Question

In: Statistics and Probability

A professor is studying to see if there is a relationship between the grades and gender...

A professor is studying to see if there is a relationship between the grades and gender of his students. He picks a random sample of his students and notes their grades (A, B, or C) and their gender (male or female). He finds that:

28% are A students.
58% are B students.
62% of his students are female.
Of those who are female, one-third are A students.
Of those who are B students, three-fifths are female students.

What percentage of the students are male and have a C grade?

Expert Solution

We know the following probabilities

28% are A students.

A randomly selected student has an A grade is

$P(\text{A grade})=0.28$

58% are B students.

A randomly selected student has a B grade is

$P(\text{B grade})=0.58$

A randomly selected student has a C grade is (Only A,B, or C grades are given)

$P(\text{C grade})=1-(P(\text{A grade})+P(\text{B grade}))=1-(0.28+0.58)=0.14$

62% of his students are female.

A randomly selected student is a female is

$P(\text{Female})=0.62$

A randomly selected student is a male is

$P(\text{Male})=1-P(\text{Female})=1-0.62=0.38$

Of those who are female, one-third are A students.

$P(\text{A Grade}\mid \text{Female})=\frac{1}{3}$

Of those who are B students, three-fifths are female students

$P(\text{Female}\mid \text{B Grade})=\frac{3}{5}$

The probability that a randomly selected student is female and has an A grade is

$\begin{align*} P(\text{A Grade}\cap \text{Female})&=P(\text{A Grade}\mid \text{Female})P(\text{Female})\quad\text{using the formula for conditional probability}\\ &=\frac{1}{3}\times 0.62\\ &= 0.2067 \end{align*}$

The probability that a randomly selected student is female and has a B grade is

$\begin{align*} P(\text{B Grade}\cap \text{Female})&=P(\text{Female}\mid \text{B Grade})P(\text{B Grade})\quad\text{using the formula for conditional probability}\\ &=\frac{3}{5}\times 0.58\\ &=0.348 \end{align*}$

Using the law of total probabilities we can get the probability that a randomly selected student is female and has a C grade using

$\begin{align*} &P(\text{Female})=P(\text{A Grade}\cap \text{Female})+P(\text{B Grade}\cap \text{Female})+P(\text{C Grade}\cap \text{Female})\\ \implies &P(\text{C Grade}\cap \text{Female})=P(\text{Female})-(P(\text{A Grade}\cap \text{Female})+P(\text{B Grade}\cap \text{Female}))\\ \implies &P(\text{C Grade}\cap \text{Female})=0.62-(0.2067+0.348)= 0.0653 \\ \end{align*}$

Using the law of total probabilities again, we can get the probability that a randomly selected student is male and has a C grade using

$\begin{align*} &P(\text{C Grade})=P(\text{C Grade}\cap \text{Male})+P(\text{C Grade}\cap \text{Female})\\ \implies &P(\text{C Grade}\cap \text{Male})=P(\text{C Grade})-P(\text{C Grade}\cap \text{Female})\\ \implies &P(\text{C Grade}\cap \text{Male})=0.14-0.0653 =0.0747 \end{align*}$

ans: The percentage of the students are male and have a C grade is 7.47%

orchestra answered 2 years ago

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A professor is studying to see if there is a relationship between the grades and gender...

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