Question

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

a.	​P(zgreater than>1.981.98​)	e.	​P(zgreater than or equals≥​0)
b.	​P(zless than<negative 1.64−1.64​)	f.	​P(negative 2.72−2.72less than or equals≤zless than or equals≤1.531.53​)
c.	​P(0.170.17less than or equals≤zless than or equals≤2.122.12​)	g.	​P(zgreater than or equals≥negative 2.63−2.63​)
d.	​P(negative 1.25−1.25less than or equals≤zless than<negative 0.65−0.65​)	h.	​P(zless than<2.632.63​)

Answer 1

a)

P(Z > 1.98) = 1 - P(Z < 1.98)

= 1 - 0.9761(From Z table)

= 0.0239

b)

P(Z < -1.64) = 1 - P(Z < 1.64)

= 1 - 0.9495 (From Z table)

= 0.0505

c)

0.17 <= Z <= 2.12) = P(Z <= 2.12) - P(Z <= 0.17)

= 0.9830 - 0.5675 (From Z table)

= 0.4155

d)

P(-1.25 <= Z <= -0.65) = P(Z <= -0.65) - P(Z <= -1.25)

= ( 1 - P(Z < 0.65) ) - ( 1 - P(Z < 1.25) )

= ( 1 - 0.7422) - ( 1 - 0.8944) (From Z table)

= 0.1522

e)

P(Z >= 0) = 1 - P(Z < 0)

= 1 - 0.5 (From Z table)

= 0.5

f)

P(-2.72 <= Z <= 1.53) = P(Z < 1.53) - P(Z < -2.72)

P(Z < 1.53) - ( 1 - P(Z < 2.72) )

= 0.9370 - ( 1 - 0.9967 ) (From Z table)

= 0.9337

g)

P(Z >= - 2.63) = P(Z < 2.63)

= 0.9957(From Z table)

h)

P(Z < 2.63) = 0.9957 (From Z table)