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  1. 1. -- B. Kedhar Guhan -- -- X D -- -- 34 --
  2. 2. In this PPT, we would first recap what we had learntin 9th.-• Histograms : SLIDE 3• Frequency polygons: SLIDE 4• Numerical representatives of ungrouped data: SLIDE 5
  3. 3. Then we sneak-a-peek on- Central Tendencies of a Grouped Data: SLIDE 7 Grouped Data : SLIDE 8 Mean of a grouped data: SLIDE 9 Direct Method: SLIDE 11 Assumed mean method: SLIDE 13 Step deviation Method: SLIDE 15 Mode: SLIDE 16 Concept of Cumulative Frequency: SLIDE 17• Median: SLIDE 18• Ogives : SLIDE 20
  4. 4.  A Histogram displays a range of values of a variable that have been broken into groups or intervals. Histograms are useful if you are trying to graph a large set of quantitative data It is easier for us to analyse a data when it is represented as a histogram, rather than in other forms.
  5. 5.  Midpoints of the interval of corresponding rectangle in a histogram are joined together by straight lines. It gives a polygon They serve the same purpose as histograms, but are especially helpful in comparing two or more sets of data.
  6. 6. 1. Arithmetic Mean: (or Average)• Sum of all observation divided by the Number of observation.• Let x1,x2,x3,x4 ….xn be obs.( thus there are „n‟ number of scores) Then Average =(x1+x2+x3+x4 ….+xn)/n
  7. 7. 2 Median: When the data is arranged in ascending or descending order, the middle observation is the MEDIAN of the data. If n is even, the median is the average of the n/2nd and (n/2+1/2 )nd observation.3 Mode: It is the observation that has the highest frequency.
  8. 8.  A grouped data is one which is represented in a tabular form with the observations (x) arranged in ascendingdescending order and respective frequencies( f ) given.
  9. 9.  To obtain the mean,1. First, multiply value of each observation(x) to its respective frequency( f ).2. Add up all the obtained values(fx).3. Divide the obtained sum by the total no. of observations. MEAN =
  10. 10.  Lets find the mean of the given data. Marks 31 33 35 40 obtained (x) No. of 2 4 2 2 students (f )Lets find the Σfx and Σf. x f fxXi 31 Fi 2 FiXi 62Xii 33 Fii 4 FiiXii 144Xiii 35 Fiii 2 FiiiXiii 70Xiv 40 Fiv 2 FivXiv 80 Σf = 2+2+2+4 = 10 Σfx = 62+144+70+80 = 356So, Mean = Σfx = 356 =35.6 Σf 10
  11. 11.  Often, we come across sets of data with class intervals, like: Class 10-25 25-40 40-55 55-70 70-85 85-100 Interval No. os 2 3 7 6 6 6 students To find the mean of such data , we need a class mark(mid-point), which would serve as the representative of the whole class. Class Mark = Upper Limit + Lower Limit 2 **This method of finding mean is known as DIRECT METHOD**
  12. 12.  Lets find the class mark of the first class of the given table.Class Mark = Upper Limit(25) + Lower Limit(10) 2 = 35 = 17.5 2 Similarly, we can all the other Class Marks and derive this following table:C.I. No. of students(f ) C.M (x) fx Now,10-25 2 17.5 35.0 mean = Σfx25-40 3 32.5 97.540-55 7 47.5 332.5 Σf55-70 6 32.5 375.0 = 186070-85 6 77.5 465.0 3085-100 6 92.5 555.0Total Σf=30 Σfx=1860.0 = 62
  13. 13.  Another method of finding MEAN:1. Choose one of the observation as the “Assumed Mean”. [select that xi which is at the centre of x1, x2,…, xn.2. Then subtract a from each class mark x to obtain the respective d value (x-a).3. Find the value of FnDn, where n is a particular class; F is the frequency; and D is the obtained value.
  14. 14. Mean of the data= mean of the deviations =
  15. 15. 1. Follow the first two steps as in Assumed Mean method.2. Calculate u = xi-a h3. Now, mean = x = a+h { }, Σ fu Σf Where h=size of the CI f=frequency of the modal class a= assumed mean
  16. 16. The class with the highest frequency iscalled the MODAL CLASSC.I. No. of C.M fx students(f ) (x) In this set of10-25 2 17.5 35.0 data, the class25-40 3 32.5 97.5 “40-55” is the40-55 7 47.5 332.555-70 6 32.5 375.0 modal class as70-85 6 77.5 465.0 it has the85-100 6 92.5 555.0 highestTotal Σf=30 Σfx=1860 frequency
  17. 17.  It‟s the „running total‟ of frequencies. It‟s the frequency obtained by adding the of all the preceding classes. When the class is taken as less than [the Upper limit of the CI], the cumulative frequencies is said to be the less than type. When the class is taken as more than [the lower limit of the CI], the cumulative frequencies is said to be the more than type.
  18. 18.  If n ( no. of classes) is odd, the median is {(n+1)/2}nd class. If n is even, then the median is the average of n/2nd and (n/2 + 1)th class. Median for a grouped data is given by Median = l{ n/2f- cf }h Where l= Lower Limit of the class n= no. of observations cf= cumulative frequency of the preceding class f= frequency h= class size
  19. 19. 3 Median = Mode + 2 Mean
  20. 20.  Cumulative frequency distribution can be graphically represented as a cumulative frequency curve( Ogive )
  21. 21. More than type Ogive: Mark the LL each class intervals on the x-axis. Mark their corresponding cumulative frequency on the y-axis. Plot the points (L.l. , c.f.) Join all the plotted points by a free hand smooth curve. This curve is called Less than type ogive
  22. 22. Less than type Ogive : Mark the UL each class intervals on the x-axis. Mark their corresponding cumulative frequency on the y-axis. Plot the points (U.l. , c.f.) Join all the plotted points by a free hand smooth curve. This curve is called Less than type ogive .
  23. 23. METHOD 1 Locate n/2 on the y-axis. From here, draw a line parallel to x-axis, cutting an ogive ( less/more than type) at a point. From this point, drop a perpendicular to x-axis. The point of intersection of this perpendicular and the x-axis determines the median of the data.
  24. 24. METHOD 2: Draw Both the Ogives of the data. From the point of intersection of these Ogives, draw a perpendicular on the x-axis. The point of intersection of the perpendicular and the x-axis determines.
  25. 25. -- B. Kedhar Guhan -- -- X D -- -- 34 --