6.5% of people in the U. S. Have A - blood type. You randomly select 6 Americans and ask them if their blood type is A-. a) Find the probability that all 6 are type A-. b) Find the probability that at most 4 of them are type A-. c) Find the mean and standard deviation.

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