Question

Find the following probability for the standard normal random variable z.

a.	​P(zgreater than>1.381.38​)	e.	​P(zgreater than>​0)
b.	​P(zless than<negative 1.11−1.11​)	f.	​P(negative 2.31−2.31less than or equals≤zless than or equals≤1.331.33​)
c.	​P(0.680.68less than or equals≤zless than or equals≤2.412.41​)	g.	​P(zgreater than or equals≥negative 2.64−2.64​)
d.	​P(negative 1.18−1.18less than or equals≤zless than<negative 0.52−0.52​)	h.	​P(zless than<2.642.64​)

Answer 1

a) P(Z > 1.38)

= 1 - P(Z < 1.38)

= 1 - 0.9162

= 0.0838

b) P(Z < -1.11)

= 0.1335

c) P(0.68 < Z < 2.41)

= P(Z < 2.41) - P(Z < 0.68)

= 0.9920 - 0.7517

= 0.2403

d) P(-1.18 < Z < -0.52)

= P(Z < -0.52) - P(Z < -1.18)

= 0.3015 - 0.1190

= 0.1825

e) P(Z > 0)

= 1 - P(Z < 0)

= 1 - 0.5 = 0.5

f) P(-2.31 < Z < 1.33)

= P(Z < 1.33) - P(Z < -2.31)

= 0.9082 - 0.0104

= 0.8978

g) P(Z > -2.64)

= 1 - P(Z < -2.64)

= 1 - 0.0041

= 0.9959

h) P(Z < 2.64)

= 0.9959