Question

Find the z-score, to the nearest hundredth, that satisfies the given condition.

0.206 square unit of the area of the standard normal distribution is to the right of z.

Answer 1

solution

Using standard normal table,

P(Z > z) = 0.206

= 1 - P(Z < z) = 0.206

= P(Z < z ) = 1 - 0.206

= P(Z < z ) = 0.794

= P(Z < 0.82) = 0.794

z =0.82 (using standard normal (Z) table )

z-score=0.82