Question

Suppose that the biologist takes a random sample of size 50. Find the probability that fewer than 35 of the sampled will be infected, using the normal approximation approach. The Population proportion is 80%.

Answer 1

Answer: 0.038528

Explanation:

Given: Sample size : n= 50

Population proportion: p = 80% = 0.80

Let x be the random variable .

Mean
`\mu= np = (50)(0.80) = 40`

Standard deviation
`\sigma= √(50* (0.80)(1-0.80))=√(50*0.80*0.2)`

`=√(8)=2.828`

The probability that fewer than 35 of the sampled will be infected =

`P(x<35)=P((x-\mu)/(\sigma)<(35-40)/(2.828))\\\\=P(Z<(-5)/(2.828))\\\\=P(Z<-1.76803)\\\\=1-P(Z<1.76803)\\\\=1-0.961472\\\\=0.038528`

Hence, the required probability = 0.038528