Bio exercise 2 - Jonathan Church JC8975 BIO 205L Exercise 2...

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Jonathan Church JC8975 BIO 205L: Exercise 2 1a. Mean length = (21.0 + 16.0 + 17.0 + 16.5 + 14.0)/5 = 16.9 r.u. Mean Width = (5.0 + 4.5 + 5.0 + 5.0 + 5.0)/5 = 4.9 r.u. 1b. Standard Deviation Length = sqrt(((21-16.9)^2 + (16-16.9)^2 + (17-16.9)^2 + (16.5-16.9)^2 + (14- 16.9)^2)/4) = 2.559 Standard Deviation Width = sqrt(((5-4.9)^2 +(5-4.9)^2 +(5-4.9)^2 +(5-4.9)^2 +(4.5-4.9)^2)/4) = 0.223 1c. Length - 2.559 x 2 = 5.118, so 95% confidence levels are 16.9 + 5.118 or 16.9 - 5.118 Width – 0.223 x 2 = 0.446, so 95% confidence levels are 4.9 + 0.446 or 4.9 – 0.446 2. To approximate the volume of a paramecium, I would use the volume of a rectangular prism. I believe the Paramecium’s shape most closely resembles a rectangular prism, however the estimated volume will be larger than the real value due to the smooth round ends of the paramecium. The equation for the volume of a rectangular prism is Volume = Length x Width x Height. When using this equation, we will assume that the paramecium has equal width and height. 16.9 r.u. x 9.8 µm/1 ru = 165.62 µm
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Bio exercise 2 - Jonathan Church JC8975 BIO 205L Exercise 2...

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