Answer:
a) 7.54189*10^-8
b) 0.99999999
c) E ( X ) = 0.39 , s ( X ) = 0.604
Explanation:
Solution:-
- We will assume the proportion of people with A- blood group is independent and remains constant for a fairly small sample of n = 6 Americans selected at random.
- We will denote a random variable X = The number of americans out of 6 that have blood group type A-.
- The random variable is assumed to follow binomial distribution.
- The probability of success is the proportion of people in U.S that have blood group type A-, p = 0.065:
X ~ Bin ( 6 , 0.065 )
Where, r represents the number of americans out of selected 6 have blood group A-. The pmf of a binomial variate is given as:
P ( X = r ) = nCr * ( p ) ^r * ( 1 - p ) ^ ( n - r )
a) Find the probability that all 6 are type A-
- We will pmf given above and set r = 6. And evaluate the resulting probability:
P ( X = 6 ) = 6C6 * ( p )^6 * ( 1 - p ) ^ ( 0 )
= p^6
= ( 0.065 )^6
= 7.54189*10^-8
b) Find the probability that at most 4 of them are type A-
- We will pmf given above and evaluate the following expression:
P ( X ≤ 4 ) = 1 - P ( X = 5 ) - P ( X = 6 )
P ( X ≤ 4 ) = 1 - 6C5 * ( p )^5 * ( 1 - p ) ^ ( 1 ) - p^6
= 1 - 6*(0.065)^5 ( 0.935 ) - 0.065^6
= 1 - 6.50923*10^-8 - 7.54189*10^-8
= 0.99999999
c) Find the mean and standard deviation.
- The mean E ( X ) of the defined random variable distributed binomially is given by:
E ( X ) = n*p = 6*0.065 = 0.39 people
- The mean s( X ) of the defined random variable distributed binomially is given by:
s ( X ) = √n*p*q = √(6*0.065*0.935) = 0.604