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?
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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