Question

Approximately 10% of people in the world are left-handed, so P(a person is left handed) = 0.1. Suppose we have a class of 15 people and UHD is interested in how many left-handed desks the class will need, so we define a random variable X, the number of left-handed people in the class.

1. What distribution family does this situation belong to? What is (are) the parameter(s)?

2. What is the probability that none of the students will be left handed, P(X=0)?

3. What is the probability that at most 7 students will be left-handed, P(X ≤ 7)?

4. What is the expected number of left-handed students in class, E(X)?

Answer 1

1)

Let X denote the number of left-handed people in the class. Then

X ~ Bin(n = 15, p = 0.1)

P(X = z) = (0.1)"(1 -0.1)15-,1 = 0,1,2,3, 4, ..., 15

2)

Required probability =

P(X = 0) = (b)(0.1)º(1 – 0.1)15-0 = 0.2059

3)

Required probability =

P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

= ()0.1)°1 – 0.125-0 + (11)(0.1)'(1 – 0.1)15-1 + (15)(0.1°(1 – -0.1)15-2+(13(0.1)*(1–0.1)15-3+(13) (0.1)*(1–0.115-4 + (15)) (0.1)*(1 – 10.115-5 + (1)(0.1)*(1 – 0.1)35-6 + (13)(0.1)(1 – 0.1)15-7

=0.999966

4)

Since

X ~ Bin(n = 15, p = 0.1)

So,

EX) = np = 15(0,1) = 1.5