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3 years ago

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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

1 answer:

densk [106]3 years ago

6 0

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

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