Lec15 Worked Examples - Worked examples from Lec15 FMV,...

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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 50-watt 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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Lec15 Worked Examples - Worked examples from Lec15 FMV,...

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