Question

A professional basketball team made 37.9% of its three-point field goals in one season. If 80 three-point field goal attempts are randomly selected from the season, what is the probability that more than

asked Mar 12, 2021 86.9k views

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Zen Of Kursat · Answer 1 · 2021-03-19T02:27:54+0000

Answer:

P(X>35) = P(X=36) +P(X=37) +...+P(X=80)

And we can use the complement rule like this:

$P(X>35) = 1-P(X \leq 35)$

And we can use the followin excel code and we got:

"=1-BINOM.DIST(35;80;0.379;TRUE)"

$P(X>35) = 1-P(X \leq 35)= 0.117$

The other way to solve the problem is with the normal approximation:

$\mu = n*p =80*0.379= 30.32$

$\sigma = √(np(1-p))= √(80*0.379*(1-0.379))= 4.339$

And we can use the z score given by:

$z = (x -\mu)/(\sigma)$

And we got:

P(X>35) = P(Z> (35-30.32)/(4.339)) = P(Z>1.079)

And using the complement rule we got:

P(Z>1.079) =1-P(Z<1.079) = 0.1402

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:

$X \sim Binom(n=80, p=0.379)$

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^(n-x)

Where (nCx) means combinatory and it's given by this formula:

nCx=(n!)/((n-x)! x!)

And we want to find this probability: