Answer:
[tex]P(X\geq 9)=P(X=9)+P(X=10)[/tex]
And using the probability mass function we got:
[tex]P(X=9)=(10C9)(0.913)^9 (1-0.913)^{10-9}=0.383[/tex]
[tex]P(X=10)=(10C10)(0.913)^{10} (1-0.913)^{10-10}=0.402[/tex]
And replacing we got:
[tex] P(X \geq 9) = 0.383 +0.402= 0.785[/tex]
Step-by-step explanation:
Let X the random variable of interest "number of students graduated", on this case we now that:
[tex]X \sim Binom(n=10, p=0.913)[/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]
We want to find the following probability:
[tex]P(X\geq 9)=P(X=9)+P(X=10)[/tex]
And using the probability mass function we got:
[tex]P(X=9)=(10C9)(0.913)^9 (1-0.913)^{10-9}=0.383[/tex]
[tex]P(X=10)=(10C10)(0.913)^{10} (1-0.913)^{10-10}=0.402[/tex]
And replacing we got:
[tex] P(X \geq 9) = 0.383 +0.402= 0.785[/tex]
