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TI83 and the normal approximation
Posted:
Feb 5, 1997 10:35 AM


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 TI83 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 TI83 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. TI83 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.



