Find expert answers and community insights on IDNLearn.com. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
Sagot :
Let's address the problem step by step:
### a) Constructing the Frequency Distribution
We are given the ticket sales in million dollars of the 15 highest grossing movies of all time. These ticket sales are:
1744, 1601, 1303, 1200, 1200, 1106, 1155, 1151, 983, 978, 939, 921, 918, 914, 900
We need to start with a first class of 877-906 and create an appropriate frequency distribution.
Here is the frequency distribution table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Sales (million \$)} & \text{Number of Movies} \\ \hline \$877-906 & 1 \\ \hline \$907-936 & 3 \\ \hline \$937-966 & 1 \\ \hline \$967-996 & 2 \\ \hline \$997-1026 & 0 \\ \hline \$1027-1056 & 0 \\ \hline \$1057-1086 & 0 \\ \hline \$1087-1116 & 1 \\ \hline \$1117-1146 & 0 \\ \hline \$1147-1176 & 2 \\ \hline \$1177-1206 & 2 \\ \hline \$1207-1236 & 0 \\ \hline \$1237-1266 & 0 \\ \hline \$1267-1296 & 0 \\ \hline \$1297-1326 & 1 \\ \hline \$1327-1356 & 0 \\ \hline \$1357-1386 & 0 \\ \hline \$1387-1416 & 0 \\ \hline \$1417-1446 & 0 \\ \hline \$1447-1476 & 0 \\ \hline \$1477-1506 & 0 \\ \hline \$1507-1536 & 0 \\ \hline \$1537-1566 & 0 \\ \hline \$1567-1596 & 0 \\ \hline \$1597-1626 & 1 \\ \hline \$1627-1656 & 0 \\ \hline \$1657-1686 & 0 \\ \hline \$1687-1716 & 0 \\ \hline \$1717-1746 & 1 \\ \hline \end{array} \][/tex]
### b) Constructing the Histogram
A histogram is a type of bar chart that is used to represent the frequency distribution of a dataset. On the x-axis, we list the class intervals. On the y-axis, we list the frequency of each class. Here's how we can construct it based on our table:
1. Create a label for each class interval on the x-axis: \[tex]$877-906, \$[/tex]907-936, \[tex]$937-966, etc. 2. Set the y-axis to represent frequencies (starting from 0 to the maximum frequency value, which is 3 in this case). 3. Draw bars for each class interval with heights corresponding to their frequency. ### c) Constructing the Frequency Polygon A frequency polygon is a graphical representation of the frequencies where points are plotted for each class interval at the midpoints and then connected with straight lines. Steps to construct the frequency polygon: 1. Calculate the midpoint for each class interval. For the interval \$[/tex]877-906, the midpoint would be [tex]\(\frac{877 + 906}{2} = 891.5\)[/tex].
2. Plot these midpoints on the x-axis.
3. Plot the frequencies on the y-axis.
4. Connect these points with straight lines.
5. Optionally, you can start and end the graph on the x-axis baseline by connecting to the midpoints just before the first class and just after the last class (both with a frequency of 0).
Remember, the visualization part would typically be something you'd construct in a graphing tool or software. But now you should have a clear idea of how both the histogram and frequency polygon should look.
### a) Constructing the Frequency Distribution
We are given the ticket sales in million dollars of the 15 highest grossing movies of all time. These ticket sales are:
1744, 1601, 1303, 1200, 1200, 1106, 1155, 1151, 983, 978, 939, 921, 918, 914, 900
We need to start with a first class of 877-906 and create an appropriate frequency distribution.
Here is the frequency distribution table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Sales (million \$)} & \text{Number of Movies} \\ \hline \$877-906 & 1 \\ \hline \$907-936 & 3 \\ \hline \$937-966 & 1 \\ \hline \$967-996 & 2 \\ \hline \$997-1026 & 0 \\ \hline \$1027-1056 & 0 \\ \hline \$1057-1086 & 0 \\ \hline \$1087-1116 & 1 \\ \hline \$1117-1146 & 0 \\ \hline \$1147-1176 & 2 \\ \hline \$1177-1206 & 2 \\ \hline \$1207-1236 & 0 \\ \hline \$1237-1266 & 0 \\ \hline \$1267-1296 & 0 \\ \hline \$1297-1326 & 1 \\ \hline \$1327-1356 & 0 \\ \hline \$1357-1386 & 0 \\ \hline \$1387-1416 & 0 \\ \hline \$1417-1446 & 0 \\ \hline \$1447-1476 & 0 \\ \hline \$1477-1506 & 0 \\ \hline \$1507-1536 & 0 \\ \hline \$1537-1566 & 0 \\ \hline \$1567-1596 & 0 \\ \hline \$1597-1626 & 1 \\ \hline \$1627-1656 & 0 \\ \hline \$1657-1686 & 0 \\ \hline \$1687-1716 & 0 \\ \hline \$1717-1746 & 1 \\ \hline \end{array} \][/tex]
### b) Constructing the Histogram
A histogram is a type of bar chart that is used to represent the frequency distribution of a dataset. On the x-axis, we list the class intervals. On the y-axis, we list the frequency of each class. Here's how we can construct it based on our table:
1. Create a label for each class interval on the x-axis: \[tex]$877-906, \$[/tex]907-936, \[tex]$937-966, etc. 2. Set the y-axis to represent frequencies (starting from 0 to the maximum frequency value, which is 3 in this case). 3. Draw bars for each class interval with heights corresponding to their frequency. ### c) Constructing the Frequency Polygon A frequency polygon is a graphical representation of the frequencies where points are plotted for each class interval at the midpoints and then connected with straight lines. Steps to construct the frequency polygon: 1. Calculate the midpoint for each class interval. For the interval \$[/tex]877-906, the midpoint would be [tex]\(\frac{877 + 906}{2} = 891.5\)[/tex].
2. Plot these midpoints on the x-axis.
3. Plot the frequencies on the y-axis.
4. Connect these points with straight lines.
5. Optionally, you can start and end the graph on the x-axis baseline by connecting to the midpoints just before the first class and just after the last class (both with a frequency of 0).
Remember, the visualization part would typically be something you'd construct in a graphing tool or software. But now you should have a clear idea of how both the histogram and frequency polygon should look.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.
Saturn Has A Diameter Of 74,898 Miles. What Is Its Diameter In Kilometers? (1
mile = 1.6 Kilometers)
