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

Step-by-step 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 $\sigma= \sqrt{50\times (0.80)(1-0.80)}=\sqrt{50\times0.80\times0.2}$

$=\sqrt{8}=2.828$

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

$P(x<35)=P(\dfrac{x-\mu}{\sigma}<\dfrac{35-40}{2.828})\\\\=P(Z<\dfrac{-5}{2.828})\\\\=P(Z<-1.76803)\\\\=1-P(Z<1.76803)\\\\=1-0.961472\\\\=0.038528$

Hence, the required probability = 0.038528