Solution Found!
A type of bacteria cell divides at a constant rate ? over
Chapter 3, Problem 203SE(choose chapter or problem)
A type of bacteria cell divides at a constant rate \(\lambda\) over time. (That is, the probability that a cell divides in a small interval of time is approximately
) Given that a population starts out at time zero with
cells of this bacteria and that cell divisions are independent of one another, the size of the population at time
, has the probability distribution
\(P[Y(t)=n]=(n-1 k-1) e^{\lambda k t}\left(1-e^{-\lambda s}\right)^{n-k}, n=k, k+1, \ldots\)
a. Find the expected value and variance of in terms of \(\lambda\) and
.
b. If, for a type of bacteria cell, \(\lambda=1\) per second and the population starts out with two cells at time zero, find the expected value and variance of the population after five seconds.
Equation transcription:
Text transcription:
P[Y(t)=n]=(n-1 k-1) e^{\lambda k t}\left(1-e^{-\lambda s}\right)^{n-k}, n=k, k+1, \ldots
lambda
lambda=1
Questions & Answers
QUESTION:
A type of bacteria cell divides at a constant rate \(\lambda\) over time. (That is, the probability that a cell divides in a small interval of time is approximately
) Given that a population starts out at time zero with
cells of this bacteria and that cell divisions are independent of one another, the size of the population at time
, has the probability distribution
\(P[Y(t)=n]=(n-1 k-1) e^{\lambda k t}\left(1-e^{-\lambda s}\right)^{n-k}, n=k, k+1, \ldots\)
a. Find the expected value and variance of in terms of \(\lambda\) and
.
b. If, for a type of bacteria cell, \(\lambda=1\) per second and the population starts out with two cells at time zero, find the expected value and variance of the population after five seconds.
Equation transcription:
Text transcription:
P[Y(t)=n]=(n-1 k-1) e^{\lambda k t}\left(1-e^{-\lambda s}\right)^{n-k}, n=k, k+1, \ldots
lambda
lambda=1
ANSWER:Solution:
Step 1 of 3:
Let Y(t) denotes the population size at time t. Y(t) has the probability distribution