MATH 110 MODULE 4 EXAM
QUESTIONS AND ANSWERS-
PORTAGE LEARNING
[Document subtitle]
[DATE]
[COMPANY NAME]
[Company address]
, Stati sti cs - Portage Online Summer 2018
Module 4 Exam
Exam Page 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is .08. What is the probability that six or more safety systems will fail during an
emergency?
f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) )
n=8
x = 6, 7, 8 (number of failures)
p = 0.08
6 failures:
n=8
x=6
p = 0.8
n-x = 8-6 = 2
( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-6
7 failures:
n=8
x=7
p = 0.8
n-x = 8-7 = 1
( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-7
8 failures:
n=8
x=8
p = 0.8
n-x = 8-8 = 0
( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-9
f(6) = 6.21 x 10^-6
This study source was downloaded by 100000819885058 from CourseHero.com on 04-18-2022 01:43:33 GMT -05:00
https://www.coursehero.com/file/32343987/Module-4-Examdocx/
QUESTIONS AND ANSWERS-
PORTAGE LEARNING
[Document subtitle]
[DATE]
[COMPANY NAME]
[Company address]
, Stati sti cs - Portage Online Summer 2018
Module 4 Exam
Exam Page 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety
systems failing is .08. What is the probability that six or more safety systems will fail during an
emergency?
f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) )
n=8
x = 6, 7, 8 (number of failures)
p = 0.08
6 failures:
n=8
x=6
p = 0.8
n-x = 8-6 = 2
( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-6
7 failures:
n=8
x=7
p = 0.8
n-x = 8-7 = 1
( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-7
8 failures:
n=8
x=8
p = 0.8
n-x = 8-8 = 0
( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-9
f(6) = 6.21 x 10^-6
This study source was downloaded by 100000819885058 from CourseHero.com on 04-18-2022 01:43:33 GMT -05:00
https://www.coursehero.com/file/32343987/Module-4-Examdocx/