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Exercise 2. A service facility charges a $20 fixed fee plus $25 per hour of service up to 6 hours, and no additional fee is charged for service exceeding 6 hours. Suppose that the service time $\tau$ is equally likely to be any number of hours in {1,2,3,4,..., 10} hours (Assume that it takes some integer number of hours). Let X represent the cost of service in the facility. (a) Find and sketch the probability mass function for X. (b) Find and sketch the cumulative distribution function for X. (c) What is the probability that you end up paying less than or equal to $70 for service? Answer this part two different ways: • Using your answer to part (b). • Finding the time (call it $\tau_0$) at which the service would cost exactly $70 and then finding the probability that $\tau \le \tau_0$. (d) What is the expected value (or mean) of X? (e) The repair shop calls after 4.5 hours and tells you that your car is not ready yet; hence, this is going to cost you at least $145. Given this information, come up with a new probability mass function for how much it will cost. ```

          Exercise 2. A service facility charges a $20 fixed fee plus $25 per hour of service up to 6 hours, and no
additional fee is charged for service exceeding 6 hours. Suppose that the service time $\tau$ is equally
likely to be any number of hours in {1,2,3,4,..., 10} hours (Assume that it takes some integer
number of hours). Let X represent the cost of service in the facility.
(a) Find and sketch the probability mass function for X.
(b) Find and sketch the cumulative distribution function for X.
(c) What is the probability that you end up paying less than or equal to $70 for service? Answer this
part two different ways:
• Using your answer to part (b).
• Finding the time (call it $\tau_0$) at which the service would cost exactly $70 and then finding the
probability that $\tau \le \tau_0$.
(d) What is the expected value (or mean) of X?
(e) The repair shop calls after 4.5 hours and tells you that your car is not ready yet; hence, this is going
to cost you at least $145. Given this information, come up with a new probability mass function for
how much it will cost.
```

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exercise 2 a service facility charges a 20 fixed fee plus 25 per hour of service up to 6 hours and no additional fee is charged for service exceeding 6 hours suppose that the service time ta 21387

Added by Martin F.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Andrew Kim

11/09/2019

Step 1

The cost of service X is given by: $X = 20 + 25\tau$ if $\tau \le 6$ $X = 20 + 25(6) = 170$ if $\tau > 6$ Since $\tau$ can take values from 1 to 10, we have: If $\tau = 1$, $X = 20 + 25(1) = 45$ If $\tau = 2$, $X = 20 + 25(2) = 70$ If $\tau = 3$, $X = 20 + 25(3) = Show more…

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Exercise 2. A service facility charges a $20 fixed fee plus $25 per hour of service up to 6 hours, and no additional fee is charged for service exceeding 6 hours. Suppose that the service time $\tau$ is equally likely to be any number of hours in {1,2,3,4,..., 10} hours (Assume that it takes some integer number of hours). Let X represent the cost of service in the facility. (a) Find and sketch the probability mass function for X. (b) Find and sketch the cumulative distribution function for X. (c) What is the probability that you end up paying less than or equal to $70 for service? Answer this part two different ways: • Using your answer to part (b). • Finding the time (call it $\tau_0$) at which the service would cost exactly $70 and then finding the probability that $\tau \le \tau_0$. (d) What is the expected value (or mean) of X? (e) The repair shop calls after 4.5 hours and tells you that your car is not ready yet; hence, this is going to cost you at least $145. Given this information, come up with a new probability mass function for how much it will cost. ```

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Transcript

00:01 All right, so we're given some information about a service company taking in service calls.

00:08 And we're given that each call can take one, two, three, or four hours, and we're also given that the different types of malfunctions occur about the same frequency.

00:20 So basically, it's equal probability that the call will take one, two, or three, or four hours.

00:26 So first up, for part a, let's develop a probability distribution.

00:30 So let x equal the number of hours.

00:34 Oops, that's not how you spell.

00:39 There we go.

00:42 Number of hours a call takes.

00:52 So x can take the values 1, 2, 3, or 4.

00:58 And the probability distribution, since we're given that the types of malfunctions that correspond to a 1 -hour call, 2 -hour call, 3 -hour call, or 4 -hour call, all happen approximately the same distribution.

01:09 Then each of these probabilities will be equal, so we get 0 .25 for all four of these.

01:20 Not my best penmanship, but you get the general idea.

01:24 For part b, we're asked to draw a graph of this.

01:26 So let's draw our axes.

01:32 This is x.

01:33 This is p of x.

01:36 And let's just take this off at 1, 2, 3, 4.

01:45 And then right here, 0 .25.

01:50 And you'll notice that i'm going to use green for this.

01:56 All these have the same distribution of 0 .25.

02:00 So the distribution will look like that, except the lines are supposed to be straight and supposed to hit the ticks.

02:14 But such as the limitations of drawing using a computer...

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