# SOLUTION: Chebychev's Theorem: Suppose the time needed to assemble a piece of furniture is not normally distributed and that the mean assembly time is 28 minutes. What is the value of the s

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 Click here to see ALL problems on Probability-and-statistics Question 599968: Chebychev's Theorem: Suppose the time needed to assemble a piece of furniture is not normally distributed and that the mean assembly time is 28 minutes. What is the value of the standard deviation if at lease 77% of the assembly times are between 24 and 32 minutes? Answer by Edwin McCravy(8879)   (Show Source): You can put this solution on YOUR website!Suppose the time needed to assemble a piece of furniture is not normally distributed and that the mean assembly time is 28 minutes. What is the value of the standard deviation if at lease 77% of the assembly times are between 24 and 32 minutes? ------------- ```The mean is 28 minutes. Therefore 24 minutes is 4 minutes below the mean and 32 minutes is 4 minutes above the mean, So we are talking about 4 minutes from the mean. We want to know what standard deviation would guarantee that at least 77% of the data would lie within 4 minutes of the mean. Chebychev's theorem says that 100(1-)% of the data lies within k standard deviations of the mean. That means that ks = 4, where s is the standard deviation. So first find k by setting 100(1-)% equal to 77% and solve for k: 100(1-)% = 77% Drop the percents: 100(1-) = 77 Divide both sides by 100 1 - = .77 <--(If your book just gives 1- you can start here) Clear of fractions by multiplying through by kČ: kČ - 1 = .77kČ Get the kČ terms on the left and the 1 on the right: kČ - .77kČ = 1 Factor out kČ on the left: kČ(1 - .77) = 1 kČ(.23) = 1 Divide both sides by .23 kČ = kČ = 4.347826087 Take the square root of both sides: k = 2.085144141 So that means that at least 77% of the data lies within 2.085144141 standard deviations of the mean. Since ks = 4, 2.085144141s = 4 minutes Divide both sides by 2.085144141 s = minutes s = 1.918332609 minutes Round off to however many decimal places your teacher told you, probably to 1.9 or 1.92 minutes. Edwin```