Standar normal distribution, given that:
We have to find the value for z=0.87
Using the z table:
P(0.87)=0.80785
Substituing:
P(c)=0.02685
Using the z table, we have to find the z-value for 0.02685.
P(c)=0.0268 approximately.
Using the z-table for nagative values:
P(-1.93)=0.0268.
Therefore, c=-1.93.