Respuesta :
Answer:
[tex]np=13*0.6=7.8 \geq 5[/tex]
[tex]n(1-p)=13*(1-0.6)=5.2 \geq 5[/tex]
We assume the independence condition on each trial, so then we can conclude that the normal approximation for this case is suitable.
[tex] X \sim N(\mu = np , \sigma = np(!-p)[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=13, p=0.6)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Solution to the problem
We need to check the conditions in order to use the normal approximation.
[tex]np=13*0.6=7.8 \geq 5[/tex]
[tex]n(1-p)=13*(1-0.6)=5.2 \geq 5[/tex]
We assume the independence condition on each trial, so then we can conclude that the normal approximation for this case is suitable.
[tex] X \sim N(\mu = np , \sigma = np(!-p)[/tex]
