ch5_processing_of_large_data_sets.studentview

N in finding the median of ungrouped data see note

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Unformatted text preview: , whether N is odd or 2 even. A detailed argument would prove this. (HOWEVER, I WILL NOT BE TESTING THE FORMULA FOR MEDIAN FOR GROUPED DATA THIS SEMESTER.) MEDIAN Example 3: Find the median for the Example Find data of Table 2. data ∴ N = 12.5 2 Solution: N=25. N=25. Since classes 1, 2 and 3 have 8 (=1+3+4) members together, the 12.5th position together, MUST be in the 4th class, MUST which is identified as the median class. median This class is of width 10 units = 10) ,(cstarting from 39.95 to 49.95. It has 6 ( f = 6) members, of which the median is in the 4.5th (12.5 - 8=4.5) position 4.5 of that class. m Class Class Boundaries Boundaries Class Class mark mark freq 1 9.95-19.95 14.95 1 2 19.95-29.95 24.95 3 3 29.95-39.95 34.95 4 4 39.95-49.95 44.95 6 5 49.95-59.95 54.95 7 6 59.95-69.95 64.95 4 X total XXX 25 m ∴ Med = 39.95 + 10 × 12.5 − (1 + 3 + 4) = 47.45 6 Formula for the median of grouped data is: Formula N − ( f1 + f 2 + .... + f m −1 ) Med = Lm + cm × 2 fm where m −1 ∑f i =1 i Lm = Left boundary of median class, cm = width of median Class, width frequency f m = frequency of median class, = f1 + .......... + f m −1 = total frequency BEFORE median class total N = number of data F: Finding the Mode for grouped data F: For grouped data, the mode is found with reference to the diagram below. Example 4: Find the mode of the data Find of Table 2. of Solution: The mode is in the class The having the highest frequency ( f = 7) . This class is called the modal class. modal Its exact position is determined by the frequency of the class to the left ( f M −1 = 6) , and the frequency of the class to the right ( f M +1 = 4) of the modal class. of The main idea is that h : k = a : b , The M h a a = ⇒ h = (h + k ) × h+k a+b a+b yielding Therefore, Mod = Left boundary + h Therefore, a = f M − f M −1 = 1 b = f M − f M +1 = 3 c = 10 Class Boundaries Class mark freq 1 9.95-19.95 14.95 1 2 19.95-29.95 24.95 3 3 29.95-39.95 34.95 4 4 39.95-49.95 44.95 6 5 49.95-59.95 54.95 7 6 59.95-69.95 64.95 4 X total XXX 25 a 1 ) = 49.95 + 10 × ( ) = 52.45 a +b 1+ 3 f M − f M −1 ( 7 − 6) = LM + c × = 49.95 + 10 × 2 f M − f M −1 − f M +1 (7 − 6) + (7 − 4) ∴ Mod = LM + c × ( where LM = Left boundary of modal class, c = Class width, where f M = frequency of modal class Proof : (THE FORMULA FOR FINDING THE MODE FOR GROUPED DATA WILL NOT BE TESTED THIS SEMESTER) GROUPED The background for the above The argument is based on some analytical geometry knowledge . We first locate the mid-points P, Q and R of the top edges of the 3 neighbouring rectangles. rectangles. Then fit a parabola to pass Then through P, Q and R. From the highest point , T, drop a vertical down to the floor, dividing the base in the ratio h : k Then h : k = a : b G: Skewness (This formula will be tested) G: * When a distribution of data is When symmetric, symmetric • Mean=Median (Figure 1). The Mean=Median skewness = 0. skewness * If Mean>Median, distribution...
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This document was uploaded on 09/28/2013.

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