Suppose you conducted an experiment to determine the number of students who skip

Question

Suppose you conducted an experiment to determine the number of students who skip INEN435. There are a total of 33 people registered for the course: 24 males and 9 females. You spent 2 weeks examining the no-show behavior of students and collecting data to estimate the likelihood that they will skip class. If a person didn't arrive within the first 10 minutes of class, you assumed the person was not coming. Based on the data collected, you determined the probability that a female student skipped class was 2/33 and the probability that a male student skipped class was 6/33. a) Define the sample space for the experiment. b) Is the sample space for this experiment a discrete or continuous sample space? (Don't just say yes or no, explain why) c) Given that a student is male, what is the probability that he will skip class? d) Given that a student is female, what is the probability that she will skip class? c) Given that a student is male, what is the probability that he will attend class? A large industrial firm uses three different warehouses (A, B, and C) to store its manufactured product. From past records, it is known that 17% of the manufactured problem are assigned to warehouse A, 54% are assigned to B, and 29% are assigned to C. It is know that 4% of the product in warehouse A are defective, 7% in warehouse B are defective, and 9% in warehouse C are defective. a) Find P (D) where D = the event the product is defective. b) Using Baye's Rule, find P (C | D).

Explanation / Answer

Ans:2)Given that

P(A)=0.17

P(B)=0.54

P(C)=0.29

P(defective/A)=0.04

P(defective/B)=0.07

P(defective/C)=0.09

a)using Law of total probability:

P(defective)=P(defective/A)*P(A)+P(defective/B)*P(B)+P(defective/C)*P(C)

=0.04*0.17+0.07*0.54+0.09*0.29=0.0707

P(defective)=0.0707

b)D=defective

P(C/D)=P(C/Defective)=P(Defective/C)*P(C)/P(defective)

=0.09*0.29/0.0707

=0.0261/0.0707

=0.369

P(C/D)=0.369

1)P(female and skipped class)=2/33=0.06

P(male and skipped class)=6/33=0.18

P(male)=24/33=0.73

P(female)=9/33=0.27

a)Sample Space=

b)Discrete sample space,as the number of males or females can be discrete only.

c)P(skip class/male)=P(male and skipped class)/P(male)=0.182/0.73=0.25

d)P(skip class/female)=P(female and skipped class)/P(female)=0.06/0.27=0.22

e)P(attend class/male)=1-P(skip class/male)=1-0.25=0.75

males females Total skipped 6 2 8 not skipped or attend 18 7 25 Total 24 9 33 males females Total skipped 0.18 0.06 0.24 not skipped or attend 0.55 0.21 0.76 Total 0.73 0.27 1.00

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