Question 142919: In this problem set we are asked to answer a series of 5 questions that pertain to the probability that the local hardware/sporting goods store in Dove Creek, Montana will gross over $850.00. In the space below we will discuss each question in detail. Before we can begin our exploration we must first answer some key questions that will help us to reach the proper conclusions. State what makes the trial, (which in our study is days the store grossed over $850.00), what is a success, (which in our study will be days the store grossed more than $850.00), what is a failure, (which in our study is days when the store did not make more than $850.00), and finally to give the values of the following:
n = will vary for each set but will be the following number of Business days for each:
n = 5 business days for 8a
n = 10 business days for 8b and 8c
n = 20 business days for 8d and 8e
p = .60%
q = 0.40%, because q = 1 – p or q = 1 – 0.60% = 0.40%
Armed with this we are then going to answer the 5 main questions as follows:
8a at least 3 out of 5 business days
To solve for this we will use the inequality found on page 224 as follows:
P (r ≥ 3)
Let’s begin by using the technique found on page 222 part b as follows:
P (r ≥ 3) = P (r = 3or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r ≥ 3). Again I will use a calculator and write the steps down for you:
P (r ≥ 3) = P (.346 + .259 + .078)
P (r ≥ 3) = .346 + .259 + .078 =
P (r ≥ 3) = 0.683 % chance that at least 3 out of 5 business days grossed over $850.00.
8b at least 6 out of 10 business days
To solve for this we will use the inequality found on page 224 as follows:
P (r ≥ 6)
Let’s begin by using the technique found on page 222 part b as follows:
P (r ≥ 6) = P (r = 6 or r = 7 or r = 8 or r = 9 or r = 10)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r ≥ 6). Again I will use a calculator and write the steps down for you:
P (r ≥ 6) = P (.251 + .215 + .121 + .040 + .006)
P (r ≥ 6) = .251 + .215 + .121 + .040 + .006 =
P (r ≥ 6) = 0.633 % chance that at least 6 out of 10 business days grossed over $850.00.
8c fewer than 5 out of 10 business days
To solve for this we will use the inequality found on page 224 as follows:
P (r < 5)
Let’s begin by using the technique found on page 222 part b as follows:
P (r < 5) = P (r = 0 or r = 1 or r = 2 or r = 3 or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r < 5). Again I will use a calculator and write the steps down for you:
P (r < 5) = P (.000 + .002 + .011 + .042 + .111 + .201)
P (r < 5) = .000 + .002 + .011 + .042 + .111 + .201 =
P (r < 5) = 0.367 % chance that fewer than 5 out of 10 business days grossed over $850.00.
8d fewer than 6 out of the next 20 business days. If this actually happened, might it shake you confidence in the statement p = 0.60? Might it make you suspect that p is less than 0.60? Explain.
8e more than 17 of the next 20 business days. If this actually happened, might you suspect that p is greater than 0.60? Explain.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 8d fewer than 6 out of the next 20 business days.
P(r < 6) = 0.0016
---------
If this actually happened, might it shake you confidence in the statement p = 0.60? Might it make you suspect that p is less than 0.60? Explain.
The chances of it happening is very close to 1 in 1000 which is pretty low.
You might begin to suspect that p is not 60%
--------------------
8e more than 17 of the next 20 business days. If this actually happened, might you suspect that p is greater than 0.60? Explain.
P(r > 17) = 0.0036...
If p = 0.6, the probability is close to 3 in 1000 that would be more than
17 in the next 20 business days.
=========================
Cheers,
san H.
|
|
|
