As we were covering IPS Ch. 5 and discussing the normal approximation to the Binomial. One of my students did a problem much differently than I and then asked a very pertinent question. Why do we need the normal approximation. Using Chapter 5 - Problem 4 from Test Bank for IPS from McCabe and McCabe "Offspring of a particular genetic cross have an undesirable trait with probability 1/8. Inheritance of this trait by separate offspring is independent. You examine 100 offspring from this cross and count the number X who have the undesirable trait. What is the probability that 10 or less of the offspring have this trait?"
Using the TI-83 normal approximation you would get .2248 from normalcdf(-999,10,12.5,sd) with the continuity correction you would get .2727 from normalcdf(-999,10.5,12.5,sd). For those of you unfamiliar with TI-83 normalcdf(lowerbound, upperbound, mean, sd) returns area under curve quickly.
The student had done simply binompdf(100,0.125) stored to L1 and then did sum(L1,1,11) and arrived with .280999. TI-83 binompdf(100,0.125) returns all probabilities from 0 to 100 for p = 1/8. He summed items 1 to 11 to get P(0) through P(10). He was, with ease at which he had done this, amazed that anyone would do it the other way. I was wondering if anyone else had had this come up in class, and wondered if AP graders are aware of this binomial summing ability.
I have really enjoyed all the posts, I have been reading since late December, and I thought it was time for me to share an idea or two.