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pav-90 [236]
3 years ago
5

Think about the many statistical scores or rankings you receive in school, such as GPA, SAT and ACT scores, test scores, and cla

ss ranking. Describe the usefulness and limitations of these pieces of data in defining who you are as a person or as a student. In what ways do they help give a clear picture? What are they not conveying?
Mathematics
1 answer:
frutty [35]3 years ago
6 0
The usefulness is that this data allows you to understand how well students perform with their instruction, based on a standardized test. The limitation is that the scores don't account for different situations that may affect the results you receive.
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Answer:

5.

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Step-by-step explanation:

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A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon
melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.  

X~ Uniform(0,100)

Then the probability mass function is given as follows.

f(x) = P(X=x) = 1/100  \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0  \,\,\,\, \text{otherwise}

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located  70 meters away from city A then the mid point between 20 and 70  is (70+20)/2 = 45 then we can represent D as follows

D(x) =\left\{ \begin{array}{ll}  x  & \mbox{if } 0\leq x \leq 20 \\  x-20 & \mbox{if } 20\leq x < 45\\                70-x & \mbox{if } 45 \leq x \leq 70\\                x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable Y = D(X), Y is a random variable as well, remember that there is a theorem that says that

E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx

Where f(x) is the probability mass function of X. Using the information of our problem

E[Y] = \int\limits_{-\infty}^{\infty}  D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx  \bigg]\\= \frac{51}{4} = 12.75

3 0
3 years ago
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