Question

Below is a probability distribution for the number of failures in an elementary statistics course. X 0 1 2 3 4 P(X=x) 0.41 0.18 ? 0.06 0.06 Determine the following probabilities: a. P(X = 2) = b. P(X < 2) = c. P(X ≤ 2) = d. P(X > 2) = e. P(X = 1 or X = 4) = f. P(1 ≤ X ≤ 4) =

Answer 1

a) P(X = 2) = 0.29

b) P(X < 2) = 0.59

c) P(X ≤ 2) = 0.88

d) P(X > 2) = 0.12

e) P(X = 1 or X = 4) = 0.24

f) P(1 ≤ X ≤ 4) = 0.59

Explanation:

We have the following probability distribution:

P(X = 0) = 0.41

P(X = 1) = 0.18

P(X = 2) = p

P(X = 3) = 0.06

P(X = 4) = 0.06

a. P(X = 2) =

The sum of all those probabilities is decimal 1. So

0.41 + 0.18 + p + 0.06 + 0.06 = 1

p = 1 - 0.71

p = 0.29

P(X = 2) = 0.29

b. P(X < 2) =

P(X < 2) = P(X = 0) + P(X = 1) = 0.41 + 0.18 = 0.59

c. P(X ≤ 2) =

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.41 + 0.18 + 0.29 = 0.88

d. P(X > 2) =

P(X > 2) = P(X = 3) + P(X = 4) = 0.06 + 0.06 = 0.12.

e. P(X = 1 or X = 4) =

P(X = 1 or X = 4) = P(X = 1) + P(X = 4) = 0.18 + 0.06 = 0.24

f. P(1 ≤ X ≤ 4) =

P(1 ≤ X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.18 + 0.29 + 0.06 + 0.06 = 0.59

Answer 2

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Explanation:

a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29

b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59

c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88

d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12

e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24

f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59