Answer:
a) x=12
b) x=8.71
c) x=10.95
d) x=3.92
e) x=5.152
Step-by-step explanation:
a) x=12.
To have a probability of 0.5, x must be the mean of X.
b) x=8.71
[tex]z=-1.6449\\\\x=\mu+z\sigma=12-1.6449*2=8.71[/tex]
c) x=10.95
[tex]P(X<12)=0.5\\\\P(X<x)=0.5-P(X<12)=0.5-0.2=0.3\\\\z=-0.5244\\\\x=12-0.5244*2=10.95[/tex]
d) x=3.92
If we define Y=X-12, then
[tex]\mu_Y=\mu_X-12\\\\\sigma_Y=\sigma_X[/tex]
For z=1.96, the probability around the mean is 0.95. Then:
[tex]x=(12-12)+1.96*2=0+3.92=3.92[/tex]
e) x=5.152
As in the question d, the area around the mean now is 0.99. This happens when z=2.576. Then:
[tex]x=(12-12)+2.576*2=0+5.152=5.152[/tex]
