Question

Records indicate that the daily high January temperatures on a tropical island tend to have a uniform distribution over the interval from 75°F to 90°F. A tourist arrives on the island on a randomly selected day in January.
a. What is the probability that the temperature will be above 80°F
b. What is the probability that the temperature will be between 80°F and 85°F 
c. What is the expected temperature?

Answer 1

Answer:

a) 0.6667 = 66.67% probability that the temperature will be above 80°F.

b) 0.3333 = 33.33% probability that the temperature will be between 80°F and 85°F.

c) The expected temperature is of 82.5ºF.

Step-by-step explanation:

Uniform distribution:

The probability of all outcomes between a and b is the same.

Uniform distribution over the interval from 75°F to 90°F.

This means that $a = 75, b = 90$

a. What is the probability that the temperature will be above 80°F?

We have that, on the uniform distribution:

$P(X > x) = \frac{b - x}{b - a}$

In this question:

$P(X > 80) = \frac{90 - 80}{90 - 75} = \frac{10}{15} = 0.6667$

0.6667 = 66.67% probability that the temperature will be above 80°F.

b. What is the probability that the temperature will be between 80°F and 85°F?

We have that:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

Then, in this question:

$P(80 \leq X \leq 85) = \frac{85 - 80}{90 - 75} = \frac{5}{15} = 0.3333$

0.3333 = 33.33% probability that the temperature will be between 80°F and 85°F.

c. What is the expected temperature?

The expected value of the uniform distribution is:

$E = \frac{a + b}{2}$

In this question:

$E = \frac{75 + 90}{2} = 82.5$

The expected temperature is of 82.5ºF.