Question #171934

A new vaccine is able to give protection to 85% of bacterial disease. a. If 10 people are given the vaccines, find the probability that at least six people will be protected from the bacterial disease. 2 marks b. If 100 people are given the vaccines, find the value of n such that the probability that at least n people will be protected from the bacterial disease is 0.94. 4 marks c. State the type of distribution use to solve question (a) and (b). Briefly explain why you choose the distribution.


1
Expert's answer
2021-03-18T13:48:30-0400

We have that

p = 85% = 0.85

a) m = 10

This follows the binomial distribution

The binomial probability is calculated by the formula:


P(X=n)=C(m,n)pn(1p)mnP(X=n)=C(m,n)\cdot p^n\cdot(1-p)^{m-n}

P(X6)=P(X=6)+P(X=7)+P(X=8)+(X=9)+P(X=10)P(X\ge6)=P(X=6)+P(X=7)+P(X=8)+(X=9)+P(X=10)

P(X=6)=C(10,6)0.856(10.85)106=0.04P(X=6)=C(10,6)\cdot 0.85^6\cdot(1-0.85)^{10-6}=0.04

P(X=7)=C(10,7)0.8570.153=0.1299P(X=7)=C(10,7)\cdot 0.85^7\cdot0.15^3=0.1299

P(X=8)=C(10,8)0.8580.152=0.2759P(X=8)=C(10,8)\cdot 0.85^8\cdot0.15^2=0.2759

P(X=9)=C(10,9)0.8590.151=0.3474P(X=9)=C(10,9)\cdot 0.85^9\cdot0.15^1=0.3474

P(X=10)=C(10,10)0.85100.150=0.1969P(X=10)=C(10,10)\cdot 0.85^{10}\cdot0.15^0=0.1969

P(X6)=0.04+0.1299+0.2759+0.3474+0.1969=0.9901P(X\ge6)=0.04+0.1299+0.2759+0.3474+0.1969=0.9901

b) m = 100

P(Xn)=0.94P(X\ge n)=0.94

Using a calculator we approximate that there are at least 80 successes such that the probability is 0.94.

c) This question follows the binomial distribution because we have a fixed number of trials and boolean-valued outcome: success (protection from the disease with probability p = 0.85) and failure (with probability 1 - p = 0.15), and the probability of success is exactly the same from one trial to another.


Answer:

a) 0.9901

b) 80

c) the binomial distribution


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