Respuesta :
The distribution of the time until the first sale of a book [tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]
The probability that no books are sold during a particular hour [tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex].
The expected number of customers who buy a book during a particular hour [tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex]
As per the details share in the above question are as follow,
Poisson process with a rate λ per hour
While customer buys a book with probability p
First we have to find the distribution of the time until the first sale of a book.
Second to find the probability that no books are sold during a particular hour.
Third to find the expected number of customers who buy a book during a particular hour.
Giving the duration till a book sells its first copy
Consider,
Following are some instances of waiting times,
[tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]
The distribution of the time until the first sale of a book [tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]
Now, to calculate the likelihood that no books will be sold at a specific time.
[tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex]
The probability that no books are sold during a particular hour [tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex].
Consider the Poisson distribution to determine the anticipated number of book purchases during a specific hour.
[tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex]
The expected number of customers who buy a book during a particular hour [tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex].
For more such question on Poisson process.
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