Question

A bus comes by every 9 minutes. The times from when a person arives at the busstop until the busarrives follows a Uniform distribution from 0 to 9 minutes. A person arrives at the bus stop at arandomly selected time. Round to 4 decimal places where possible.a. The mean of this distribution isb. The standard deviation isc The probability that the person will wait more than 6 minutes isd. Suppose that the person has already been waiting for 2.4 minutes. Find the probability thatthe person's total waiting time will be between 4.8 and 6.8 minutes.e. 37% of all customers wait at least how long for the train?minutes

Answer 1

a. The mean of this distribution is 4.5 minutes. b. The standard deviation is approximately 2.5981 minutes. c. The probability of waiting more than 6 minutes is 0.6667. d. The probability of the total waiting time being between 4.8 and 6.8 minutes is approximately 0.2222. e. 37% of all customers wait at least 3.33 minutes for the bus.

Step-by-step explanation:

a. The mean of a uniform distribution is the average of the minimum and maximum values. In this case, the minimum value is 0 minutes and the maximum value is 9 minutes. Therefore, the mean is (0 + 9) / 2 = 4.5 minutes.

b. The standard deviation of a uniform distribution can be found using the formula (max - min) / sqrt(12), where max is the maximum value and min is the minimum value. In this case, the standard deviation is (9 - 0) / sqrt(12) ≈ 2.5981 minutes.

c. To find the probability that the person will wait more than 6 minutes, we subtract the probability of waiting less than or equal to 6 minutes from 1. Since the distribution is uniform, this probability is equal to (6 - 0) / (9 - 0) = 6 / 9 = 2 / 3 = 0.6667.

d. Since the person has already been waiting for 2.4 minutes, the remaining waiting time can be modeled as a uniform distribution from 0 to (9 - 2.4) = 6.6 minutes. Therefore, we can find the probability of the total waiting time being between 4.8 and 6.8 minutes by subtracting the probability of waiting less than 4.8 minutes from the probability of waiting less than 6.8 minutes. This probability is equal to ((6.8 - 0) - (4.8 - 0)) / (9 - 0) = 2 / 9 ≈ 0.2222.

e. To find the time at which 37% of all customers wait, we can use the inverse cumulative distribution function (CDF) of the uniform distribution. The CDF of the uniform distribution is given by (x - min) / (max - min), where x is the time and min and max are the minimum and maximum values of the distribution. By setting this equal to 0.37 and solving for x, we find that x = 0.37 * (9 - 0) + 0 = 3.33 minutes.

Answer 2

a. The mean of this distribution

Mean of a uniform distribution is given as

`\begin{gathered} (b+a)/(2) \\ =(9+0)/(2) \\ =(9)/(2) \\ =4.5 \end{gathered}`

Hence, the mean is 4.5

b. The standard deviation

This is given as

`\begin{gathered} \frac{b-a}{\sqrt[]{12}} \\ =\frac{9-0}{\sqrt[]{12}} \\ =\frac{9}{\sqrt[]{12}} \\ =2.5981 \end{gathered}`

Hence the standard deviation is 2.5981

c. The probability that the person will wait more than 6 minutes

This is

`\begin{gathered} P(x>6)=(9-6)/(9-0) \\ =(3)/(9) \\ =0.3333 \end{gathered}`

Hence, the answer is 0.3333

d. Suppose that the person has already been waiting for 2.4 minutes. Find the probability that the person's total waiting time will be between 4.8 and 6.8 minutes.

This can be written as

`P(4.82.4)`

From here,

`f(x)=(1)/(9-2.4)=(1)/(6.6)`

So,

`\begin{gathered} P(4.82.4)=(6.8-4.8)*(1)/(6.6) \\ =2*(1)/(6.6) \\ =0.3030 \end{gathered}`

Hence the answer is 0.3030

e. 37% of all customers wait at least how long for the train?

This becomes

100 - 37 = 63

`k((1)/(9))=0.63`

Then we find k

`\begin{gathered} k((1)/(9))=0.63 \\ k=(0.63)/((1)/(9)) \\ k=5.67\text{ minutes } \end{gathered}`

Hence, the answer is 5.67 minutes