Question

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

a.	​P(zgreater than>1.321.32​)	d.	​P(negative 1.79−1.79less than or equals≤zless than<negative 0.61−0.61​)
b.	​P(zless than<negative 1.96−1.96​)	e.	​P(zgreater than>​0)
c.	​P(0.690.69less than or equals≤zless than or equals≤2.592.59​)	f.	​P(negative 2.52−2.52less than or equals≤zless than or equals≤1.011.01​)

​(Round to three decimal places as​ needed.)

Answer 1

a) P( Z > 1.32) = 1- P( Z < 1.32)

= 1- 0.90658

=0.09342

b) P( Z < -1.96)= 1- P( Z < 1.96)

= 1- 0.97500

= 0.025

c) P(0.69 ≤ Z ≤ 2.59)= P( Z <2.59) - P( Z < 0.69)

= 0.99520 - 0.75490

= 0.2403

d)P( -1.79 ≤ Z< -0.61)= P( z < -0.61) - P( z < -1.79)

= [ 1- P(z < 0.61)] - [ 1- P( z < 1.79)]

= P( z < 1.79) - P( z < 0.61)

= 0.96327 - 0.72907

= 0.2342

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

= 1- 0.5

= 0.5

f) P( -2.52 ≤ Z ≤ 1.01 )= P( z < 1.01) - P( z < -2.52)

= P( z < 1.01) -[1- P( z < 2.52)]

= P( z < 1.01) - 1 + P( z < 2.52)

= 0.84375 - 1 + 0.99413

= 0.83788