Seeds from the buckeye sprout with proababiltity 0.07. If you attempt to germinate 4 of the seeds, what is the chance that exactly one will sprout? Solution Let X : denoting number of seeds that will sprout X follows Binomial distribution because either seed will sprout or not Let p : probability that seed will sprout = 0.07 q :probability that seed will not sprout = 1 - 0.07 = 0.93 Here n = 4 , x = 1 P(X = x) = nCx (q)^(n-x) *(p)^(x) P(X= 1) = 4C1 (0.93)^(4-1) *(0.07) = 4*(0.93)^(3) *0.07 =0.225.

1. "Seeing Statistics" Please respond to the following:
Go to http://www.seeingstatistics.com/seeing1999/resources/opening.html. Using the
"Contents" button in the upper left, access and read sections 10.0 through 10.4. Interact with the
various graphs.
Solution
The application of t-tests had the greatest impact on me because you got a first-hand
view of how to calculate it using the formulas then use computer programs to verify the results.
It was a good way to actually interpret the data and know what it means. It was easy to insert
your own data and move the graphs around for interpretation. Nothing much was challenging
except that certain some information was a bit vague and could of gone in more in depth.