Essays on Quantitavtive Techniques Assignment

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BLACKBURN PCTAgesFrequencyFMid xfxClass limitst = (x-55.23)t 2f t 2cf25-3480129.523629.524.5- 34.5-25.73662.053026280135-44113739.544911.534.5-44.5-15.73247.4281293.8        193845-54120549.559647.544.5-54.5  -5.7332.839524314355-64138259.58222954.5-64.54.2718.2    25152.4454265-74114969.579855.564.5-74.5  14.27203.6233936.4567475-8469179.554934.574.5-84.524.27589.02 406999636585-9418589.516557.584.5-94.534.271174.4217270.16550Ef =6550Efx=361765Ef t 2 = 1538437.7Question 1Estimated mode = 59.5. Estimated but the actual will be given by the formulae;   Mode = L + ((A)/(A+B))I.                  L = 54.5                    A = 1382-1205=177                    B = 1382-1149 = 233Mode = 54.5 + (177)/(177+ 233))10Mode therefore becomes 58.82Question twoMean = Efx EfMean = 361765 6550 =55.2312Variance = (1/N) summation of f(X---mean) 2Standard deviation = square root of varianceVariance = 1538437.7/6550V = 234.88Standard deviation = 15.33SALFORD PCTAgesFrequencyFMid xfxClass limitst = (x-52.38)t 2f t 2cf25-34157929.546580.524.5- 34.5-22.88523.5826606.65157935-44191839.57576134.5-44.5-12.88165.9318196.2349745-54221849.510979144.5-54.5-2.888.318409.4571555-64258459.515374854.5-64.57.12 50.718398.08829965-74224969.5156305.564.5-74.517.1217.1238502.881054875-84134979.566775.574.5-84.527.12735.5992189.51189785-9438289.53418984.5-94.537.121377.9526357.812279Ef =12279Efx=643150.5Ef t 2 = 2,738,660.51Question 1Estimated mode = 59.72Question 1Estimated mode = 59.5Mode = L + ((A)/(A+B))I. Where A is the difference between the frequency of the modal and theupper frequency and B is the difference between the frequency of themodal class and the next upper class.

L is the lower class boundary ofthe modal class and I is the class interval of the modal class.              A= 2584 – 2218                  =366               B = 2584 – 2249                  = 335               L =54.5Mode therefore becomes 54.5 + (366/(335+366))10                  = 54.5 + 5.22                  = 59.72.Question twoMean = Efx/EfMean = 643150.5 /12279        =52.378Variance = (1/N) summation of f(X---mean) 2          = 2,738,660.51/(12279)         = 223.036Standard deviation = square root of the varianceStandard deviation = 14.93Estimated median = 55.9

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