequencies. Grouping can be in the  form of a discrete frequency distribution or a continuous frequency  distribution. Whatever may be the type of distribution, cumulative frequencies  have to be calculated to know the total number of items.
Cumulative frequency (cf)
                  Cumulative frequency of each class  is the sum of the frequency of the class and the frequencies of the pervious  classes, ie adding the frequencies successively, so that the last cumulative  frequency gives the total number of items. 
  Discrete Series
                  Step1: Find cumulative frequencies.
  ![]()
                  Step3: See in the cumulative frequencies  the value just greater than ![]()
                  Step4: Then the corresponding value of x  is median.
  Example 6
                  The following data pertaining to the  number of insects per plant. Find median number of insects per plant.
Number of insects per plant (x)  | 
                      1  | 
                      2  | 
                      3  | 
                      4  | 
                      5  | 
                      6  | 
                      7  | 
                      8  | 
                      9  | 
                      10  | 
                      11  | 
                      12  | 
                    
No. of plants(f)  | 
                      1  | 
                      3  | 
                      5  | 
                      6  | 
                      10  | 
                      13  | 
                      9  | 
                      5  | 
                      3  | 
                      2  | 
                      2  | 
                      1  | 
                    
Solution
                  Form  the cumulative frequency table
x  | 
                      f  | 
                      cf  | 
                    
1  | 
                      1  | 
                      1  | 
                    
2  | 
                      3  | 
                      4  | 
                    
3  | 
                      5  | 
                      9  | 
                    
4  | 
                      6  | 
                      15  | 
                    
5  | 
                      10  | 
                      25  | 
                    
6  | 
                      13  | 
                      38  | 
                    
7  | 
                      9  | 
                      47  | 
                    
8  | 
                      5  | 
                      52  | 
                    
9  | 
                      3  | 
                      55  | 
                    
10  | 
                      2  | 
                      57  | 
                    
11  | 
                      2  | 
                      59  | 
                    
12  | 
                      1  | 
                      60  | 
                    
  | 
                      60  | 
                      
  | 
                    
Median  = size of  ![]()
                  Here the number of observations is even.  Therefore median = average of (n/2)th item and (n/2+1)th item.  
                  =  (30th item +31st item) / 2 = (6+6)/2 = 6
                  
                  Hence  the median size is 6 insects per plant.
Continuous Series
The steps given below are followed for  the calculation of median in continuous series.
Step1: Find cumulative frequencies.
Step2: Find  ![]()
Step3: See in the cumulative frequency  the value first greater than 
,  Then the  corresponding class interval is called the Median class. Then apply the formula
Median = 
where              l = Lower limit of the medianal class
m = cumulative  frequency preceding the medianal class
c = width of  the class
f =frequency  in the median class.
n=Total frequency.
Example  7
For the frequency distribution of weights  of sorghum ear-heads given in table below. Calculate the median.
Weights of ear heads ( in g)  | 
                      No of ear heads (f)  | 
                      Less than class  | 
                      Cumulative frequency (m)  | 
                    
60-80  | 
                      22  | 
                      <80  | 
                      22  | 
                    
80-100  | 
                      38  | 
                      <100  | 
                      60  | 
                    
100-120  | 
                      45  | 
                      <120  | 
                      105  | 
                    
120-140  | 
                      35  | 
                      <140  | 
                      140  | 
                    
140-160  | 
                      24  | 
                      <160  | 
                      164  | 
                    
