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# Median

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### Transcript

• 1. Median is called the middle most score. It is the score that is found midway of all the scores in a score distribution. The median score divides the group into two, the higher group those which are found above the median score and lower group those which are found just below the median score.
• Characteristics of the Median
• 1. An ordinal statistics.
• 2. A rank or position average.
• 3. Value is the middlemost score.
• 4. Not affected by extreme values.
• 5. Can be subjected only to few mathematical computations.
• 6. Less widely used than the mean.
• 2.
• When to use the Median
• 1. When the exact midpoint of the distribution is wanted, 50% point.
• 2. When the extremes score which would markedly affect the Mean.
• 3. When it is desired that certain scores should influence the central tendency, but all that is known about them is that they are above or below the median.
Finding the median from the ungrouped data. 1. When N is odd the middle most score is the median when the score are arranged from highest to lowest. N+1/2
• 3. Illustrative Example Example: 1. Find the median for the following; 67, 89, 76, 77, 90, 78, 77, 86, 84, Solution: Arrange the score from highest to lowest. 90, 89, 86, 84, 78, 77, 77, 76, 67 . Counting from 90 the middle most score is N + 1 = 9 + 1 = 5, 2 2 the middle most score is the 5th score counting from 90 or 67 . The 5th score is 78 , the median score.
• 4. 2. 78, 74, 98, 77 83, 80, 75, 82, 76, 82 Solution: N is even. There are two middle most score the 5th score counting from the highest or the lowest score. The median score is the sum of the two-middle most score divided by two. 98, 83, 82, 82, 80, 78, 77, 76, 75, 74 Counting the 5th score from 98 is 80 and counting the 5th score form 74 is 78, to get the median add the two 80 + 78 = 79 2
• 5. Finding the median from grouped data . Formula: Median = Where; = median Ll =lower limit of assumed median determine by the expression N/2 cfb = cumulative freq. just below the assumed median f = corresponding frequency ci = class interval
• 6. Steps: 1. Prepare a frequency distribution table with 3 columns. ( class marks, class frequency, less than cumulative frequency) 2. Determine the lower limit of the assumed median which is lies in N/2. 3. Substitute data to the formula.
• 7. Illustrative Example: Compute for the median for the following data . 2 11 - 13 2 14 - 16 2 17 - 19 2 20 - 22 6 23 - 25 12 26 - 28 6 29 - 31 4 32 - 34 2 35 - 37 3 38 - 40 4 41 - 43 3 44 - 46 2 47 - 49 f ci
• 8.
• Solution:
• Prepare a frequency distribution table with 3 columns. ( class marks, class frequency, less than cumulative frequency)
• 2. Determine the lower limit of the assumed median which is lies in N/2.
• 3. Substitute data to the formula.
• 9. 2 2 12 11 - 13 4 2 15 14 - 16 6 2 18 17 - 19 8 2 21 20 - 22 14 6 24 23 - 25 26 12 27 26 - 28 32 6 30 29 - 31 36 4 33 32 - 34 38 2 36 35 - 37 41 3 39 38 - 40 45 4 42 41 - 43 48 3 45 44 - 46 50 2 48 47 - 49 Less than Cf Frequency (f) Class marks X Class Interval
• 10. Md = = = 30.08 Interpretations: The computed median score is more than half of the highest score. The students could be concluded to perform good in the quiz. Solution: