Respuesta :
Answer:
[tex] P(X=15) = (40C15)(0.2)^{15} (1-0.2)^{40-15} = 0.00498[/tex]
We need to check if we can use the normal approximation , the conditions are:
[tex]np=40*0.2=8>5[/tex] and [tex]n(1-p)=40*(1-0.2)=32>5[/tex]
So then we can apply the normal approximation to the binomial distribution in our case:
[tex]X \sim N(\mu=8,\sigma=2.529)[/tex]
But if we find [tex] P(X=15) [/tex] using the normla approximation we have a value of 0 since in a continuous random variable the area below the curve is 0.
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".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=40, p=0.2)[/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]
If we use the exact distribution we have this:
[tex] P(X=15) = (40C15)(0.2)^{15} (1-0.2)^{40-15} = 0.00498[/tex]
We need to check if we can use the normal approximation , the conditions are:
[tex]np=40*0.2=8>5[/tex] and [tex]n(1-p)=40*(1-0.2)=32>5[/tex]
So then we can apply the normal approximation to the binomial distribution in our case:
[tex]X \sim N(\mu=8,\sigma=2.529)[/tex]
But if we find [tex] P(X=15) [/tex] using the normla approximation we have a value of 0 since in a continuous random variable the area below the curve is 0.
