Question

Somebody speculates that a person will test COVID 19 positive with probability 0.2. Then for finding the probability for exactly 100 patients testing positive in a sample of 400 patients, how do you do continuity correction? In the above question, what is the required probability using normal approximation to binomial? What is the probability that at least 70 patients test COVID 19 positive?

Answer 1

Let X be the number of patients testing positive

X follow Binomail with n= 400 , p=0.2

Using Normal approximation to Binomial

X follow Normal wuth mean = np = 400*0.2 = 80

and standard deviation =

We have to find P( X=100)

Since for using Normal distribution we need continuous data , we need to do continuity correction for X=100

Continuity correction for X=100 is 99.5 < X < 100.5

P(X=100)

=P( 99.5 < X < 100.5)

then

P( 99.5 < X < 100.5)

= P( 2.44 < z < 2.56)

= P( z < 2.56) - P( z < 2.44)

= 0.9948 - 0.9927 ( from z table)

= 0.0021

Probability that exactly 100 patients tested positive = 0.0021

To find

P( X > 70)

= P( z >  -1.25)

= 0.8944 ( from z table )

Probability that at least 70 patients test positive = 0.8944