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08 2308 13 computation of mean and standard deviation

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Unformatted text preview: 1 - f 0 ×C Mode = l + 2f1 - f 0 - f 2 l =20, f1=21, f0=13, f2=16, C=5 21 − 13 Mode = 20 + ×5 2 × 21 − 13 − 16 8×5 = 20 + 42 − 29 40 = 20 + = 20 + 3.08 = 23.08 13 Computation of Mean and Standard deviation X Mid – Frequen Deviations fd d= cy point f M m − 22.5 5 -8 -4 2 2.5 0-5 -15 -3 5 7.5 5-10 -14 -2 7 12.5 10-15 -13 -1 13 17.5 15-20 0 0 21 22.5 20-25 16 1 16 27.5 25-30 16 2 8 32.5 30-35 9 3 3 37.5 35-40 N=75 ∑fd = -9 178 d 2 16 9 4 1 0 1 4 9 fd’ 2 32 45 28 13 0 16 32 27 ∑fd 2= 193 ∑ fd ×c N −9 = 22.5 + ×5 75 45 = 22.5 – 75 = 22.5 – 0.6 = 21.9 Mean = A+ ∑ fd 2 ∑ fd − ×c N N 2 σ= 193 − 9 − ×5 75 75 2 = 2.57 − 0.0144 × 5 = 2.5556 × 5 = 1.5986 × 5 = 7.99 Karl – Pearson’ s coefficient of skewness Mean - Mode = S .D. 21.9 − 23.08 = 7.99 − 1.18 = = −0.1477 7.99 7.10.2 Bowley’ s Coefficient of skewness: In Karl – Pearson’ s method of measuring skewness the whole of the series is needed. Prof. Bowley has suggested a formula based on relative position of quartiles. In a symmetrical distribution, the quartiles are equidistant from the value of the median; ie., Median – Q1 = Q3 – Median. But in a skewed distribution, the quartiles will not be equidistant from the median. Hence Bowley has suggested the following formula: Q + Q1 − 2 Median Bowley’ s Coefficient of skewness (sk) = 3 Q 3 − Q1 = 179 Example 21: Find the Bowley’ s coefficient of skewness for the following series. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 Solution: The given data in order 2, 4, 6, 10, 12, 14, 16, 18, 20, 22 n + 1 th Q1 = size of item 4 11 + 1 th = size of item 4 = size of 3rd item = 6 n + 1 th Q3 = size of 3 item 4 11 + 1 th = size of 3 item 4 = size of 9th item = 18 n + 1 Median = size of th item 2 11 + 1 = size of th item 2 = size of 6th item = 12 Q + Q1 − 2 Median Bowley’ s coefficient skewness = 3 Q 3 − Q1 18 + 6 − 2 × 12 =0 = 18 − 6 Since sk = 0, the given series is a symmetrical data. Example 22: Find Bowley’ s coefficient of skewness of the following series. Size : 4 4.5 5 5.5 6 6.5 7 7.5 8 f: 10 18 22 25 40 15 10 8 7 180 Solution: Size 4 4.5 5 5.5 6 6.5 7 7.5 8 Q1 f 10 18 22 25 40 15 10 8 7 c.f 10 28 50 75 115 130 140 148 155 N + 1 th = Size of item 4 155 + 1 th = Size o...
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This note was uploaded on 01/18/2014 for the course BUS 100 taught by Professor Moshiri during the Winter '08 term at UC Riverside.

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