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Assume that X is a binomial random variable with n = 27 and p = 0.90. Calculate the following probabilities. (Do not round intermediate calculations. Round your final answers to 4 decimal places.)

a. P(X = 26) :
b. P(X = 25) :
c. P(X ≥ 25) :

Respuesta :

Answer:

a) [tex] P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744[/tex]

b) [tex] P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520[/tex]

c) [tex] P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520[/tex]

[tex] P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744[/tex]

[tex] P(X=27) = 27C27 (0.9)^{27} (1-0.9)^{27-27}= 0.0582[/tex]

And addind the values we got:

[tex]P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)=0.2520+0.1744+0.0582=0.4846[/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".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=27, p=0.9)[/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]  

Part a

We want this probability:

[tex]P(X=26)[/tex]

And we can use the probability mass function and we got:

[tex] P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744[/tex]

Part b

We want this probability:

[tex]P(X=25)[/tex]

And we can use the probability mass function and we got:

[tex] P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520[/tex]

Part c

We want this probability:

[tex]P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)[/tex]

And we can use the probability mass function and we got:

[tex] P(X=25) = 27C25 (0.9)^{25} (1-0.9)^{27-25}= 0.2520[/tex]

[tex] P(X=26) = 27C26 (0.9)^{26} (1-0.9)^{27-26}= 0.1744[/tex]

[tex] P(X=27) = 27C27 (0.9)^{27} (1-0.9)^{27-27}= 0.0582[/tex]

And addind the values we got:

[tex]P(X\geq 25)= P(X=25)+P(X=26)+P(X=27)=0.2520+0.1744+0.0582=0.4846[/tex]

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