Question

Suppose 4% of all people inoculated with a vaccine develop a reaction.If 1,000 people are inoculated A: What is the expected number of peoplewho develop a reaction B. What is the standard deviation C. What is the probablity that there are at least 50 people who develop a reaction D. What is the probablity that there is less than 30 people who develop a reaction

Answer 1

X ~ Binomial (n,p)

Where n = 1000 , p = 0.04

a)

E(X) = n p

= 1000 * 0.04

= 40

b)

Standard deviation = Sqrt ( np(1-p) )

= sqrt(1000 * 0.04 * 0.96)

= 6.1968

c)

Using normal approximation,

P( X >= x) = P( Z > x-0.5 - mean / SD)

So,

P(X >= 50) = P( Z > 49.5 - 40 / 6.1968)

= P( Z > 1.5330)

= 1 - P( Z < 1.5330)

= 1 - 0.9374

= 0.0626

d)

P( X < x) = P( Z < x-0.5 - Mean / Standard deviation)

So,

P( X <30) = P( Z < 29.5 - 40 / 6.1968)

= P( Z < -1.6944)

= 1 - P( Z < 1.6944)

= 1 - 0.9549

= 0.0451