Question

1. The Sky Train from the terminal to the rental car and long-term parking center is supposed to arrive every 12 minutes. The waiting times for the train are known to follow a uniform distribution. What is the average waiting time (in minutes)?2. The Sky Train from the terminal to the rental car and long-term parking center is supposed to arrive every 16 minutes. The waiting times for the train are known to follow a uniform distribution. Find the 20th percentile for the waiting times (in minutes).3. The Sky Train from the terminal to the rental car and long-term parking center is supposed to arrive every 18 minutes. The waiting times for the train are known to follow a uniform distribution. What is the probability of waiting more than 16 minutes given a person has waited more than 9 minutes?

Answer 1

Answer:

a

  $\mu = 6$

b

  $x= 3.2$

c

 $P( X> 16 | X > 9 ) = 0.222$

Step-by-step explanation:

Considering question 1

From the question we are told that  

   The waiting times for the train are known to follow a uniform distribution

i.e  

                $X  \~ \ \ \  Uniform(a =0  , b= 12)$

Generally the cumulative distribution function for uniform distribution is mathematically represented as

   $F(x) = \left \{ {{0 \ \ \ \ \ \ \\ \ \ \ for x < a } \atop {\frac{x - a}{b-a} \ \ \ \ \ \ for  a \le x \ le b }} \atop { 1 \ \ \ \ \ \ \ \ \ \ \  for  x >  b}\right.$

Generally the average time is mathematically represented as

       $\mu = \frac{a+ b}{2}$

=>    $\mu = \frac{0+ 12}{2}$

=>    $\mu = 6$

Considering question 2

     The waiting times for the train are known to follow a uniform distribution

i.e  

                $X  \~ \ \ \  Uniform(a =0  , b= 16)$

Generally the 20th percentile of the waiting time is mathematically represented as

               $f(x) = \frac{x- a }{b- a} = 0.2$

=>            $f(x) = \frac{x- 0 }{16- 0} = 0.2$

=>           $x= 3.2$

Considering question 3

        The waiting times for the train are known to follow a uniform distribution

i.e  

                $X  \~ \ \ \  Uniform(a =0  , b= 18)$  

Generally the probability of waiting more than 16 minutes given a person has waited more than 9 minutes

       $P( X> 16 | X > 9 ) = \frac{P( X > 16 \ \ n \ \ X > 9 )}{P( X > 9 )}$

=>     $P( X> 16 | X > 9 ) = \frac{ P(X > 16 )}{P( X > 9 )}$

=>       $P( X> 16 | X > 9 ) = \frac{\frac{18 - 16}{18} }{\frac{18 - 9}{9 } }$

=>       $P( X> 16 | X > 9 ) = 0.222$

Answer 2

The probability shows that the average waiting time is 6 minutes.

How to solve the probability?

The average time will be calculated thus:

= (a + b)/2

= (0 + 12)/2

= 12/2

= 6 minutes

The 20th percentile for the waiting times will be calculated thus:

f(x) = (x - 0)/(16 - 0) = 0.2

x = 0.2 × 16

x = 3.2

Therefore, the 20th percentile for the waiting times is 3.2 minutes.

The probability of waiting more than 16 minutes given a person has waited more than 9 minutes will be calculated thus:

= [(18 - 16)/18] / [(18 - 9)/9]

= 0.222

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