Mean, Variance, and Standard Deviation for the binomial distribution

							Mean, Variance, and Standard Deviation for the binomial distribution
For a binomial distribution,

  n p   n pq
2

  n pq

Example
Randomly guessing on a 100 question multiple-choice test, where each question has 4 possible answers,

  100  1 4  25  2  100  1 4  3 4  18.8   100  1 4  3 4  4.3

Example
Using our range rule of thumb,

  2  25  2  4.3  16 .4   2  25  2  4.3  33 .6
It would be unusual to get less than 17 questions correct, or more than 33 question correct if you were randomly guessing on the test.

Another Example
You presume that people are evenly split (50/50) on their preference between two candidates. You poll 80 people, and find 45 prefer candidate A. Is this unexpected? We could find P(45 or more). Alternatively, using our presumed 50/50 preference, find mean=40, stddev=4.5. This suggests that outcomes of 31 to 49 would be usual occurrences. Since 45 falls in that range, we shouldn’t consider it unusual.

Another Example
Now suppose you poll 800 people, and 450 prefer candidate A. Is this unexpected? Using our presumed 50/50 preference, mean = 400, stddev = 14.1. Again using the range rule of thumb, this suggests that outcomes of 372 to 428 would be usual.

Since the outcome of 450 is unusual, we should question whether our presumption of a 50/50 split is accurate.

Homework
4.4: 1, 5, 7, 11, 13