Statistics - Portage Online Summer
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=8x=6p=
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=8x=7p=
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=8x=8p=
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
f(7) = 1.54 x 10^-7 f(8) =
1.68 x 10^-9
(6.21 x 10^-6) + (1.54 x 10^-7) + (1.68 x 10^-9) = 6.355x10^-6
Statistics - Portage Online Summer Module 4 Exam
, Statistics - Portage Online Summer
Probability of 6,7, and 8 failing during an emergency = 6.36 x 10^-6
Answer Key
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?
Exam Page 2
Find each of the following probabilities:
a. Find P(Z ≤ 1.27) .
P(Z ≤ 1.27) = 0.89796
b. Find P(Z ≥ -.73) .
1 - p(z ≥ -0.73) subtract from 1 cause greater than probability
-0.73 = 0.2327
1 - 0.23270 = 0.7673
P(Z ≥ -0.73) = 0.7673
c. Find P(-.09 ≤ Z ≤ .86).
Statistics - Portage Online Summer Module 4 Exam
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=8x=6p=
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=8x=7p=
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=8x=8p=
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
f(7) = 1.54 x 10^-7 f(8) =
1.68 x 10^-9
(6.21 x 10^-6) + (1.54 x 10^-7) + (1.68 x 10^-9) = 6.355x10^-6
Statistics - Portage Online Summer Module 4 Exam
, Statistics - Portage Online Summer
Probability of 6,7, and 8 failing during an emergency = 6.36 x 10^-6
Answer Key
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?
Exam Page 2
Find each of the following probabilities:
a. Find P(Z ≤ 1.27) .
P(Z ≤ 1.27) = 0.89796
b. Find P(Z ≥ -.73) .
1 - p(z ≥ -0.73) subtract from 1 cause greater than probability
-0.73 = 0.2327
1 - 0.23270 = 0.7673
P(Z ≥ -0.73) = 0.7673
c. Find P(-.09 ≤ Z ≤ .86).
Statistics - Portage Online Summer Module 4 Exam
