(1 point) Find the z-score such that:
(a) The area under the standard normal curve to its left is 0.8164
z =
(b) The area under the standard normal curve to its left is 0.6192
z =
(c) The area under the standard normal curve to its right is 0.1941
z =
(d) The area under the standard normal curve to its right is 0.1055
z =

(a) The value of z is 0.90.

(b) The value of z is 0.30.

(c) The value of z is 0.86.

(d) The value of z is 1.25.

Step-by-step explanation:

If , then , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, .

(a)

Compute the value of z for P (Z < z) = 0.8164 as follows:

P (Z < z) = 0.8164

The probability is approximately true for z = 0.90.

Thus, the value of z is 0.90.

(b)

Compute the value of z for P (Z < z) = 0.6192 as follows:

P (Z < z) = 0.6192

The probability is approximately true for z = 0.30.

Thus, the value of z is 0.30.

(c)

Compute the value of z for P (Z > z) = 0.1941 as follows:

P (Z > z) = 0.1941

P (Z < z) = 1 - 0.1941

             = 0.8059

The probability is approximately true for z = 0.86.

Thus, the value of z is 0.86.

(d)

Compute the value of z for P (Z > z) = 0.1055 as follows:

P (Z > z) = 0.1055

P (Z < z) = 1 - 0.1055

             = 0.8945

The probability is approximately true for z = 1.25.

Thus, the value of z is 1.25.

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !