Question

A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.
a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X.

d. How many of the 12 students do we expert to attend the festivities?

e. Find the probability that at most four students will attend.

f. Find the probability that more than two students will attend.

Answer 1

Answer:

a) Let X the random variable of interest "number of students who attend the Tet festivities",

b) The possible values for the random variable X are : 0,1,2,3,4,5,6,7,8,9,10,11,12

c) The distribution for X on this case is given by:

$X \sim Binom(n=12, p=0.18)$

d) $E(X) = np = 12*0.18=2.16$

e) $P(X \leq 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=0.9511$

f) $P(X>2) =1-P(X\leq 2) =1-[P(X=0)+P(X=1)+P(x=2)]=1-[0.09242+0.24345+0.29392]=0.3702$

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

Part a

Let X the random variable of interest "number of students who attend the Tet festivities",

Part b

The possible values for the random variable X are : 0,1,2,3,4,5,6,7,8,9,10,11,12

Part c

The distribution for X on this case is given by:

$X \sim Binom(n=12, p=0.18)$

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:

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

Part d

The expected value for the binomial distribution is given by:

$E(X) = np = 12*0.18=2.16$

Part e

For this case we want this probability:

$P(X \leq 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

$P(X=0)=(12C0)(0.18)^0 (1-0.18)^{12-0}=0.09242$

$P(X=1)=(12C1)(0.18)^1 (1-0.18)^{12-1}=0.24345$

$P(X=2)=(12C2)(0.18)^2 (1-0.18)^{12-2}=0.29392$

$P(X=3)=(12C3)(0.18)^3 (1-0.18)^{12-3}=0.21506$

$P(X=4)=(12C4)(0.18)^4 (1-0.18)^{12-4}=0.10622$

$P(X \leq 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=0.9511$

Part f

For this case we want this probability:

$P(X>2)$

And we can use the complement rule to find this like that:

$P(X>2) =1-P(X\leq 2) =1-[P(X=0)+P(X=1)+P(x=2)]$

$P(X>2) =1-P(X\leq 2) =1-[P(X=0)+P(X=1)+P(x=2)]=1-[0.09242+0.24345+0.29392]=0.3702$