Solution 34514: Demonstrating Probability Distributions on the TI-84 Plus Family of Graphing Calculators.
How can I compute the probability distributions in a binomial distribution on the TI-84 Plus family of graphing calculators?
The example below shows how to compute different probabilities in a binomial distribution.
For Example:The production of an electronic component has a large 20% defective rate. If a random selection of six components is taken,
What is the probability of getting exactly two defectives?
What is the probability of getting at most two defectives?
What is the probability of getting at least two defectives?
What is the probability of getting from two to four defectives?
Please Note: The example below requires you to turn your STAT DIAGNOSTICS off by following the steps below:
1) Press [MODE]
2) Press [↓] until the cursor is in front of STAT DIAGNOSTICS with OFF highlighted and press [ENTER].
3) Press [2ND] [QUIT] to return to the home screen and follow the steps below.
Solution:
First, generate the complete distribution by following the steps below:
1) Press [2nd] then [Vars] to access the DISTR menu.
2) Press [ALPHA] then [Math] to access the command A:binompdf(.
3) Input [6] [,] [.] [2] [)].
4) Press [STO->] [2ND] [1] [ENTER] to store the answer into L1
Next, enter the values 0 to 6 into L2 to clearly identify the different probabilities by following the steps below:
1) Press [STAT].
2) Press 1: Edit.
3) Scroll over to the L2 list and press 0 [Enter] 1 [Enter] 2 [Enter] 3 [Enter] 4 [Enter] 5 [Enter] 6 [Enter]
To solve the different questions, please follow the steps below:
Question 1
The probability of getting exactly two defectives is P (1) =.39322. This value can be found in L1. To calculate, please follow the steps below:
1) Press [2nd] then [Vars] followed by [ALPHA] then [Math] to access the command A:binompdf(
2) Press [6] [,] [.] [2] [,] [1] [)].
3) Press [ENTER].
Question 2
The probability of getting at most two defectives is P (0) + P (1) +P (2) =.90112. To calculate, please follow the steps below:
1) Press [2nd] then [Vars] followed by [ALPHA] then [apps] to access the command B:binomcdf(
2) Press [6] [,] [.] [2] [,] [2] [)].
3) Press [ENTER].
Question 3
The probability of getting at least two defectives is P(2)+P(3)+P(4)+P(5)+P(6)=1-(P(0)+P(1))= .34464. To calculate, please follow the steps below:
1) Press [1] [-].
2) Press [2nd] then [Vars] followed by [ALPHA] then [apps] to access the command B:binomcdf(
3) Press [6] [,] [.] [2] [,] [1] [)].
4) Press [ENTER].
Question 4
The probability of getting from two to four defectives is P(2)+P(3)+P(4)=.34304 =(P(0)+P(1)+P(2)+ P(3)+P(4))-(P(0)+P(1)).To calculate, please follow the steps below:
1) Press [2nd] then [Vars] followed by [ALPHA] then [apps] to access the command B:binomcdf(
2) Press [6] [,] [.] [2] [,] [4] [)] [-].
3) Press [2nd] then [Vars] followed by [ALPHA] then [apps] to access the command B:binomcdf(
4) Press [6] [,] [.] [2] [,] [1] [)].
5) Press [ENTER].
The same result can be obtained by entering sum(binompdf(6, .2,{2,3,4})) from the home screen.
Please see the TI-84 Plus Family guidebooks for additional information.
Last updated: 7/24/2024