Theoretical probabilities: From the given table of theoretical distribution values, \(P (X = 3) = 0.2503.\) Now, \(P (1 < X < 4)\)
\(= P (2 \leq X \leq 3)\)
\(= P (X = 2) + P (X = 3)\)
\(= 0.2816 + 0.2503\)
\(= 0.5319.\)
\(\text{Thus}, P (1 < X < 4) = 0.5319.\) Again, \(P (X \geq 8)\)
\(= P (X = 8) + P (X = 9) + P (X = 10)\)
\(= 0.0004 + 0.00003 + 0.000001\)
\(= 0.000431.\) Thus, the \(P (X \geq 8) = 0.000431.\) Probabilities from Organize the Data section: The probability (P) can be defined as the relative frequency (RF) in the long run. From the given image of the histogram, it appears that the length of the bar corresponding to the horizontal axis value 3, is approximately 0.25 on the vertical axis. Thus, \(RF (X = 3) \approx 0.25.\) Now, \(RF (1 < X < 4)\)
\(= RF (2 \leq X \leq 3)\)
\(= RF (X = 2) + RF (X = 3).\) From the histogram, the length of the bar corresponding to the horizontal axis value 3 is somewhere between vertical axis values 0.25 and 0.3, being slightly closer to 0.3. Thus, it can be said that, \(RF (X = 2) \approx 0.28.\) As a result, \(RF (1 < X < 4)\)
\(\approx 0.28 + 0.25\)
\(= 0.53.\)
\(\text{Thus}, RF (1 < X < 4) \approx 0.53.\) Again, \(RF (X \geq 8)\)
\(= RF (X = 8) + RF (X = 9) + RF (X = 10).\) From the histogram, it appears that the length of the bar for each of the horizontal axis values 8, 9, and 10 is 0. Thus, the \(RF (X \geq 8) \approx 0.\) Discussion Question 1. The probability, \(P (X = 3) = 0.5319\) is approximately equal to the relative frequency likely to be from the simulation distribution, \(RF (X = 3) = 0.53\), when rounded to two decimal places. The probability, \(P (1 < X < 4) = 0.2503\) is approximately equal to the corresponding relative frequency likely to be from the simulation distribution, \(RF (1 < X < 4) = 0.25\), when rounded to two decimal places. The probability, \(P (X \geq 8) = 0.000431\) is approximately equal to the relative frequency likely to be from the simulation distribution, \(RF (X \geq 8) = 0\), when rounded to three decimal places. These are some of the similarities between the graph of likely to be the simulation distribution, and the theoretical distribution.
Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.
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The following table represents the Frequency Distribution and Cumulative Distributions for this data set: 12, 13, 17, 18, 18, 24, 26, 27, 27, 30, 30, 35, 37, 41, 42, 43, 44, 46, 53, 58
\(\begin{array}{|c|c|} \hline \text{Class}&\text{Frequency}&\text{Relative Frequency}&\text{Cumulative Frequency}\\ \hline \text{10 but les than 20}&5\\ \hline \text{20 but les than 30}&4\\ \hline \text{30 but les than 40}&4\\ \hline \text{40 but les than 50}&5\\ \hline \text{50 but les than 60}&2\\ \hline \text{TOTAL}\\ \hline \end{array}\)
What is the Relative Frequency for the class: 20 but less than 30? State you answer as a value with exactly two digits after the decimal. for example 0.30 or 0.35
This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function
\(\displaystyle{p}{\left({x}\right)}={\frac{{{1}}}{{\sqrt{{{2}\pi^{\sigma}}}}}}{e}-\frac{{\left({x}-\mu\right)}^{{2}}}{{{2}\sigma^{{2}}}}\)
where \(\pi\) = 3,14159265 ... and \(\sigma\) and \(\mu\) are constants called the standard deviation and the mean, respectively. Its graph (for \(\sigma=1\) and \(\mu=2)\) is shown in the figure.
With \(\displaystyle\sigma={\color{red}{{5}}}\) and \(\mu=0\), approximate \(\displaystyle{\int_{{{0}}}^{{+\infty}}}{p}{\left({x}\right)}{\left.{d}{x}\right.}\).(Round your answer to four decimal places.)
Presenting data in the form of table. For the data set shown by the table, Solve,
a) Create a scatter plot for the data.
b) Use the scatter plot to determine whether an exponential function, a logarithmic function, or a linear function is the best choice for modeling the data. (If applicable, you will use your graphing utility to obtain these functions.)
\(\begin{array}{|c|c|}\hline \text{Intensity (wattd per}\ meter^{2}) & \text{Loudness Level (decibels)} \\ \hline 0.1\text{(loud thunder)} & 110 \\ \hline 1\text{(rock concert, 2 yd from speakers)} & 120 \\ \hline 10 \text{(jackhammer)} & 130 \\ \hline 100 \text{(jet take off, 40 yd away)} & 140 \\ \hline \end{array}\)
The following table shows the average yearly tuition and required fees, in thousand of dollars, charged by a certain private university in the school year beginning in the given year.
\(\begin{array}{|c|c|}\hline \text{Year} & \text{Average tuition} \\ \hline 2005 & $17.6 \\ \hline 2007 & $18.1 \\ \hline 2009 & $19.5 \\ \hline 2011 & $20.7 \\ \hline 2013 & $21.8 \\ \hline \end{array}\)
What prediction does the formula modeling this data give for average yearly tuition and required fees for the university for the academic year beginning in 2019?
Let's say the widget maker has developed the following table that shows the highest dollar price p. widget where you can sell N widgets. Number N Price p \(200 53.00\)
\(250 52.50\)
\(300 52.00\)
\(35051.50\)
(a) Find a formula for pin terms of N modeling the data in the table.
(b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in month as a function of the number N of widgets produced in a month. \(R=\) Is Ra linear function of N?
(c) On the basis of the tables in this exercise and using cost, \(C= 35N + 900\), use a formula to express the monthly profit P, in dollars, of this manufacturer asa function of the number of widgets produced in a month \(p=\) ?
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a) can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. [Note: A normal probability plot of data exhibits a reasonably linear pattern.]
b) Calculate and interpret a confidence interval with a 99% confidence level for true average lifetime. [Hint: mean \(= 1191.6, s = 506.6\).]
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5). \(\begin{array}{|c|c|}& P(x)\\ Left & 0.636 \\ Right & 0.304 \\ No\ preference & 0.060\end{array}\)
