Given that z is a standard normal random variable, what is the value of z if the area to the left of z is 0.9382: a. 1.8
b. 1.54
c. 2.1
d. 1.77

1.54

Step-by-step explanation:

Given that :

Z = normal random variable

Area to the left of z = 0.9382

For a random normal distribution ;

P(Z < x) = 0.9382

Using the Z table (standard normal distribution table) ;

Locating the probability value (0.9382) on the table and picking the intersecting z value both horizontally and vertically :

Horizontal value = 1.5

Vertical value = 0.04

Horizontal + vertical = (1.5 + 0.04) = 1.54

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3