Question

For the standard normal random variable z, find z for each situation. If required, round your answers to two decimal places. For those boxes in which you must enter subtractive or negative numbers use a minus sign. (Example: -300)'

a. The area to the left of z is 0.1827. z =

b. The area between −z and z is 0.9830. z =

c. The area between −z and z is 0.2148. z =

d. The area to the left of z is 0.9997. z =

e. The area to the right of z is 0.6847. z=

Answer 1

Solution :

Using standard normal table,

a)

P(Z < z) = 0.1827

P(Z < -0.91) = 0.1827

z = -0.91

(b)

P(-z < Z < z) = 0.9830

P(Z < z) - P(Z < z) = 0.9830

2P(Z < z) - 1 = 0.9830

2P(Z < z) = 1 + 0.9830

2P(Z < z) = 1.9830

P(Z < z) = 1.9830 / 2 = 0.9915

P(Z < 2.39) = 0.9915

z = 2.39

(c)

P(-z < Z < z) = 0.2148

P(Z < z) - P(Z < z) = 0.2148

2P(Z < z) - 1 = 0.2148

2P(Z < z) = 1 + 0.2148

2P(Z < z) = 1.2148

P(Z < z) = 1.2148 / 2

P(Z < 0.27) = 0.6074

z = 0.27

d)

P(Z < z) =0.9997

P(Z < 3.43) = 0.9997

z = 3.43

e)

P(Z > z) = 0.6847

1 - P(Z < z) = 0.6847

P(Z < z) = 1 - 0.6847 = 0.3153

P(Z < -0.48) = 0.3153

z = -0.48