Answer:
i) [tex]P(X=12)=0.0355[/tex]
ii) [tex]P(X<10)=0.7553[/tex]
iii) [tex]P(X\geq 5)=0.9490[/tex]
Step-by-step explanation:
Let's start by defining the random variable ⇒
[tex]X:[/tex] '' Number of students that will get 7 marks out of 10 in first attempt of a certain Quiz while attempting Quiz on BlackBoard LMS ''
[tex]X[/tex] is a discrete random variable.
The probability of a randomly selected student getting 7 marks out of 10 is 0.4
(This is a data from the question).
Now, if we assume independence between the students while they are doing the Quiz and also we assume that this probability remains constant , we can modelate [tex]X[/tex] as a binomial random variable ⇒
[tex]X[/tex] ~ Bi (n,p)
Where ''n'' and ''p'' are the parameters of the variable.
''n'' is the number of students attempting the Quiz and ''p'' is the probability that a student will get 7 marks out of 10 which is 0.4 ⇒
[tex]X[/tex] ~ Bi (20, 0.4) in the question.
The probability function for [tex]X[/tex] is
[tex]p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}[/tex] (I)
Where [tex]\left(\begin{array}{c}n&x\end{array}\right)[/tex] is the combinatorial number define as
[tex]\left(\begin{array}{c}n&x\end{array}\right)=\frac{n!}{x!(n-x)!}[/tex]
Replacing the parameters in the equation (I) ⇒
[tex]P(X=x)=\left(\begin{array}{c}20&x\end{array}\right)(0.4)^{x}(0.6)^{20-x}[/tex] (II)
For i) we need to find [tex]P(X=12)[/tex]
Then, we only need to replace by [tex]x=12[/tex] in equation (II) ⇒
[tex]P(X=12)=\left(\begin{array}{c}20&12\end{array}\right)(0.4)^{12}(0.6)^{8}=0.0355[/tex]
For ii) we need to calculate [tex]P(X<10)[/tex]
This probability is equal to ⇒ [tex]P(X<10)=P(X\leq 9)[/tex] and to calculate it we need to sum [tex]P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)[/tex]
[tex]+P(X=6)+P(X=7)+P(X=8)+P(X=9)[/tex]
We can do it summing each term or either using any program.
The result is [tex]P(X<10)=P(X\leq 9)=0.7553[/tex]
Finally for iii) we need to find ⇒ [tex]P(X\geq 5)[/tex]
This probability is equal to ⇒ [tex]P(X\geq 5)=1-P(X\leq 4)[/tex] ⇒
[tex]1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)][/tex]
Again we can find each term by using the equation (II) or either using a program. The result is [tex]P(X\geq 5)=0.9490[/tex]
