ch6sometechniquesofcalculation-2.studentview

6 10 hence the mean sd of the 250 new marks can be

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Unformatted text preview: y = 60 − (50 − x) 10 x ≥ 50 x < 50 Find the mean and standard deviation of the 250 revised marks. revised Solution: Solution For the 200 students whose old mark are x ≥ 50 , 4 Mean(Y ) = 60 + (70 − 50) = 76 5 4 SD (Y ) = × 8 = 6.4 5 Similarly, for the 50 students with x < 50 , Similarly, Mean(Y ) = 60 − 11 (50 − 40) = 49 10 SD(Y ) = 11 × 6 = 6.6 10 Hence the mean & s.d. of the 250 new marks can be Hence found from the frequency table: found x f 76-6.4= 76+6.4= 49-6.6= 49+6.6= 69.6 100 82.4 100 42.4 25 55.6 25 N 250 ∴ µ = 70.6, σ = 12.5748 Simple Exercise: Simple Data set : x , x ,........, x Data Mean=12, Median=14, Mode=15 and CV=30%. Define new data set: yi = 80 − 4 xi i = 1,2,........,80 1 2 80 Find the mean, median, mode, s.d, CV and Sk for the yi ' s . Sk Solution: Solution RULES: RULES Y = a + cX then µY = a + cµ X MedY = a + c ⋅ Med X Mo...
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