PROBABILITY DISTRIBUTIONS

Example 1 :

Find the value of each of the following probabilities by using the calculator.

(a) P(Z > 0.934)

SD

Mode casio 570

shift + 3dist P( Q( R(

1 2 3

Choose P(z) if P(z<)

Choose Q(z) if P( <z<0) or P( 0<z< )

Choose R(z) if P(z>)

= R(0.934)

= 0.1752

Example 2 :

Find the value of each of the following probabilities by using the calculator.

(b) P(Z <1.25)

Solution P(z< 1.25) = 1 – P(Z > 1.25) = 1 – 0.1057

= 0.8944

1.25 1.25

Or P(1.25)

= 0.8944

(c) P(Z < - 1.512)

-1.512 1.512

P(Z < - 1.512) = P(Z > 1.512) = 0.06527

or

P(-1.512) = R(1.512)

= 0.06527

(d) P(Z > - 0.23)

-0.23 0.23

P(z > - 0.23) = 1 – P(Z < - 0.23) = 1 – P(Z > 0.23) = 1 – 0.40905

= 0.59095

orR(-0.23)

=0.59095

(e) P(0.4 < Z < 1.2)

P(0.4 < Z < 1.2) = P(Z > 0.4) – P(Z > 1.2) = 0.3446 – 0.1151

= 0.2295

0.4 1.20.4 1.2

Or R(0.4) – R (1.2) = 0.3446 – 0.1151

= 0.2295

(f) P(- 0.828 < Z < - 0. 555)

-0.828 -0.555 0.555 0.828 0.555 0.828

P(-0.828 < z < - 0. 555) = P(0.555 < z < 0. 828)

= P(Z > 0.555) – P(Z > 0.828) = 0.28945 – 0.20384 = 0.08561

R(0.555) – R(0.828)P(-0.555) – P(-0.828) =

= 0.08561

(g) P(- 0.255 < Z < 0.13)

-0.255 0.13 0.13-0.255

P(-0.255< Z < 0.13) = 1 – P(Z < - 0.255) – P(Z > 0.13) = 1 – P(Z > 0.255) – P(Z > 0.13) = 1 – 0.39936 – 0.44828 = 0.1524

or

Q(-0.255) + Q(0.13)

= 0.10064 + 0.05172

= 0.1524

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