a random variable is normally distributed with a mean of 25 and a standard deviation of 5.a. what value will be exceeded 10% of the time. b.what value will be exceeded 85% of the time
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Answer:
P(z>z')=0.15\to P(z<z')=0.85P(z>z
′
)=0.15→P(z<z
′
)=0.85
From z-table:
z'=1.04z
′
=1.04
x'=20+(1.04)4=24.16x
′
=20+(1.04)4=24.16
b)
P(z>z')=0.75\to P(z<z')=0.25P(z>z
′
)=0.75→P(z<z
′
)=0.25
From z-table:
z'=-0.67z
′
=−0.67
x'=20+(-0.67)4=17.32x
′
=20+(−0.67)4=17.32
c)
P(z<-z')=0.2, P(z>z')=0.2P(z<−z
′
)=0.2,P(z>z
′
)=0.2
From z-table:
P(z<-0.84)=0.2005P(z<−0.84)=0.2005
z'=0.84z
′
=0.84
x'_1=20+(-0.84)4=16.64x
1
′
=20+(−0.84)4=16.64
x'_2=20+(0.84)4=23.36x
2
′
=20+(0.84)4=23.36
d)
P(z<z')=0.25P(z<z
′
)=0.25
Thus, this is the same value as in the part b).
x'=17.32x
′
=17.32
Step-by-step explanation:
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