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

Answer:

4

Step-by-step 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