Lori passes 80% of her tests. In the next week, Lori has 4 tests. She wants to determine the probability that she passes all 4 o
2 answers:
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Solution :
a).
Given : Number of times, n = 25
Sigma, σ = 0.200 kg
Weight, μ = 13 kg
Therefore the hypothesis should be tested are :
b). When the value of
Test statics :
= 45.5
P-value = 2 x P(Z > 45.5)
= 2 x 1 -P (Z < 45.5) = 0
Reject the null hypothesis if P value < α = 0.01 level of significance.
So reject the null hypothesis.
Therefore, we conclude that the true mean measured weight differs from 13 kg.
Answer:
The probability of an operating system error given that a failure has ocurred is P(O/E)= 0.19
Step-by-step explanation:
Hello!
There are three different sources of error when a computer system fails:
D: disk drive error. ⇒ P(D)= 0.50
M: computer memory error. ⇒P(M)= 0.20
O: operating systems error. ⇒ P(O)= 0.30
According to the component perdormance standard:
The probability of failure "E", given that there is a disk drive error is P(E/D)= 0.40
The probability of failure "E", given that there is a computer memory error is P(E/M)= 0.6
The probability of failure "E", given that there is a operating system error is P(E/O)= 0.25
You need to calculate the probability of an operating system error given that a failure has ocurred, symbolically:
P(O/E)
To reach the probability of the marginal E you have to add all intersections between the event "E" and the events "D", "M" and "O"
D M O Total
E: P(E∩D); P(E∩M); P(E∩O); P(E)
Using the formula of the conditional probability I'll clear all three intersections using the known probabilities:
General formula: ⇒
P(E∩D)= P(E/D)*P(D)= 0.4*0.5= 0.2
P(E∩M)= P(E/M)*P(M)= 0.6*0.2= 0.12
P(E∩O)= P(E/O)*P(O)= 0.25*0.3= 0.075
P(E)= P(E∩D) + P(E∩M) + P(E∩O)= 0.2+0.12+0.075= 0.395
I hope it helps!