Total  | 
                      164  | 
                      
  | 
                      
  | 
                    
Solution
                  Median = 
  
=![]()
                  It lies between 60 and 105.  Corresponding to 60 the less than class is 100 and corresponding to 105 the  less than class is 120. Therefore the medianal class is 100-120. Its lower  limit is 100.
                  Here  
100, n=164 , f = 45 , c = 20,   m =60
                  Median = ![]()
   Merits of Median 
                  1. Median is not influenced by extreme  values because it is a positional average.
                  2. Median can be calculated in case of  distribution with open-end intervals.
                  3. Median can be located even if the data  are incomplete.
Demerits of Median 
                  1. A slight change in the series may  bring drastic change in median value.
                  2. In case of even number of items or  continuous series, median is an estimated value other than   any  value in the series.
                  3. It is not suitable for further  mathematical treatment except its use in calculating mean deviation.
                  4. It does not take into account all the observations.
Mode 
                  The mode refers to that value in a  distribution, which occur most frequently. It is an actual value, which has the  highest concentration of items in and around it. It shows the centre of  concentration of the frequency in around a given value. Therefore, where the  purpose is to know the point of the highest concentration it is preferred. It  is, thus, a positional measure.  
                  Its importance is very great in  agriculture like to find typical height of a crop variety, maximum source of  irrigation in a region, maximum disease prone paddy variety. Thus the mode is  an important measure in case of qualitative data.
                  Computation  of the mode
  Ungrouped or Raw Data
                  For ungrouped data or a series of  individual observations, mode is often found by mere inspection.
  Example 8
                  Find the mode for the following seed weight
                  2 , 7, 10, 15, 10, 17, 8, 10, 2 gms
                  \Mode = 10
                  In some cases the mode may be absent  while in some cases there may be more than one mode.
  Example 9
                  (1) 12, 10, 15, 24, 30 (no mode)
                  (2) 7, 10, 15, 12, 7, 14, 24, 10, 7, 20,  10
                  the modal values are 7 and 10 as both  occur 3 times each.
Grouped Data
                  For Discrete distribution, see the  highest frequency and corresponding value of x is mode.
                  Example: 
                  Find the mode for the following
Weight of sorghum in gms (x)  | 
                      No. of ear head(f)  | 
                    
50  | 
                      4  | 
                    
65  | 
                      6  | 
                    
75  | 
                      16  | 
                    
80  | 
                      8  | 
                    
95  | 
                      7  | 
                    
100  | 
                      4  | 
                    
Solution
                  The maximum frequency is 16. The corresponding  x value is 75. 
                  \ mode = 75 gms.
  Continuous distribution 
                  Locate the highest frequency the class corresponding  to that frequency is called the modal class. Then apply the formula.
                  Mode = ![]()
                  Where 
= lower limit of the model class
                  
= the frequency of the class preceding the model class
                  
= the frequency of the class succeeding the model class
                  and     c = class interval           
Example 10
                  For the frequency distribution of weights  of sorghum ear-heads given in table below. Calculate the mode
Weights of ear heads (g)  | 
                      No of ear heads (f)  | 
                      
  | 
                    
60-80  | 
                      22  | 
                      
  | 
                    
80-100  | 
                      
  | 
                      
  | 
                    
100-120  | 
                      
  | 
                      f  | 
                    
120-140  | 
                      
  | 
                      
  | 
                    
140-160  | 
                      20  | 
                      
  | 
                    
Total  | 
                      160  | 
                      
  | 
                    
Solution
                  Mode = ![]()
                  Here 
100, f = 45, c = 20, m =60, 
=38, 
=35
                  Mode = ![]()
                  =![]()
                  = 109.589
  Geometric mean
                  The geometric mean of a series containing  n observations is the nth root of the product of the values. 
                  If x1, x2…, xn are observations then
                  G.M=![]()
                  = ![]()
                  Log GM = ![]()
                  = ![]()
                  =![]()
                  GM = Antilog ![]()
                  For grouped data
                  GM = Antilog 
                  GM is used in studies like bacterial  growth, cell division, etc.
Example 11
                  If the weights of sorghum ear heads  are 45, 60, 48,100, 65 gms. Find the Geometric mean for the following data
Weight of ear head x (g)  | 
                      Log x  | 
                    
45  | 
                      1.653  | 
                    
60  | 
                      1.778  | 
                    
48  | 
                      1.681  | 
                    
100  | 
                      2.000  | 
                    
65  | 
                      1.813  | 
                    
Total  | 
                      8.925  | 
                    
Solution
                  Here n = 5
                  GM = Antilog ![]()
                  = Antilog 
 
                  = Antilog ![]()
                  = 60.95 
  Grouped Data
  Example  12
                  Find the Geometric mean for the following
Weight of sorghum (x)  | 
                      No. of ear head(f)  | 
                    
