The bus you take every morning always arrives anywhere from 2 minutes early to 15 minutes late and it is equally likely that it arrives during any of those minutes. suppose that you arrive at the bus stop five minutes early. what is the probability that the bus is more than ten minutes late?

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anaguerrero0505 anaguerrero0505 · Answer 1 · 2019-10-18T00:34:22+02:00

Answer:

5/17

Step-by-step explanation:

To find the probability, you have to consider the probability as [tex]\frac{possibilites}{total}[/tex]

Therefore, the possibility that the bus arrives more than 10 minutes late is given by the maximum waiting time (15 minutes) minus 10 minutes, 10-15=5

The total number of possibilities for the bus to arrive is given by the 15 minutes, plus the 2 minutes it could arrive early, therefore 15+2=17

This expressed as the probability would be 5/17

joaobezerra joaobezerra · Answer 2 · 2021-09-30T13:08:20+02:00

Using the uniform distribution, it is found that there is a 0.2941 = 29.41% probability that the bus is more than ten minutes late.

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The uniform distribution has two bounds, a and b.

The probability of finding a value above x is:

[tex]P(X > x) = \frac{b - x}{b - a}[/tex]

In this problem:

Anywhere between 2 minutes early and 15 minutes late, thus between -2 and 15, meaning that [tex]a = -2, b = 15[/tex].
The probability of being more than 10 minutes late is P(X > 10), thus:

[tex]P(X > 10) = \frac{15 - 10}{15 - (-2)} = \frac{5}{17} = 0.2941[/tex]

0.2941 = 29.41% probability that the bus is more than ten minutes late.

A similar problem involving the uniform distribution is given at https://brainly.com/question/15855314