Question

Today, the waves are crashing onto the beach every 4.8 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.8 seconds. 61% of the time a person will wait at least how long before the wave crashes in?

Answer 1

Answer:

61% of the time a person will wait at least 1.872 seconds before the wave crashes in.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

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

Uniform distribution from 0 to 4.8 seconds.

This means that $a = 0, b = 4.8$

61% of the time a person will wait at least how long before the wave crashes in?

This is the 100 - 61 = 39% percentile, which is x for which $P(X \leq x) = 0.39$. So

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

$0.39 = \frac{x - 0}{4.8 - 0}$

$x = 4.8*0.39$

$x = 1.872$

61% of the time a person will wait at least 1.872 seconds before the wave crashes in.