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Identify the parameter n in the following binomial distribution scenario. A basketball player has a 0.479 probability of
making a free throw and a 0.521 probability of missing. If the player shoots 17 free throws, we want to know the probability
that he makes more than 9 of them. (Consider made free throws as successes in the binomial distribution.)

Respuesta :

mathmate

Answer:

n = 17

Step-by-step explanation:

Assuming

- probability of success (making free throw) does not vary

We have

n = 17 (trials)

p = 0.479

x > 9

The answer is "[tex]\bold{p(x>9)=0.2550319}[/tex]"

[tex]\to X:[/tex] Number of creating free throws in a set [tex]\bold{17\ \ x \sim bin(17,0.479)}[/tex]

Know we calculating the P(makes more than 9 of them)

[tex]=\bold{9(X>9)=1-P(Z<=9)}[/tex]

Using the R-code:

[tex]\to \bold{1-p\ binom(9,17,0.479)}\\\\\to \bold{[1]0.2550319}\\\\\bold{\therefore}\\\\ \to \bold{p(x>9)=0.2550319}[/tex]

Learn more:

binomial distribution: brainly.com/question/9065292

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