It is given that,
[tex]P(zHere 'z' is the standard normal variate.CASE-1 Assuming that 'c' lies on right side of z=0 line.
Simplify the expression as,
[tex]P(z<0)+P(0We know that the probability of z being less than zero, is 0.5,[tex]0.5+P(0Observe that the probability comes out to be negative, which is not possible.CASE-2 Assuming that 'c' lies on the left side of z=0 line.
Then,
[tex]\begin{gathered} P(zFrom the Standard Normal Distribution Table, we see that the probability 0.1316 corresponds to the score close to 0.34 i.e.[tex]\varnothing(0.34)=0.1331[/tex]Thus, the value of the variable 'c' is 0.34 approximately.
