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$X.X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

d. How many of the $12$students do we expect 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.

Short answer

a. number of students who attend the Tet festivities

b. $0,1,2,3,4,5,6,7,8,9,10,11,12$

c. $X~\mathrm{Binom}(n=12,p=0.18)$

d. $E\left(X\right)=np=12\ast 0.18=2.16$

e. probability that at most four students will attend $=0.9511$

f. Probability that more than two students will attend$=0.3702$

Step-by-step solution

Given information

Total number of students surveyed $12$

probability of students attended Tet festive $0.18$

Explanation (part a)

$X$the random variable of interest "number of students who attend the Tet festivities".

Explanation (part b)

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

Explanation (part c)

The distribution for $X$on this case is given by:

$X~\mathrm{Binom}(n=12,p=0.18)$

Explanation (part d)

The number of students expected to join in $12$random variables of $18\%$of students is given by,

$E\left(X\right)=np=12\ast 0.18=2.16$

Explanation (part e)

the probability that at most four students will attend:

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

Explanation (part f)

the probability that more than two students will attend:

$\begin{array}{r}P(X>2)=1-P(X\le 2)=1-\left[P\right(X=0)+P(X=1)+P(x=2\left)\right]=\\ 1-[0.09242+0.24345+0.29392]=0.3702\end{array}$