Question

1. Let the random variable X represent the number of walks you take a day. Also, assume that the number of walks you
take is uniformly distributed between 0 and 3; that is, you are equally likely to take 0, 1, 2, or 3 walks a day.

a.What type of distribution is the random variable X?

b. What is the probability that on a randomly chosen day, you walk exactly 0 walks?
c. What is the probability that on a randomly chosen day, you walk 2 or 3 days?

2. Every two seconds someone in America needs blood. It is important that hospitals have the right types of quality
blood to meet the patient needs. Blood donation centers help hospitals meet this demand. There are four main
blood groups: A, B, AB, and O, of which O is the most common. One of the less common blood types is B+. In
the United States, only 9% of the population has B+ blood type.
Assume that 5 donors arrive to a blood donation center and you are interested in the number of these donors that
have B+ blood type. Therefore, let X be the random variable that represents the number of donors with B+ blood
type among the 5 donors that came to the center.

a) What type of distribution is the random variable X?

b.) What is the probability that exactly 2 of the donors have B+ blood type?
c.)What is the probability that fewer than 2 donors have B+ blood type?
d.)What is the probability that 3 or more of the donors have B+ blood type?

Answer 1

1. a) The random variable X follows Uniform distribution i.e. Uniform(0,3).

b) The probability that walk exactly 0 walks on a randomly chosen day is 1/3 = 0.333.

c) The probability is P(2) + P(3) - P(2) P(3) = (1/3) + (1/3) - (1/3)*(1/3) = 0.556.

2. a) The random variable X follows Binomial distribution i.e. Binomial(5, 0.09).

b) The probability that exactly 2 of the donors have B+ blood type is

(5!/(2!3!)) * 0.09^2 * 0.91^3 = 0.061

c) The probability that fewer than 2 of the donors have B+ blood type is

(5!/(0!5!)) * 0.09^0 * 0.91^5 + (5!/(1!4!)) * 0.09^1 * 0.91^4 = 0.9326

d) The probability that 3 or more of the donors have B+ blood type is

1 - [(5!/(0!5!)) * 0.09^0 * 0.91^5 + (5!/(1!4!)) * 0.09^1 * 0.91^4 + (5!/(2!3!)) * 0.09^2 * 0.91^3] = 0.0063