Question

During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X = the number of shots that scored points.

a. What is the probability distribution for X?

b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.

c. Use your calculator to find the probability that DeAndre scored with 60 of these shots.

d. Find the probability that DeAndre scored with more than 50 of these shots

Short answer

a. The probability distribution of $X~B(80,0.613)$.$X~B(80,0.613)$

b. (i) The mean of \(X\) is$49.04$

(ii) The standard deviation of \(X\) is $4.3564$

c. The probability that DeAndre scored with $0.0036$of these shots is $0.0036$

d. The probability that DeAndre scored with more than $0.3718$of these shots is

Step-by-step solution

Introduction

The proportion of times an event occurs out of a large number of trials is the likelihood of the event.

Explanation (Part a)

The binomial dispersion for a discrete arbitrary variable with two potential results.

A random variable is said to have a binomial dispersion assuming the likelihood mass capacity is:

$P(X=x)=\frac{n!}{(n-x)!x!}{p}^x(1-p{)}^{n-x}$

$X~B(n,p)$

the probability distribution for \(X\) is,

$X~B(80,0.613)$

Explanation (Part b)

i. Calculating the mean

we know, the binomial distribution is $X~B(n,p)$

Mean \(=np\)

$=80\times 0.613=49.04$

ii. The standard deviation is,

$SD=\sqrt{np(1-p)}$

$\Rightarrow \sqrt{80\times 0.613\times (1-0.613)}=4.3564$

Explanation (Part c)

Using calculator,

First, we press $2nd$and then press VARS and Scroll down to \(binompdf(n,p,x)\)

n = no. of trials

p = success probability

x = required success

$\text{binompdf}(80,0.613,60)=0.0036$

$P(x=60)=0.0036$

Explanation (Part d)

As we know,$P(x>50)=1-P(x\le 50)$

First, we press $2nd$and then press VARS and Scroll down to \(binompdf(n,p,x)\)

Using,

$1-\text{binomcdf}(80,0.613,50)=0.3718$

Hence,$P(x>50)=0.3718$