Given that z is a standard normal random variable, find z for each situation (to 2 decimals).

a. the area to the left of z is .9738
b. the area between 0 and z is .4738 ( is positive).
c. the area to the left of z is .8531
d. the area to the right of z is .1251
e. the area to the left of z is .6700
f. the area to the right of z is .3300

a. 1.94

b. 1.94

c. 1.05

d. 1.15

e. 0.44

f. 0.44

Step-by-step explanation:

z-table is used to calculate the values

a. The area to the left of z is .9738

z= 1.94

b. The area between 0 and z is .4738 ( is positive)

z= 1.94

c. The area to the left of z is .8531

z=1.05

d. The area to the right of z is .1251

z=1.15

e. The area to the left of z is .6700

z=0.44

f. The area to the right of z is .3300

z=0.44

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14