Answer each question: The probability distribution for a discrete random variable: 3.8) A single cell can either die, with probability 0.1, or split into two cells, with probability 0.9, producing a...

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Answer each question:
The probability distribution for a discrete random variable:
3.8)
A single cell can either die, with probability 0.1, or split into two cells, with probability 0.9, producing a new generation of cells. Each cell in the new generation dies or splits into two cells independently with the same probabilities as the initial cell. Find the probability distribution for the number of cells in the next generation.
The expected value of a random variable or a function of a random variable:
3.12)
Let Y be a random variable with p(y) given in the accompanying table. Find E(Y), E(1/Y), E(Y2-1), and V(Y).
Y 1 2 3 4
p(y) 0.4 0.3 0.2 0.1
3.22)
A single fair die is tossed once. Let Y be the number facing up. Find the expected value and variance of Y.
The binomial probability distribution:
3.40)
The probability that a patient recovers from a stomach disease is 0.8. Suppose 20 people are know to have contracted this disease. What is the probability that

  1. Exactly 14 recover?

  2. At least 10 recover?

  3. At least 14 but not more than 18 recover?

  4. At most 16 recover?



3.44)
A new surgical procedure is successful with a probability of p. Assume that the operation is performed five times and the results are independent of one another. What is the probability that

  1. All five operations are successful if p= 0.8?

  2. Exactly four are successful if p= 0.6?

  3. Less than two are successful if p= 0.3?



3.52)
The taste test for PTC (phenylthiocarbamide) is a favorite exercise in beginning human genetics classes. It has been established that a single gene determines whether or not an individual is a “taster.” If 70% of Americans are “tasters” and 20 Americans are randomly selected, what is the probability that

  1. At least 17 are “tasters”?

  2. Fewer than 15 are “tasters”?



The poisson probability distribution:
3.124)
Approximately 4% of silicon wafers produced by a manufacturer have fewer than two large flaws. If Y, the number of flaws per wafer, has a Poisson distribution, what proportion of the wafers have more than five large flaws?
Moments and moment-generating functions:
3.146)
Differentiate the moment-generating function m(t) = (pet
+q)n
, where q= 1-p to find E(Y) and E(Y2). Then find V(Y).
Probability-generating functions (optional):
3.170)
The U.S. mint produces dimes with an average diameter of 0.5 inch and standard deviation 0.01 Using Tchebysheff’s theorem, find a lower bound for the number of coins in a lot of 400 coins that are expected to have a diameter between 0.48 and .052.
The uniform probability distribution:
4.50)
If a point is randomly located in an interval (a,b) and if Y denotes the location of the point, the Y is assumed to have a uniform distribution over (a,b). A plant efficiency expert randomly selects a location along a 500 foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

  1. Within 25 feet of the end of the line?

  2. Within 25 feet of the beginning of the line?

  3. Closer to the beginning of the line than to the end to the line?



The covariance of two random variables:
5.90)
We determine that the joint probability distribution of Y1, the number of married executives, and, Y2, the number of never married executives, is given by
P(y1,y2) =

------------------------


Where y1
and y2
are integers, 0 y1
3, 0 y2
3, and 1 y1
+ y2
3. Find Cov (Y1,Y2).
5.92)
We determine that f(y1,y2) =

Is a valid joint probability density function. Find Cov (Y1,Y2). Are Y1
and Y2
independent?
The expected value and variance of linear functions of random variables:
5.104)
We determine that the joint probability distribution of Y1, the number of married executives, and Y2, the number of never married executives, is given by
P(y1,y2) =

------------------------


Where y1
and y2
are integers, 0 y1
3, 0 y2
3, and 1 y1
+ y2
3.

  1. Find E(Y1
    +Y2) and V(Y1
    + Y2) by first finding the probability distribution of Y1
    +Y2.

  2. We determine that Cov(Y1,Y2) = -1/3. Find E(Y1
    + Y2) and V(Y1
    + Y2) by using theorem of the expected value and variance of linear functions of random variables.



Answered Same DayDec 21, 2021

Answer To: Answer each question: The probability distribution for a discrete random variable: 3.8) A single...

Robert answered on Dec 21 2021
125 Votes
3.8) The probability distribution for the numbers of dead and alive cells will
correspond to this proportion:
p/(p+q*2^n) : (q*2^n)/(p+q*2^n)
Where,
p is the probability to die and q is the probability to survive.
The probability distribution for the numbers of dead cells and alive cells after
splitting;
For first generation
= 0.1 / 0.9
For second generation
= 1: 0.05 /0.95
3.12) E(Y) = 1*0.4+2*0.3+3*0.2+4*0.1 = 2
E(1/Y) = (1/Y) p(Y)
= (1/1)*0.4+(1/2)*0.3+(1/3)*0.2+(1/4)*0.1 = 0.675
E(Y2-1) = E(Y2) – 1
= (12)*0.4+(22)*0.3+(32)*0.2+(42)*0.1 -1
=4
V(Y) = E(Y2) – [E(Y)]2
=5-22 = 1
3.22) Probability of coming 1 = 1/6
Probability of coming 2 = 1/6
Probability of coming 3 = 1/6
Probability of coming 4 = 1/6
Probability of coming 5 = 1/6
Probability of coming 6 = 1/6
E(Y) = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5
E(Y2) = (12)*(1/6) + (22)*(1/6) + (32)*(1/6) + (42)*(1/6)+ (52)*(1/6)+ (62)*(1/6)
= 15.17
V(Y) = E(Y2) – [E(Y)]2
= 15.17 – 3.52 = 2.92
3.40) a) Probability = 20C14 (0.8)
14 (0.2)6 = 0.109
b) Probability = 20C10 (0.8)
10 (0.2)10+20C11 (0.8)
11 (0.2)9+------------------------
-------------------+20C20 (0.8)
20 (0.2)0
= 0.999
c) Probability = 20C14 (0.8)
14 (0.2)6+-----------------+20C18 (0.8)
18 (0.2)2
= 0.844
d) Probability = 20C0 (0.8)
20 (0.2)0+-----------------+20C16 (0.8)
16 (0.2)4
= 0.589
3.44) a) Probability = 5C5 (0.8)
5(0.2)0 = 0.328
b) Probability =...
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