Revision as of 00:56, 26 June 2024 by Admin (Created page with "'''Solution: D''' The number of persons that show up for the flight equals <math>N</math> and has a binomial distribution with <math>n=100, p = 0.96 </math> and we are interested in the probability that <math>N \leq 98 </math>. This equals <math display = "block"> 1-P(N=99)-P(N=100) = 1- 100 \cdot 0.96^{99} \cdot 0.04 - 0.96^{100} = 0.9128. </math>")
Exercise
ABy Admin
Jun 26'24
Answer
Solution: D
The number of persons that show up for the flight equals [math]N[/math] and has a binomial distribution with [math]n=100, p = 0.96 [/math] and we are interested in the probability that [math]N \leq 98 [/math]. This equals
[[math]]
1-P(N=99)-P(N=100) = 1- 100 \cdot 0.96^{99} \cdot 0.04 - 0.96^{100} = 0.9128.
[[/math]]