50  | 
                      4  | 
                    
65  | 
                      6  | 
                    
75  | 
                      16  | 
                    
80  | 
                      8  | 
                    
95  | 
                      7  | 
                    
100  | 
                      4  | 
                    
Solution
Weight of sorghum (x)  | 
                    No. of ear head(f)  | 
                    Log x  | 
                    f x log x  | 
                  
50  | 
                    5  | 
                    1.699  | 
                    8.495  | 
                  
63  | 
                    10  | 
                    10.799  | 
                    17.99  | 
                  
65  | 
                    5  | 
                    1.813  | 
                    9.065  | 
                  
130  | 
                    15  | 
                    2.114  | 
                    31.71  | 
                  
135  | 
                    15  | 
                    2.130  | 
                    31.95  | 
                  
Total  | 
                    50  | 
                    9.555  | 
                    99.21  | 
                  
Here n= 50
                  GM = Antilog 
                  = Antilog ![]()
= Antilog 1.9842 = 96.43
Continuous distribution 
                  Example 13
For the frequency distribution of  weights of sorghum ear-heads given in table below. Calculate the Geometric mean
Weights of ear heads ( in g)  | 
                      No of ear heads (f)  | 
                    
60-80  | 
                      22  | 
                    
80-100  | 
                      38  | 
                    
100-120  | 
                      45  | 
                    
120-140  | 
                      35  | 
                    
140-160  | 
                      20  | 
                    
Total  | 
                      160  | 
                    
Solution
Weights of ear heads ( in g)  | 
                      No of ear heads (f)  | 
                      Mid x  | 
                      Log x  | 
                      f log x  | 
                    
60-80  | 
                      22  | 
                      70  | 
                      1.845  | 
                      40  | 
                    
80-100  | 
                      38  | 
                      90  | 
                      1.954  | 
                      74.25  | 
                    
100-120  | 
                      45  | 
                      110  | 
                      2.041  | 
                      91.85  | 
                    
120-140  | 
                      35  | 
                      130  | 
                      2.114  | 
                      73.99  | 
                    
140-160  | 
                      20  | 
                      150  | 
                      2.176  | 
                      43.52  | 
                    
Total  | 
                      160  | 
                      
  | 
                      
  | 
                      324.2  | 
                    
Here n = 160
                  GM = Antilog 
                  = Antilog ![]()
                  = Antilog ![]()
                  = 106.23
  Harmonic mean (H.M)
                  Harmonic mean of a set of  observations is defined as the reciprocal of the arithmetic average of the  reciprocal of the given values. If x1, x2…..xn are n observations,
  
                  For a frequency distribution
  
                  H.M is used when we are dealing with speed, rates, etc.
Example 13
                  From the given data 5, 10,17,24,30 calculate  H.M.
X  | 
                      
  | 
                    
5  | 
                      0.2000  | 
                    
10  | 
                      0.1000  | 
                    
17  | 
                      0.0588  | 
                    
24  | 
                      0.0417  | 
                    
30  | 
                      0.4338  | 
                    
= 11.526
                    Example 14
                  Number of tomatoes per plant are given below.  Calculate the harmonic mean.
Number of tomatoes per plant  | 
                      20  | 
                      21  | 
                      22  | 
                      23  | 
                      24  | 
                      25  | 
                    
Number of plants  | 
                      4  | 
                      2  | 
                      7  | 
                      1  | 
                      3  | 
                      1  | 
                    
Solution
Number of tomatoes per plant (x)  | 
                      No of plants(f)  | 
                      
  | 
                      
  | 
                    
20  | 
                      4  | 
                      0.0500  | 
                      0.2000  | 
                    
21  | 
                      2  | 
                      0.0476  | 
                      0.0952  | 
                    
22  | 
                      7  | 
                      0.0454  | 
                      0.3178  | 
                    
23  | 
                      1  | 
                      0.0435  | 
                      0.0435  | 
                    
24  | 
                      3  | 
                      0.0417  | 
                      0.1251  | 
                    
