Answer:
(1). y = x ~ Exp (1/3).
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
Step-by-step explanation:
Kindly check the attachment to aid in understanding the solution to the question.
So, from the question, we given the following parameters or information or data;
(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.
(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "
(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."
(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.
Answer:
A AND D
Step-by-step explanation:
Answer:
P=2.326
Step-by-step explanation:
Raw Score (X):=1000
Population Mean (μ):=460
Standard Deviation (σ): =Sqrt(npq)
Where n=1000, p=460/1000=0.46 and q=1-0.46=0.54
Sqrt(npq)=sqrt(460X0.54)=15.7607
Z = (X - μ) / σ
Z = (1000 - 460) / 15.7607
Z = 34.26244
For p-value 0.01
P(x=0.01) = 2.32635
Hence, P=2.326 t 3 decimal places.
The answer is 2 miles or 3520 yards. (they are the same answer.)
If this helps, can you give me the brainliest answer? I need it...
Thank you very much.
Two events are said to be Disjoint or Mutually Exclusive if the two events can not happen at the same time.For example when we throw a die getting an even number is disjoint to getting an odd number.
I.e Probability(A∩B)=0
Let me explain this concept through venn diagram.
Pr[A∪B]=0.7, Pr[A]=0.25
Since events are Disjoint
Pr[A∩B]=0
Pr[A∪B]=Pr[A] + Pr[B]
0.7=0.25 +Pr[B]
0.7-0.25=Pr[B]
⇒Pr[B]=0.45=45/100=9/20
Now events are said to be independent if Pr[A and B]=Pr[A] ×Pr[B]
Events are said to be independent if occurrence of one is not affected by occurrence of other.For example getting multiple of 2 as one event and getting multiple of 3 as second event when we throw a die.
Pr[A∪B]=0.7, Pr[A]=0.25
Pr[A∪B]= Pr[A]+ Pr[B]-Pr[A∩B]
But Pr[A∩B]= Pr[A] ×Pr[B]
⇒Pr[A∪B]= Pr[A]+ Pr[B]- Pr[A] ×Pr[B]
⇒0.7=0.25+p-0.25×p
⇒0.7-0.25=p- 0.25 p
⇒0.45=0.75 p
⇒p= 0.45/0.75
⇒p =3/5