Solution 40501: Demonstrating Probability Distributions on the TI-84 Evo.
How can I compute the probability distributions in a binomial distribution on the TI-84 Evo?
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?
Solution:
First, generate the complete distribution by following the steps below:
1) Press [alpha] then [stat] to access the DISTR menu.
2) Press [4] the [1] to access the command 1:binompdf
3) Input [6] [enter] then [.] [2] and then press [window] or [zoom] to paste
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] to select 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 [mode] to exit the stat menu.
2) Press [alpha] then [stat] followed by [4] then [1] to access the command 1:binompdf
3) Press [6] [enter] [.] [2] [enter] [1] and then press [window] or [zoom] to paste
4) Press [enter] to get the result 0.393216
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 [alpha] then [stat] followed by [4] then [2] to access the command 2:binomcdf
2) Press [6] [enter] [.] [2] [enter] [2] and then press [window] or [zoom] to paste
3) Press [enter] to get the result 0.90112
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 [alpha] then [stat] followed by [4] then [2] to access the command 2:binomcdf
3) Press [6] [enter] [.] [2] [enter] [1] and then press [window] or [zoom] to paste.
4) Press [enter] to get the result 0.34464
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 [alpha] then [stat] followed by [4] then [2] to access the command 2:binomcdf
2) Press [6] [enter] [.] [2] [enter] [4] and then press [window] or [zoom] to paste.
3) Press [-]
4) Press [alpha] then [stat] followed by [4] then [2] to access the command 2:binomcdf
5) Press [6] [enter] [.] [2] [enter] [1] and then press [window] or [zoom] to paste.
6) Press [enter] to get the result of 0.34304
