Question

in a population distribution a score of x=28 corresponds to z=-1.00 and a score of x=34 what is the mean and standard deviation

Answer 1

$x=28\implies z=\dfrac{28-\mu}\sigma\implies \mu-\sigma=28$

$x=34\implies z=\dfrac{34-\mu}\sigma\implies \mu+\sigma z=34$

We need the exact value of $z$ to find a proper solution, but a general one can still be found. Subtracting the first equation from the second gives

$(\mu+\sigma z)-(\mu-\sigma)=34-28\implies (z+1)\sigma=6\implies \sigma=\dfrac6{z+1}$

Plug this into either equation and you get

$\mu-\dfrac6{z+1}=28\implies \mu=28+\dfrac6{z+1}$

(It's guaranteed that $z\neq-1$ in this case because that already corresponds to $x=28$.)