Question

The chances that a student will get 7 marks out of 10 in first attempt of a certain Quiz while attempting Quiz on BlackBoard LMS is 40%. Suppose that 20 students are selected at random then find a probability that i)Exactly 12 students will get 7 marks out 10 in the Quiz. ii)Fewer than 10 students will get 7 marks out 10 in the Quiz. iii)At least 5 students will get 7 marks out 10 in the Quiz.

Answer 1

i)
`P(X=12)=0.0355`

ii)
`P(X<10)=0.7553`

iii)
`P(X\geq 5)=0.9490`

Explanation:

Let's start by defining the random variable ⇒

`X:` '' Number of students that will get 7 marks out of 10 in first attempt of a certain Quiz while attempting Quiz on BlackBoard LMS ''

`X` 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
`X` as a binomial random variable ⇒

`X` ~ 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 ⇒

`X` ~ Bi (20, 0.4) in the question.

The probability function for
`X` is

`p_(X)(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^(x)(1-p)^(n-x)` (I)

Where
`\left(\begin{array}{c}n&x\end{array}\right)` is the combinatorial number define as

`\left(\begin{array}{c}n&x\end{array}\right)=(n!)/(x!(n-x)!)`

Replacing the parameters in the equation (I) ⇒

`P(X=x)=\left(\begin{array}{c}20&x\end{array}\right)(0.4)^(x)(0.6)^(20-x)` (II)

For i) we need to find
`P(X=12)`

Then, we only need to replace by
`x=12` in equation (II) ⇒

`P(X=12)=\left(\begin{array}{c}20&12\end{array}\right)(0.4)^(12)(0.6)^(8)=0.0355`

For ii) we need to calculate
`P(X<10)`

This probability is equal to ⇒
`P(X<10)=P(X\leq 9)` and to calculate it we need to sum
`P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)`

`+P(X=6)+P(X=7)+P(X=8)+P(X=9)`

We can do it summing each term or either using any program.

The result is
`P(X<10)=P(X\leq 9)=0.7553`

Finally for iii) we need to find ⇒
`P(X\geq 5)`

This probability is equal to ⇒
`P(X\geq 5)=1-P(X\leq 4)` ⇒

`1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)]`

Again we can find each term by using the equation (II) or either using a program. The result is
`P(X\geq 5)=0.9490`