Binomial Probability Distribution on TI-89
Copyright © 2007–2013 by Stan Brown, Oak Road Systems
Copyright © 2007–2013 by Stan Brown, Oak Road Systems
Summary: With your TI-89/92, you can do all types of probability calculations for a binomial probability distribution.
See also: TI-83/84 users can use the program in MATH200A Program part 3 or the calculator procedure in Binomial Probability Distribution on TI-83/84 to compute binomial probability.
To compute the binomial probability for
one particular number of successes, use the binompdf
function.
| TI-89 Home Screen | TI-89 Stats/List Editor | |
|---|---|---|
| Keystrokes | [CATALOG] [F3] [plain ( makes B] [▼] |
[F5] [ALPHA ( makes B] |
| Format | binompdf(n, p, x) | (dialog box) |
Example 1: Larry’s batting average is .260. If he’s at bat four times, what is the probability that he gets exactly two hits?
Solution:
n = 4, p = 0.26, x = 2
Note: Some textbooks use r for number of successes, rather than x.
binompdf(4,.26,2) = 0.2221
To compute the binomial probability for a
range of numbers of successes from xlow to
xhigh, use the
binomcdf function.
| TI-89 Home Screen | TI-89 Stats/List | |
|---|---|---|
| Keystrokes | [CATALOG] [F3] [plain ( makes B] |
[F5] [ALPHA ) makes C] |
| Format | binomcdf(n, p, xlow, xhigh) | (dialog box) |
Example 2: Larry’s batting average is .260. If he’s at bat six times, what is the probability that he gets two to four hits?
Solution:
n = 6, p = 0.26, 2 ≤ x ≤ 4
binomcdf(6,.26,2,4) = 0.4840
Example 3: Suppose 65% of the registered voters in Dryden are Republicans. In a random sample of ten registered voters, what’s the probability of fewer than six Republicans?
Solution: “Fewer than six” is zero through five.
n = 10, p = 0.65, 0 ≤ x ≤ 5
binomcdf(10, .65, 0, 5) = 0.2485
There’s about one chance in four of getting fewer than six Republicans in a random sample of ten registered voters.
Example 4: With the same data, what’s the probability of getting eight or more Republicans?
Solution: “Eight or more” means eight to ten since there are only ten trials.
binomcdf(10,.65,8,10) = 0.2616
Example 5: A fair die has a 1/6 chance of rolling a 2. In 24 rolls, what's the probability of getting no more than three 2’s?
Solution: “No more than three” means zero to three.
n = 24, p = 1/6, 0 ≤ x ≤ 3
binomcdf(24,1/6,0,3) = 0.4155
This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.
For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/