Question

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

Answer 1

4

Explanation:

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

`x_(1)`, follows binomial distribution : x∼B(N, 0.73)

`\geq`0.99

1-p(
`x_(1)`
`\geq`1)
`\geq`0.99

1-0.99
`\geq`p(
`x_(1)`=0)

p(
`x_(1)`=0)
`\leq`0.01

`{n}_C_(0)` ×
`0.73^(0)`×
`0.27^(n)`
`\leq`0.01

`0.27^(n)`
`\leq` 0.01

nln0.27
`\leq`ln0.01

n
`\geq`ln0.01/ln0.27

n
`\geq`3.52 ≈ 4