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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/12/2008 for the course BIO 205L taught by Professor Hanson during the Fall '07 term at University of Texas at Austin.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online