Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabilities. x 0 1 2 3 4 5 P(X=x) 0.1 0.15 0.25 0.25 0.15 …

Question

asked Oct 2, 2022 35.5k views

1 Answer

Camabeh · Answer 1 · 2022-10-08T01:44:50+0000

Answer:

a.
$P(2 \leq X \leq 3) = 0.5$

b.
$P(X \geq 1) = 0.9$

c.
P(X < 2) = 0.25

d.
P(X > 3) = 0.25

Explanation:

We are given the following distribution:

P(X = 0) = 0.1

P(X = 1) = 0.15

P(X = 2) = 0.25

P(X = 3) = 0.25

P(X = 4) = 0.15

P(X = 5) = 0.1

a. Two or three bars

$P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) = 0.25 + 0.25 = 0.5$

Thus:

$P(2 \leq X \leq 3) = 0.5$

b. At least one bar

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1 = 0.9$

Thus:

$P(X \geq 1) = 0.9$

c. Fewer than two bars

P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.15 = 0.25

Thus:

P(X < 2) = 0.25

d. More than three bars

P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.1 = 0.25

Thus:

P(X > 3) = 0.25