(Mixing Uniform and Binomial Distributions) A person arrives at a certain bus stop each morning. The waiting time, in minutes, for a bus to arrive is uniformly distributed on the interval (0, 15). (a) [1 pts] Find the mean waiting time. Show all work. (b) [1 pts] Find the standard deviation of the waiting times. Show all work. (c) [4 pts] Find the probability that the waiting time is between 5 and 11 minutes. Show all work. (d) [6 pts] Suppose that waiting times on different mornings are independent. What is the probability that the waiting time is less than 5 minutes on exactly 4 of 10 mornings

Answers

Answer 1

Answer:

a

[tex]\mu_x  =7.5 \ minutes [/tex]

b

[tex]\sigma_{x} = 4.33 \  minutes [/tex]

c

[tex]P(5 <  X  <  11 ) = 0.4 [/tex]

d

[tex]P(W = 4 ) =  0.228[/tex]

Step-by-step explanation:

From the question we are told that

  The  interval is [tex](w , y) =  (0 , 15)[/tex]

Considering question a

    Generally the mean waiting time is mathematically represented as

              [tex]\mu_x  = \frac{w + y}{2}[/tex]

=>          [tex]\mu_x  = \frac{0 + 15}{2}[/tex]

=>          [tex]\mu_x  =7.5 \ minutes [/tex]

Considering question b

Generally the standard deviation is mathematically represented as

      [tex]\sigma_{x} = \frac{y - w}{\sqrt{\sqrt{12} } }[/tex]

=>    [tex]\sigma_{x} = \frac{15 - 0}{\sqrt{\sqrt{12} } }[/tex]

=>    [tex]\sigma_{x} = 4.33 \  minutes [/tex]

Considering question c

Generally the probability density function for a random variable  X is  

     [tex]f(x) =  \left \{ {{ \frac{1}{y} \ \ \ \ \  0 <  X  <  y } \atop {0} \\ \ \ \  Otherwise} \right.[/tex]

=>       [tex]f(x) =  \left \{ {{ \frac{1}{15} \ \ \ \ \  0 <  X  <  15 } \atop {0} \\ \ \ \  Otherwise} \right.[/tex]

Generally the probability that the waiting time is between 5 and 11 minutes is mathematically represented as

     [tex]P(5 <  X  <  11 )= \int\limits^{11}_{5} {f(x)} \, dx[/tex]

=>    [tex]P(5 <  X  <  11 )= \int\limits^{11}_{5} {\frac{1}{15}} \, dx[/tex]

=>   [tex]P(5 <  X  <  11 ) =  \frac{1}{15}  x | \left \  11 } \atop {5}} \right.[/tex]

=>   [tex]P(5 <  X  <  11 ) =  \frac{1}{15} [ 11 - 5][/tex]

=>   [tex]P(5 <  X  <  11 ) = 0.4 [/tex]

Considering question d

      Generally the probability the  waiting time is less than 5 minutes is mathematically represented as

     [tex]P(X < 5) =  \int\limits^5_{-\infty} {\frac{1}{15} } \, dx[/tex]

Given that time cannot be negetive

     [tex]P(X < 5) =  \int\limits^5_{0} {\frac{1}{15} } \, dx[/tex]

    [tex]P(X < 5) = \frac{1}{15} *  x  | \left \  5 } \atop {0}} \right.[/tex]

    [tex]P(X < 5) =0.33.[/tex]  

Generally the probability that the waiting time is less than 5 minutes on exactly 4 of 10 mornings follows a binomial  distribution because  the number of trial is finite i.e 10  the probability of success is probability that the waiting time is less than 5 minutes and probability of failure is probability that the waiting time is not  less than 5 minutes(i.e there is  only two outcome )

Hence for a random variable K representing the number of times in the total of 10 mornings that the  waiting is less than 5 minutes then

    The probability the random variable(W) is exactly 4 is mathematically represented as

    [tex]P(W = 4 ) =  ^{10}C_4 * (0.33)^4 *  (1- p)^{10 -4}[/tex]

Here C denotes combination

So

     [tex]P(W = 4 ) =  0.228[/tex]

   


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======================================================

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