Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. P(x=3)=? P(1&lt;x&lt;4)=? P(x≥8)=? Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. RF(x=3)=? RF(1&lt;x&lt;4)=? RF(x≥8)=? Discussion Questions 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical, empirical, and simulation distributions. Use complete sentences.

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Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places. P(x=3)=? P(1&amp;lt;x&amp;lt;4)=? P(x≥8)=? Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places. RF(x=3)=? RF(1&amp;lt;x&amp;lt;4)=? RF(x≥8)=? Discussion Questions 1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical, empirical, and simulation distributions. Use complete sentences.

falhiblesw · Accepted Answer

Theoretical probabilities: From the given table of theoretical distribution values, P(X=3)=0.2503. Now, P(1&amp;lt;X&amp;lt;4) =P(2≤X≤3) =P(X=2)+P(X=3) =0.2816+0.2503 =0.5319. Thus,P(1&amp;lt;X&amp;lt;4)=0.5319. Again, P(X≥8) =P(X=8)+P(X=9)+P(X=10) =0.0004+0.00003+0.000001 =0.000431. Thus, the P(X≥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)≈0.25. Now, RF(1&amp;lt;X&amp;lt;4) =RF(2≤X≤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)≈0.28. As a result, RF(1&amp;lt;X&amp;lt;4) ≈0.28+0.25 =0.53. Thus,RF(1&amp;lt;X&amp;lt;4)≈0.53. Again, RF(X≥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≥8)≈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&amp;lt;X&amp;lt;4)=0.2503 is approximately equal to the corresponding relative frequency likely to be from the simulation distribution, RF(1&amp;lt;X&amp;lt;4)=0.25, when rounded to two decimal places. The probability, P(X≥8)=0.000431 is approximately equal to the relative frequency likely to be from the simulation distribution, RF(X≥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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