Question

- Given that z is a Standard Normal random variable, find z for each situation:

(**9 marks**)

- the area to the left of z is 0.68

- the area to the right of z is 0.68

the total area to the left of –z and to the right of z is 0.32

Answer #1

solution

Using standard normal table,

P(Z < z) = 0.68

= P(Z < z) = 0.68

= P(Z < 0.47) = 0.68

z = 0.47

(B)

Using standard normal table,

P(Z > z) = 0.68

= 1 - P(Z < z) = 0.68

= P(Z < z ) = 1 - 0.68

= P(Z < z ) = 0.32

= P(Z < -0.47 ) = 0.32

z = -0.47 (using standard normal (Z) table )

(c)

P(-z < Z < z) = 0.32

P(Z < z) - P(Z < -z) = 0.32

2 P(Z < z) - 1 = 0.32

2 P(Z < z) = 1 + 0.32 = 1.32

P(Z < z) = 1.32 / 2 = 0.66

P(Z <0.41 ) = 0.66

z ± 0.41

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