# A personality test is administered to a large group of

A personality test is administered to a large group of subjects. Five scores are shown below, in original units and in standard units. Fill in the bla
.
$\left[\begin{array}{ccccc}79& 64& 52& 72& -\\ 1.8& 0.8& -& -& -1.4\end{array}\right]$
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The standard normal variable is z
$z=\frac{x-\mu }{\sigma }$
When $x=79,z=1.8i.e.$
$1.8=\frac{79-\mu }{\sigma }⇒\mu +1.8\sigma =78$
When $x=64,z=0.8i.e.$
$0.8=\frac{64-\mu }{\sigma }⇒\mu +0.8\sigma =64$
Solving the above two equations give
$\sigma =14$
Now, $\mu +1.8\left(14\right)=78⇒\mu =52.8$
When $x=52$
$z=\frac{52-52.8}{14}=-0.06$
When $z=-1.4$
$-1.4=\frac{x-52.8}{14}$

$⇒=33.2$

###### Not exactly what you’re looking for?
usaho4w
Use z-score formula to find the missing values.
$z=\frac{x-mean}{sd}$
$sd=$ standard deviation
First you have to figure out the mean and standard deviation.
Lets