(9 marks)
the total area to the left of –z and to the right of z is 0.32
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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