In a large population, 73% of the population are short-sighted. A samples of 7 people is randomly chosen from the population. A random sample of n people is chosen instead from the same population. Find the minimum number n such that there is at least 99% chance of having one or more people who are short-sighted

4

Step-by-step explanation:

Let x , be the number of short sighted people among n people.

, follows binomial distribution : x∼B(N, 0.73)

0.99

1-p(1)0.99

1-0.99p(=0)

p(=0)0.01

× ×0.01

0.01

nln0.27ln0.01

nln0.01/ln0.27

n3.52 ≈ 4

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step-by-step explanation:

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answer: y=0 if that is what u r looking for