25  | 
                      1  | 
                      0.0400  | 
                      0.0400  | 
                    
  | 
                      18  | 
                      
  | 
                      0.8216  | 
                    

![]()
                    Merits of H.M 
                  1. It is rigidly defined.
                  2. It is defined on all observations.
                  3. It is amenable to further algebraic  treatment.
                  4. It is the most suitable average when  it is desired to give greater weight to smaller observations and less weight to  the larger ones.
  Demerits of H.M 
                  1. It is not easily understood.
                  2. It is difficult to compute.
                  3. It is only a summary figure and may  not be the actual item in the series
                  4. It gives greater importance to small  items and is therefore, useful only when small items have 
                  to be given greater weightage.
                  5. It is rarely used in grouped data.
Percentiles 
                  The percentile  values divide the distribution into 100 parts each containing 1 percent of the  cases. The xth percentile is that value below which x percent of  values in the distribution fall. It may be noted that the median is the 50th  percentile.
For raw data,  first arrange the n observations in increasing order. Then the xth  percentile is given by 
                    ![]()
                  For a frequency distribution the xth  percentile is given by 
  ![]()
                  Where 
                  
= lower limit of the percentile calss which contains the xth  percentile value (x. n /100)
              
= cumulative frequency uotp ![]()
                  
= frequency of the percentile class
                  C=  class interval
                  N=  total number of observations
  Percentile for Raw Data or Ungrouped  Data 
  Example 15 
                  The  following are the paddy yields (kg/plot) from 14 plots: 30,32,35,38,40.42,48,49,52,55,58,60,62,and  65 ( after arranging in ascending order). The computation of 25th  percentile (Q1) and 75th  percentile (Q3) are given below: 
  ![]()
  ![]()
                  = 3rd item  + (4th item – 3rd item)![]()
                  = 35  + (38-35) ![]()
                  = 35  + 3
 = 37.25 kg
![]()
                    ![]()
                  = 11th  item + (12th item – 11th item)![]()
                  = 55 +(58-55) ![]()
                  = 55 + 3
 = 55.75 kg
Example 16
                  The frequency distribution of  weights of 190 sorghum ear-heads are given below. Compute 25th  percentile and 75th percentile.
Weight of ear-heads (in g)  | 
                      No of ear heads  | 
                    
40-60  | 
                      6  | 
                    
60-80  | 
                      28  | 
                    
80-100  | 
                      35  | 
                    
100-120  | 
                      55  | 
                    
120-140  | 
                      30  | 
                    
140-160  | 
                      15  | 
                    
160-180  | 
                      12  | 
                    
180-200  | 
                      9  | 
                    
Total  | 
                      190  | 
                    
Solution
Weight of ear-heads (in g)  | 
                    No of ear heads  | 
                    Less than class  | 
                    Cumulative frequency  | 
                  
40-60  | 
                    6  | 
                    < 60  | 
                    6  | 
                  
60-80  | 
                    28  | 
                    < 80  | 
                    
 47.5  | 
                  
80-100  | 
                    35  | 
                    <100  | 
                    
  | 
                  
100-120  | 
                    55  | 
                    <120  | 
                    
 142.5  | 
                  
120-140  | 
                    30  | 
                    <140  | 
                    154  | 
                  
140-160  | 
                    15  | 
                    <160  | 
                    169  | 
                  
160-180  | 
                    12  | 
                    <180  | 
                    181  | 
                  
180-200  | 
                    9  | 
                    <200  | 
                    190  | 
                  
Total  | 
                    190  | 
                    
  | 
                    
  | 
                  
or P25, first find out 
, and for 
, and proceed as in the case of median. For P25,  we have 
= 47.5.
                  The  value 47.5 lies between 34 and 69. Therefore, the percentile class is 80-100.  Hence,
  ![]()
  ![]()
  ![]()
                  = 80 +7.71 or 87.71 g.
                  Similarly,
  
Class
  ![]()
  ![]()
                  = 120 +14.33 =134.33 g.
Quartiles
                  The quartiles  divide the distribution in four parts. There are three quartiles. The second  quartile divides the distribution into two halves and therefore is the same as  the median. The first (lower).quartile (Q1) marks off the first one-fourth, the  third (upper) quartile (Q3) marks off the three-fourth. It may be noted that  the second quartile is the value of the median and 50th percentile.
Raw or ungrouped data
                  First arrange  the given data in the increasing order and use the formula for Q1 and Q3 then  quartile deviation, Q.D is given by
  ![]()
                  Where 
item and 
item
  Example 18
                  Compute quartiles for the data given  below (grains/panicles) 25, 18, 30, 8, 15, 5, 10, 35, 40, 45
Solution
                  5, 8, 10, 15, 18, 25, 30, 35, 40, 45 
  ![]()
                  
 
         = (2.75)th item 
                  = 2nd item + 
(3rd item – 2nd item)
                  = 8+
(10-8)
                  = 8+
x 2
                  = 8+1.5
                  = 9.5
  ![]()
                  = 3 x (2.75) th item 
                  = (8.75)th item
                  = 8th item +
(9th item – 8th item)
                  = 35+
(40-35)
                  = 35+1.25
                  = 36.25
Discrete Series
Step1: Find cumulative frequencies.
Step2: Find ![]()
Step3: See in the cumulative frequencies,  the value just greater than 
 , then the  corresponding value of x is Q1
Step4: Find ![]()
Step5: See in the cumulative frequencies,  the value just greater than  
 ,then the  corresponding value of x is Q3
Example 19
                  Compute quartiles for the data given  bellow (insects/plant).
X  | 
                      5  | 
                      8  | 
                      12  | 
                      15  | 
                      19  | 
                      24  | 
                      30  | 
                    
f  | 
                      4  | 
                      3  | 
                      2  | 
                      4  | 
                      5  | 
                      2  | 
                      4  | 
                    
Solution
x  | 
                    f  | 
                    cf  | 
                  
5  | 
                    4  | 
                    4  | 
                  
8  | 
                    3  | 
                    7  | 
                  
12  | 
                    2  | 
                    9  | 
                  
15  | 
                    4  | 
                    13  | 
                  
19  | 
                    5  | 
                    18  | 
                  
24  | 
                    2  | 
                    20  | 
                  
![]()
                    
=18.75th item \Q1= 8; Q3=24
Continuous series 
                  Step1: Find cumulative frequencies
                  Step2: Find ![]()
                  Step3: See in the cumulative frequencies,  the value just greater than
, then the corresponding class interval is called first  quartile class.
                  Step4: Find 
See in the  cumulative frequencies the value just greater than 
then the  corresponding class interval is called 3rd quartile class. Then  apply the respective formulae
  
  
                  Where l1 = lower limit  of the first quartile class
  f1 = frequency of the first quartile class
  c1 = width of the first quartile class
  m1 = c.f. preceding the first quartile class
  l3 = 1ower limit of the 3rd quartile class
  f3 = frequency of the 3rd quartile class
  c3 = width of the 3rd quartile class
  m3 = c.f. preceding the 3rd quartile class
Example 20: The following series relates to the marks secured by students in an examination.
Marks  | 
                      No. of Students  | 
                    
0-10  | 
                      11  | 
                    
10-20  | 
                      18  | 
                    
20-30  | 
                      25  | 
                    
30-40  | 
                      28  | 
                    
40-50  | 
                      30  | 
                    
50-60  | 
                      33  | 
                    
60-70  | 
                      22  | 
                    
70-80  | 
                      15  | 
                    
80-90  | 
                      12  | 
                    
90-100  | 
                      10  | 
                    
Find the quartiles
                    Solution
C.I  | 
                      f  | 
                      cf  | 
                    
0-10  | 
                      11  | 
                      11  | 
                    
10-20  | 
                      18  | 
                      29  | 
                    
20-30  | 
                      25  | 
                      54  | 
                    
30-40  | 
                      28  | 
                      82  | 
                    
40-50  | 
                      30  | 
                      112  | 
                    
50-60  | 
                      33  | 
                      145  | 
                    
60-70  | 
                      22  | 
                      167  | 
                    
70-80  | 
                      15  | 
                      182  | 
                    
80-90  | 
                      12  | 
                      194  | 
                    
90-100  | 
                      10  | 
                      204  | 
                    
  | 
                      204  | 
                      
  | 
                    
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