Respuesta :
Answer:
Therefore the chance that a particular square inch is not by any drops during a given 10 second period is [tex]e^{-5a}[/tex].
Step-by-step explanation:
Poisson distribution:
[tex]P(X=x)= \frac{e^{-\lambda t} (\lambda t)^x}{x!}[/tex]
Given that rain is falling at an average rate 30 drops per square inches per minutes.
A reasonable choose of this distribution is Poisson distribution [tex](\lambda t)[/tex].
Let the area be a square inch.
Here [tex]\lambda[/tex] = average rate of rain drop × area
=(30×a)= 30 a
t = 10 second [tex]= \frac{10}{60} \ minutes[/tex] [tex]=\frac16\ minutes[/tex].
Therefore,
[tex]P(\textrm {no rain drop in }\frac16 \ minutes) =P(X=0)[/tex]
[tex]=\frac{e^{-\frac{30a}{6}}.(\frac{30a}{6})^0}{0!}[/tex]
[tex]=e^{-5a}[/tex]
Therefore the chance that a particular square inch is not by any drops during a given 10 second period is [tex]e^{-5a}[/tex].
Using the Poisson distribution, and considering that intervals are independent, it is found that there is a 0.0067 = 0.67% probability that a particular square inch is not hit by any drops during a given 10-second period.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, the rate is of 30 drops per minute, hence, for 10 seconds, considering intervals are independent, the mean is:
[tex]\mu = 10\frac{30}{60} = 10(0.5) = 5[/tex]
The probability that a particular square inch is not hit by any drops during a given 10-second period is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}5^{0}}{(0)!} = 0.0067[/tex]
0.0067 = 0.67% probability that a particular square inch is not hit by any drops during a given 10-second period.
A similar problem is given at https://brainly.com/question/14829339
