A) The cost to send a package that weighs 3.2 pounds is $4.13. Since this weight exceeds 3 pounds but remains below 4 pounds, we have to refer to the pricing that applies to 4-pound packages (see the attached document for pricing details).
b) To illustrate the Media Mail shipping costs based on the weight of the books, a line graph is appropriate. In this graph, the weight in pounds is represented on the x-axis and the shipping costs on the y-axis.
c) The graph depicting the Media Mail shipping costs as a function of book weight will be represented by the equation: f(x) = 2.69 + 0.48(x-1)
Answer:
The chance of completing the entire package installation in under 12 minutes is 0.1271.
Step-by-step explanation:
We define X as a normal distribution representing the time taken in seconds to install the software. According to the Central Limit Theorem, X is approximately normal, where the mean is 15 and variance is 15, giving a standard deviation of √15 = 3.873.
To find the probability of the total installation lasting less than 12 minutes, which equals 720 seconds, each installation should average under 720/68 = 10.5882 seconds. Thus, we seek the probability that X is less than 10.5882. To do this, we will apply W, the standard deviation value of X, calculated via the formula provided.
Utilizing 
, we reference the cumulative distribution function of the standard normal variable W, with values found in the attached file.
Given the symmetry of the standard normal distribution density function, we ascertain 
.
Consequently, the probability that the installation process for the entire package is completed within 12 minutes is 0.1271.
Answer:
There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.
Step-by-step explanation:
Problems dealing with normally distributed samples can be addressed using the z-score formula.
For a set with the mean 
and a standard deviation 
, the z-score for a measure X is calculated by
Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.
For this case
Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus 
What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?
That translates to 
Thus
has a p-value of 0.7549.
<pthis indicates="" that="" src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%2050%29%20%3D%200.7549" id="TexFormula10" title="P(X \leq 50) = 0.7549" alt="P(X \leq 50) = 0.7549" align="absmiddle" class="latex-formula">.
Additionally, we have that
There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.
</pthis>
Short answer: 195
First, you multiply 65 by 3, yielding 195 miles.
Then, multiplying 55 by 2 gives you 110 miles.
By summing both distances, you arrive at 305.
From 500, if you subtract 305, you find =195.
