Find the following z values for the standard normal variable Z. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) |
a. | P(Z ≤ z) = 0.9477 | |
b. | P(Z > z) = 0.6300 | |
c. | P(−z ≤ Z ≤ z) = 0.85 | |
d. | P(0 ≤ Z ≤ z) = 0.2034 | |
Solution:
Z values can be found out from Z table
Solution(a)
P(Z<=z) = 0.9477
if p-value = 0.9477 so z = 1.62
Solution(b)
P(Z>z) = 0.6300
this is a right tailed test so p-vlaue = (1-0.6300) = 0.37
so from Z table we found z = -0.33
Solution(b)
P(-z<=Z<=z) = 0.85
this is two tailed test so p-value = (1-0.85)/2 = 0.075
from Z table we found z = -1.44 to z = +1.44
Solution(d)
P(0<=Z<z) = 0.2034
p-value for this = (0.2034+0.5) = 0.7034
Z score = 0 and 0.53
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