Question

In a recent survey in a Statistics class, it was determined that only 60% of the students attend class on Fridays. From past data it was noted that 98% of those who went to class on Fridays pass the course, while only 20% of those who did not go to class on Fridays passed the course.
a. What percentage of students is expected to pass the course?
b. Given that a person passes the course, what is the probability that he/she attended classes on Fridays?

Answer 1

Answer:

a. 66.8%

b. 0.855

Step-by-step explanation:

In a recent survey in a Statistics class,

Percentage of students that attended class on Friday = 60%

Percentage of students that did not attend class on Friday = 100 - 60% = 40%

Percentage of those who went to class on Fridays and passed the course = 98%

Percentage of those who did not go to class on Fridays and passed the course = 20%

a. What percentage of students is expected to pass the course?

This is calculated as:

= 60% × 98% + 40% × 20%

= 0.668

Converting to percentage

= 0.668 × 100

= 66.8%

b. Given that a person passes the course, what is the probability that he/she attended classes on Fridays?

We solve using Baye's Theorem

= [60 × 98%]/[60% × 98% + 40% × 20%]

= 0.588 /0.688

= 0.8546511628

Approximately = 0.855

Answer 2

Part(a): The required pass percentage is 66.8%

Part(b) The required probability is $P(Friday/Pass)=0.8802$

Given that,

P(Friday) = 0.60, P(No Friday) = 0.40

$P(Pass/Friday)=0.98\\P(Pass/No Friday)=0.98$

Part(a): The percentage of students expected to pass the course is 0.668.

$P(Pass)=P(Friday)\times P(Pass/Friday)+P(No friday)\times P(Pass/No Friday)\\=0.60\times 0.98+0.40\times 0.20\\=0.588+0.08\\P(Pass)=0.668$

Part(b):

The probability that he/she attended classes on Fridays is 0.8802

$P(Friday/Pass)=\frac{P(friday)\times P(Pass/Friday)}{P(Friday\times P(pass/Friday+P(No friday)\times P(pass/No friday)}\\=\frac{0.60\times 0.98}{0.60\times 0.98+0.40\times 0.20} \\=0.8802$

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