jacob7542
30.11.2019 •
Mathematics
Suppose a basketball player is practicing shooting, and has a prob-ability .95 of making each of his shots. also assume that his shots are in-dependent of one another. using the poisson distribution, approximate theprobability that there are at most 2 misses in the first 100 attempts
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Ответ:
0.082
Step-by-step explanation:
Number of attempts = n = 100
Since there are only two outcomes and in-dependent of each other, the probability of missing a shot = 1 - Probability of making each shot
p = 1 - 0.95 = 0.05
Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missing
λ = 100 x 0.05 = 5
Define X such that X = Number of misses and X ≅ Poisson (λ = 5)
P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2]
P [X ≤ 2] = e⁻⁵ + e⁻⁵ x 5 + e⁻⁵ x 5²/2!
P [X ≤ 2] = e⁻⁵ [1 + 5 + 5²/2!]
P [X ≤ 2] = e⁻⁵ x 12.25 = 0.082
The required probability that there are at most 2 misses in the first 100 attempts is 0.082
Ответ: