J
jonbon
Member
Hi Everyone!
Could you kindly help me please. This has confused me forever!
For a standard normal distribution with
x p(x)
0 0.5
0.33 0.6293
0.67 0.74857
1 0.84135
1.33 0.90824
1.67 0.95254
2 0.97725
how is
1) z=-1 a 5 in a 6 chance?
2) z=4/3 a 1 in 11 chance
3) z=2/3 a 1 in 4 chance?
Is there a table of this chances?
Many thanks!
Jon.
Could you kindly help me please. This has confused me forever!
For a standard normal distribution with
x p(x)
0 0.5
0.33 0.6293
0.67 0.74857
1 0.84135
1.33 0.90824
1.67 0.95254
2 0.97725
how is
1) z=-1 a 5 in a 6 chance?
2) z=4/3 a 1 in 11 chance
3) z=2/3 a 1 in 4 chance?
Is there a table of this chances?
Many thanks!
Jon.