# For a population of exam scores, a score of x=58 corresponds to z=+0.50 and a score of x=46 corresponds to z=-1.00. Find the mean and the standard deviation for the population.

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To solve apply the formula:

`z=(x-nu)/(sigma)`

where

z -- z-score ,

x -- raw score,

`nu` -- mean and

`sigma` -- standard deviation

So, plug-in z=0.5, x=58 to the formula.

`0.5=(58-nu)/(sigma)`

`0.5sigma=58-nu` (Let this be EQ1.)

Also, plug-in z=-1 and x=46 to the formula too.

`-1=(46-nu)/(sigma)`

`-sigma=46-nu`

`sigma=-46+nu` (Let this be EQ2.)

Then, substitute EQ2 to EQ1 in order to express EQ1 with nu only.

`0.5sigma=58-nu`

`0.5(-46+nu)=58-nu`

`-23+0.5nu=58-nu`

`-81+0.5nu=-nu`

`-81=-1.5nu`

`54=nu`

Then, plug-in this value to EQ2.

`sigma=-46+nu`

`sigma=-46+54`

`sigma=8`

**Hence, the mean is 54 and the standard deviation is 8.**