Respuesta :
Answer:
a) the probability of having exactly 10 days of precipitation in the month of April
= P(x = 10) = [tex]\frac{e^{-12.2}(12.2)^{10} }{10!} = 0.1013[/tex]
b) the probability of having less than 3 days of precipitation in the month of April
= P(x < 3) = p( x = 1) + p(x = 2)
= [tex]\frac{e^{-12.2}(12.2)^{1} }{1!} + \frac{e^{-12.2}(12.2)^{2} }{2!} =\hspace{0.1cm}[/tex] [tex]0.000374 + \hspace{0.1cm} 0.0000614 \hspace{0.1cm} = \hspace{0.1cm} 0.000436[/tex]
c.) the probability of having more than 15 days of precipitation in
the month of April
= P (x > 15) = 1 - P( x≤15) = 1 - [tex]\frac{e^{-12.2}(12.2)^{15} }{15!} = 0.076[/tex]
Step-by-step explanation:
i) Here λ = 12.2
Poisson distribution, P(x = X) = [tex]\dfrac{e^{-\lambda} \lambda^{x} }{x!}[/tex]
a) the probability of having exactly 10 days of precipitation in the month of April
= P(x = 10) = [tex]\frac{e^{-12.2}(12.2)^{10} }{10!} = 0.1013[/tex]
b) the probability of having less than 3 days of precipitation in the month of April
= P(x < 3) = p( x = 1) + p(x = 2)
= [tex]\frac{e^{-12.2}(12.2)^{1} }{1!} + \frac{e^{-12.2}(12.2)^{2} }{2!} =\hspace{0.1cm}[/tex] [tex]0.000374 + \hspace{0.1cm} 0.0000614 \hspace{0.1cm} = \hspace{0.1cm} 0.000436[/tex]
c.) the probability of having more than 15 days of precipitation in
the month of April
= P (x > 15) = 1 - P( x≤15) = 1 - [tex]\frac{e^{-12.2}(12.2)^{15} }{15!} = 0.076[/tex]
