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Worked examples from Lec15
FMV, 9/30/2010
Example 1: Note that in the lecture I used the term
X
instead of X
average
to designate the
average of the sample, here I use X
average
throughout.
Let X
1
…X
100
denote the actual net output of 100 randomly selected 50watt rated
photovoltaic panels.
Let X
average
be the average of the sample of 100 panels.
a.
If the expected output of each panel is 50 watts and the variance is 1, calculate
P(49.9<
X
average
<
50.1) (approximately) using the central limit theorem.
b.
If the expected output is 49.8 watts instead of 50 watts, so that on average
panels perform slightly below the 50 watt nominal rating, calculate
P(49.9<
X
average
<
50.1)
Solutions:
Part a. we first calculate mean and standard deviation for X
average
:
1
.
0
100
1
50
=
=
=
=
=
n
average
average
X
X
σ
μ
We then use this information to calculate the desired probability:
%
68
~
6826
.
0
)
1
1
(
)
1
.
50
1
.
50
1
.
50
9
.
49
(
)
1
,
0
(
~
);
1
.
50
9
.
49
(
)
1
.
50
9
.
49
(
=
=
≤
≤

=

≤
≤

=

≤
≤

=
≤
≤
Z
P
Z
P
N
Z
Z
P
X
P
average
Part b. Mean for X
average
is now 49.8, standard deviation is still 1:
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 '08
 Stedinger

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