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The amount of time it takes Isabella to wait for the bus is continuous and uniformly distributed between 3 minutes and 15 minutes. What is the probability that it takes Isabella more than 11 minutes to wait for the bus?

Respuesta :

Answer:

33.33% probability that it takes Isabella more than 11 minutes to wait for the bus

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.

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

For this problem, we have that:

Uniformly distributed between 3 minutes and 15 minutes:

So [tex]a = 3, b = 15[/tex]

What is the probability that it takes Isabella more than 11 minutes to wait for the bus?

Either she has to wait 11 or less minutes for the bus, or she has to wait more than 11 minutes. The sum of these probabilities is 1. So

[tex]P(X \leq 11) + P(X > 11) = 1[/tex]

We want P(X > 11). So

[tex]P(X > 11) = 1 - P(X \leq 11) = 1 - \frac{11 - 3}{15 - 3} = 0.3333[/tex]

33.33% probability that it takes Isabella more than 11 minutes to wait for the bus

Cricetus

The probability that it takes Isabella more than the given time to wait will be "0.3333".

Let,

  • [tex]x \sim Uniform(3,15)[/tex]

and,

  • [tex]3 \leq x \leq 15[/tex]

then,

  • [tex]f(x) = \frac{1}{15-3}[/tex]

Now,

The probability will be:

→ [tex]P(x >11) = \int\limits^{15}_{11} {f(x)} \, dx[/tex]

                   [tex]= \int\limits^{15}_{11} {(\frac{1}{15-3} )} \, dx[/tex]

                   [tex]= \frac{1}{12} [x]^{15}_{11}[/tex]

                   [tex]= \frac{1}{12} [15-11][/tex]

                   [tex]= \frac{4}{12}[/tex]

                   [tex]=0.3333[/tex]

Thus the above approach is appropriate.

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