**Answer:**

a) 7.54189*10^-8

b) 0.99999999

c) E ( X ) = 0.39 , s ( X ) = 0.604

**Step-by-step 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**