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Solutions for 02402 exam 15. December 20111

Solutions for 02402 exam 15. December 20111

Solutions for 02402 exam 15. December 20111

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counts the ones corresponding to the event in question. And <strong>for</strong> the latter one of eachhas to be chosen:( ) ( ) ( )2 3 41 1 1P 2 =) = 6 21(93Hence the correct answer is 5.Exercise VIIn a store chain with separate bakery departments it is required, that the breads in allthe chain’s bakery shops are reasonably homogeneous. For the weight of a given breadthe following requirements are specified:1% of the breads are allowed to weigh below 600 g.5% of the breads are allowed to weigh above 650 g.Data are assumed to follow a normal distributionQuestion VI.1 (13) The mean and standard deviation (µ, σ) to be used <strong>for</strong> the productionto meet these requirements are? First we find the relevant percentiles of thestandard normal (Z) distribution:P (Z < z 0.99 ) = 0.01 and P (Z > z 0.05 ) = 0.05Hence these values can be found from R as qnorm(0.01) and qnorm(0.95) or in Table3:z 0.99 = −2.326 and z 0.05 = 1.645We need to find the µ and the σ that will give the same probabilities in a non-standardnormal X, so we can use the basic standardization <strong>for</strong>mula Z = X−µ twice and solveσ2 equations with two unknowns:−2.326 = x 0.99 − µand 1.645 = x 0.05 − µσσthat is:−2.326 = 600 − µ and 1.645 = 650 − µσσThe solutions is the one given as answer 5: µ = 629.3g and σ = 12.6gExercise VIIFour medical drugs with the same active ingredient are compared in terms of sideeffects. 200 persons participate in a study where the side effects <strong>for</strong> the individualperson is categorized as none / light / serious. The results of the investigation isshown in the table below:7

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