Question

The fatality rate is estimated at 2% across all age groups. Suppose a random sample of 200 infected persons is selected. (round probabilities to 4 decimal places)

What is the probability that between 4 and 8 infected people die?

Would it be considered unusual if less 2 infected people die? Justify your answer.

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 200 * 0.02 ) = 4
Variance = n * P * Q = ( 200 * 0.02 * 0.98 ) = 3.92
Standard deviation = √(variance) = √(3.92) = 1.9799

P ( 4 <= X <= 8 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 4 - 0.5 < X < 8 + 0.5 ) = P ( 3.5 < X < 8.5 )

X ~ N ( µ = 4 , σ = 1.9799 )
P ( 3.5 < X < 8.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 3.5 - 4 ) / 1.9799
Z = -0.25
Z = ( 8.5 - 4 ) / 1.9799
Z = 2.27
P ( -0.25 < Z < 2.27 )
P ( 3.5 < X < 8.5 ) = P ( Z < 2.27 ) - P ( Z < -0.25 )
P ( 3.5 < X < 8.5 ) = 0.9884 - 0.4013
P ( 3.5 < X < 8.5 ) = 0.5871

P ( X < 2 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 2 - 0.5 ) = P ( X < 1.5 )

X ~ N ( µ = 4 , σ = 1.9799 )
P ( X < 1.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.5 - 4 ) / 1.9799
Z = -1.26
P ( ( X - µ ) / σ ) < ( 1.5 - 4 ) / 1.9799 )
P ( X < 1.5 ) = P ( Z < -1.26 )
P ( X < 1.5 ) = 0.1038
No, since the probability is greater than 5% i.e > 0.05.