Question

Provided N(0,1) and without using the LSND program find

P(−3<Z<1)

P(Z>−1)

P(Z<−1 OR Z>0)

P(0≤Z<1)

P(Z≥0)

P(Z<−2 OR Z>3)

Answer 1

Solution

Using standard normal table

a ) P ( −3 < Z < 1)

P ( Z < 1) - P ( Z < −3 )

= 0.8413 - 0.0013

= 0.8000

Probability = 0.8000

b) P ( Z > - 1 )

1 - P ( Z < - 1 )

= 1 - 0.1587

= 0.8413

Probability = 0.8413

c ) P ( Z <- 1 )

= 0.1587

P ( Z > 0)

1 - P ( Z < 0 )

= 1 - 0.5000

= 0.5000

Probability = 0.5000 + 0.1587 =0.6587

d) P ( 0 < Z < 1)

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

= 0.8413 - 0.5000

= 0.3413

Probability = 0.3413

e ) P ( Z > 0)

1 - P ( Z < 0 )

= 1 - 0.5000

= 0.5000

Probability = 0.5000

f ) P ( Z < - 2 )

= 0.0228

P ( Z > 3 )

1 - P ( Z < 3 )

= 1 - 0.9987

= 0.0013

Probability = 0.5000 + 0.0013 =0.5013