Suppose that Z ∼ N(0, 1), find
(a) P(Z = −0.93)
(b) P(Z > −0.93)
(c) P(Z < 0.25)
(d) P(−2.13 ≤ Z ≤ 1.93)
We will use the standard normal tables to find these probabilities.
a.
"p(Z=-0.93)=0"
This probability is equal to zero because it is related to finding the area under a point(-0.93) which is zero.
b.
"p(Z\\gt-0.93)=1-p(Z\\lt-0.93)=1-0.1762= 0.8238"
c.
"p(Z\\lt0.25)=0.5987"
d.
"p(-2.13\\leqslant Z\\leqslant1.93)=\\phi(1.93)-\\phi(-2.13)=0.9732-0.0166= 0.9566"
Comments
Leave a comment