the length of an instant message conversation is normally distributed with a mean of 2 minutes and a sd of .5 minutes. what is t
1 answer:
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Answer:
Explained below.
Step-by-step explanation:
The complete question is:
Find the value of the probability of the standard normal variable Z corresponding to this area for problems 1-3.
1. P(Z < 1.62)
2. P(Z > -1.57)
3. P(-1.41 < Z < 0.63)
Solution:
Use Excel to solve the problems.
(1)
P(Z < 1.62) =NORM.S.DIST(1.62,TRUE)
= 0.9474
(2)
P(Z > -1.57) = P (Z < 1.57)
=NORM.S.DIST(1.57,TRUE)
= 0.9418
(3)
P(-1.41 < Z < 0.63) = P (Z < 0.63) - P (Z < -1.41)
=NORM.S.DIST(0.63,TRUE) - NORM.S.DIST(-1.41,TRUE)
= 0.7357 - 0.0793
= 0.6564
