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?

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

Part(b) The required probability is

Given that,

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

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

Part(b):

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

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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

part a

in a city ice cream is mostly only bought when the temperature is very high as ice cream cools down our

so we can derive the conclusion that as the temperature increases the number of ice creams sold also increases

part b

y intercept is 5

slope is 2

happy to

pls mark as brainliest

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8