Question 599968
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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<pre>
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-{{{1/k^2}}})% of the data lies within
k standard deviations of the mean.  That means that k<font face = "symbol">s</font> = 4, where <font face = "symbol">s</font> is 
the standard deviation.

So first find k by setting 100(1-{{{1/k^2}}})% equal to 77% and solve 
for k:

100(1-{{{1/k^2}}})% = 77%

Drop the percents:

100(1-{{{1/k^2}}}) = 77

Divide both sides by 100

1 - {{{1/k^2}}} = .77    <--(If your book just gives 1-{{{1/k^2}}} 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 = {{{1/.23}}}

             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 k<font face = "symbol">s</font> = 4,

   2.085144141<font face = "symbol">s</font> = 4 minutes

Divide both sides by 2.085144141

              <font face = "symbol">s</font> = {{{4/2.085144141}}} minutes

              <font face = "symbol">s</font> = 1.918332609 minutes

Round off to however many decimal places your teacher told you,
probably to 1.9 or 1.92 minutes.

Edwin</pre>