Question

In the Brazilian population the share of individuals with blood from the O + group is 36% and O- is 4%.

Let X be the number of people with blood type O, in a random sample of 800 Brazilian citizens.

Determine the narrowest interval of type A = [a, b], such that P {X€A}> 0.95.

B. In a random sample of 100 Brazilians, what is the probability that exactly 30 of them

are from group O? and 2 of group O-?

Answer 1

a.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 100 * 0.36
= 36
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 100 * 0.36 * 0.64
= 23.04
III.
standard deviation = sqrt( variance ) = sqrt(23.04)
=4.8
a.
In a random sample of 100 Brazilians, the probability that exactly 30 of them
are from group O
P( X = 30 ) = ( 100 30 ) * ( 0.36^30) * ( 1 - 0.36 )^70
= 0.0389

b.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 100 * 0.04
= 4
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 100 * 0.04 * 0.96
= 3.84
III.
standard deviation = sqrt( variance ) = sqrt(3.84)
=1.9596

In a random sample of 100 Brazilians, the probability that exactly 30 of them
are from 2 of group O-
P( X = 30 ) = ( 100 30 ) * ( 0.04^30) * ( 1 - 0.04 )^70
= 0