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Question #235425
Random sample of size 25 from a normally distributed population with the mean of 4 and variance of 16. Find the point estimate of the population mean if P(X>x)=0.9798?
Expert's answer
P
(
X
<
x
)
=
1
−
P
(
Z
≤
x
−
μ
σ
/
n
)
P(X<x)=1-P(Z\leq\dfrac{x-\mu}{\sigma/\sqrt{n}})
P
(
X
<
x
)
=
1
−
P
(
Z
≤
σ
/
n
x
−
μ
)
=
1
−
P
(
Z
≤
x
−
4
16
/
25
)
=
0.9798
=1-P(Z\leq\dfrac{x-4}{\sqrt{16}/\sqrt{25}})=0.9798
=
1
−
P
(
Z
≤
16
/
25
x
−
4
)
=
0.9798
P
(
Z
≤
1.25
(
x
−
4
)
)
=
0.0202
P(Z\leq1.25(x-4))=0.0202
P
(
Z
≤
1.25
(
x
−
4
))
=
0.0202
1.25
(
x
−
4
)
≈
−
2.04963
1.25(x-4)\approx-2.04963
1.25
(
x
−
4
)
≈
−
2.04963
x
=
2.36
x=2.36
x
=
2.36
x
=
2.36
x=2.36
x
=
2.36
The point estimate of the population mean is 4.
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