The image above represents standard normal variable.
To compute for standard normal variable, three essential parameters are needed and these parameters are value (x), mean (μ) and standard deviation (σ).
The formula for calculating standard normal variable:
z = (x – μ) ⁄ σ
Where;
z = Standard Normal Variable
x = Value
μ = Mean
σ = Standard Deviation
Let’s solve an example;
Find the standard normal variable when the value is 4, the mean is 20 and the standard deviation is 26.
This implies that;
x = Value = 4
μ = Mean = 20
σ = Standard Deviation = 26
z = (x – μ) ⁄ σ
z = (4 – 20) ⁄ 26
z = (-16) ⁄ 26
z = -0.615
Therefore, the standard normal variable is -0.615.
To compute for standard deviation, three essential parameters are needed and these parameters are Number of possible outcomes in any single trial (n), Probability of a success in any single trial (p) and Probability of a failure in any single trial (q).
The formula for calculating standard deviation:
σ = √(npq)
Where;
σ = Standard deviation
n = Number of Possible Outcomes in Any Single Trial
p = Probability of a Success in Any Single Trial
q = Probability of a Failure in Any Single Trial
Let’s solve an example;
Find the standard deviation when the number of possible outcomes in any single trial is 14, the probability of a success in any single trial is 0 and the probability of a failure in any single trial is 1.
This implies that;
n = Number of Possible Outcomes in Any Single Trial = 14
p = Probability of a Success in Any Single Trial = 0
q = Probability of a Failure in Any Single Trial = 1
σ = √(npq)
σ = √((14)(0)(1))
σ = √(0)
σ = 0
Therefore, the standard deviation is 0.
Calculating for Number of Possible Outcomes in Any Single Trial when the Standard Deviation, the Probability of a Success in Any Single Trial and the Probability of a Failure in Any Single Trial is Given.
n = σ2 / pq
Where;
n = Number of Possible Outcomes in Any Single Trial
σ = Standard deviation
p = Probability of a Success in Any Single Trial
q = Probability of a Failure in Any Single Trial
Let’s solve an example;
Given that standard deviation is 5, the probability of a success in any single trial is 1 and the probability of a failure in any single trial is 1. Find the number of possible outcomes in any single trial?
This implies that;
σ = Standard deviation = 5
p = Probability of a Success in Any Single Trial = 1
q = Probability of a Failure in Any Single Trial = 1
n = σ2 / pq
n = 52 / 1 x 1
n = 25 / 1
n = 25
Therefore, the number of possible outcomes in any single trial is 25.
To compute for mean, two essential parameters are needed and these are number of possible outcomes in any single trial (n) and probability of success in any single trial (p).
The formula for calculating mean:
μ = np
Where;
μ = Mean
n = Number of Possible Outcomes in Any Single Trial
p = Probability of success in any single trial
Let’s solve an example;
Find the mean when the number of possible outcomes in any single trial is 8 and the probability of success in any single trial is 1.
This implies that;
n = Number of Possible Outcomes in Any Single Trial = 8
p = Probability of success in any single trial = 1
μ = np
μ = (8)(1)
μ = 8
Therefore, the mean is 8.
Calculating for the Number of Possible Outcomes in Any Single Trial when the Mean and the Probability of Success in Any Single Trial is Given.
n = μ / p
Where;
n = Number of Possible Outcomes in Any Single Trial
μ = Mean
p = Probability of Success in Any Single Trial
Let’s solve an example;
Find the number of possible outcomes in any single trial when the mean is 15 and the probability of success in any single trial is 1.
This implies that;
μ = Mean = 15
p = Probability of Success in Any Single Trial = 1
n = μ / p
n = 15 / 1
n = 15
Therefore, the number of possible outcomes in any single trial is 15.
The image above represents Poisson probability distribution.
To compute for Poisson probability distribution, two essential parameters are needed and these parameters are Mean of the Theoretical Distribution (μ) and Number of successes of event A (r).
The formula for Poisson probability distribution:
P(x = r) = e-μ(μr) ⁄ r!
Where;
P(x = r) = Poisson Probability Distribution
r = Number of Successes of Event A
μ = Mean of the Theoretical Distribution
Let’s solve an example;
Find the poisson probability distribution when the mean of the theoretical distribution is 12 and the number of successes of event A is 6.
This implies that;
r = Number of Successes of Event A = 6
μ = Mean of the Theoretical Distribution = 12
