Question

Annie is a basketball player who makes, on average, 65% of her freethrows. Assume each shot is independent and the probability ofmaking any given shot is .65.
What's the probability that she will miss her first three freethrows before she makes one?
During her season, she took 140 free throws. Whats the exactbinomial probability and the normal approximation the she will makeat least 100 of them? Are these conditions able to permit to usenormal approximation?

Answer 1

Answer:

a) For this case we need that she fails the first 3 throws and the last one would be successful, so then if p represent the probability of success and 1-p the probability of fail we have this:

$(1-p)^3 p = (1-0.65)^3 (0.65)= 0.0279$

b) "=1-BINOM.DIST(99,140,0.65,TRUE)"

$P(X \geq 100) = 1-P(X$

c) $\mu = np = 140*0.65= 91$

$\sigma = \sqrt{np(1-p)}= \sqrt{140*0.65(1-0.65)}=5.644$

And we want this probability:

$P(X \geq 100) = 1-P(X$

Step-by-step explanation:

Part a

For this case we need that she fails the first 3 throws and the last one would be successful, so then if p represent the probability of success and 1-p the probability of fail we have this:

$(1-p)^3 p = (1-0.65)^3 (0.65)= 0.0279$

Part b

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".

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

$X \sim Binom(n=140, p=0.65)$

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:

$nCx=\frac{n!}{(n-x)! x!}$

For this case we want this probability:

$P(X \geq 100) = 1-P(X$

And we can use the following excel code:

"=1-BINOM.DIST(99,140,0.65,TRUE)"

$P(X \geq 100) = 1-P(X$

Part c

We need to check the conditions in order to use the normal approximation.

$np=140*0.65=91 \geq 10$

$n(1-p)=140*(1-0.65)=49 \geq 10$

So then we can use the normal approximation and we can find the mean and deviation like this:

$\mu = np = 140*0.65= 91$

$\sigma = \sqrt{np(1-p)}= \sqrt{140*0.65(1-0.65)}=5.644$

And we want this probability:

$P(X \geq 100) = 1-P(X$