Question

A study shows that 78% of adults need eye correction. If 12 adults are randomly selected, find the probability that atleast 11 of them need correction for their eyesight

Answer 1

Answer: 0.2224

Explanation:

Given: The proportion of adults need eye correction: p= 78%=0.78

Let X be a binomial variable that represents the number of adults who need eye correction.

Sample size of adults: n= 12

Binomial ditsribution formula:

`P(X=x)=^nC_xp^x(1-p)^(n-x)`

Now, the probability that at least 11 of them need correction for their eyesight will be

`P(X\geq11)=P(X=11)+P(X=12)\\\\=^(12)C_(11)(0.78)^(11)(1-0.78)^(1)+^(12)C_(12)(0.78)^(12)(1-0.78)^(0)\\\\=(12)(0.78)^(11)(0.22)+(1)(0.78)^(12)(1)\\\\\approx0.1716503+0.05071486=0.22236516\approx0.2224`

Hence, the required probability = 0.2224