Question

1) Find the following probabilities:
a) Pr{Z < 0.63}

b)Pr{Z ≥ −0.63}

c) Pr{−2.12 < Z < 2.12}
d) Pr{−2.12 < Z < 0.0}

e) Pr{−4.00 < Z < 0.0}
f) Pr{Z < −1.32 or Z > 1.32} (you want the probability that Z is outside the range −1.32 to 1.32)

g) Pr{−1.32 < Z < 1.32} h) Add (f) and (g). Are you surprised? Why or why not?

Answer 1

a)

P(z < 0.63) = 0.7357

b)

P(Z >= -0.63) = P(Z < 0.63)

= 0.7357

c)

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

= 0.9830 - 0.0170

= 0.9660

d)

P( -2.12 < Z < 0) = P(Z < 0) - P(Z < -2.12)

= 0.5 - 0.0170

= 0.4830

e)

P(-4 < Z < 0) = P(Z < 0) - P(Z < -4.00)

= 0.5 - 0

= 0.5

f)

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

= 1 - [ P(Z < 1.32) - P(Z < -1.32) ]

= 1 - [ 0.9066 - 0.0934 ]

= 0.1868

g)

P( -1.32 < Z < 1.32) = P(Z < 1.32) - P(Z < -1.32)

= 0.9066 - 0.0934

= 0.8132

h)

f + g = 0.1868 + 0.8132

= 1

yes, Since total area between z-score of -1.32 and z-score of 1.32 and outside of -1.32 and 1.32 is

whole area of the curve.

It should be 1.