Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean µ = 0 and standard deviation σ = 1. (12 pts.)
a. The area to the left of z is 15%.
b. The area to the right of z is 65%.
c. The area to the left of z is 10%.
d. The area to the right of z is 5%
e. The area between –z and z is 95%. (Hint: draw a picture and figure out the area to the left of the –z)
f. The area between –z and z is 99%
USING EXCEL.
Ans:
a)
P(Z<z)=0.15
z=normsinv(0.15)=-1.036
b)
P(Z>z)=0.65
P(Z<z)=1-0.65=0.35
z=normsinv(0.35)=-0.385
c)
P(Z<z)=0.1
z=normsinv(0.1)
z=-1.282
d)
P(Z>z)=0.05
P(Z<z)=1-0.05=0.95
z=normsinv(0.95)=1.645
e)Tails area=1-0.95=0.05
half tail area=0.05/2=0.025
z=normsinv(0.025) or normsinv(1-0.025)
z=+/-1.96
f)
Tails area=1-0.99=0.01
half tail area=0.01/2=0.005
z=normsinv(0.005) or normsinv(1-0.005)
z=+/-2.576
Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal...
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