document.write( "Question 1184005: Use the given frequency distribution to find the
\n" ); document.write( "Class width
\n" ); document.write( "Class midpoints
\n" ); document.write( "Class boundaries
\n" ); document.write( "Temperature 35-39,40-44,45-49,50-54,55-59,60-64,65-69
\n" ); document.write( "Frequency 1,3,5,11,7,7,7,1 \n" ); document.write( "
Algebra.Com's Answer #849887 by CPhill(1959)\"\" \"About 
You can put this solution on YOUR website!
Here's the breakdown of how to find the class width, midpoints, and boundaries for the given frequency distribution:\r
\n" ); document.write( "\n" ); document.write( "**1. Class Width:**\r
\n" ); document.write( "\n" ); document.write( "The class width is the difference between the upper and lower limits of consecutive class intervals. It should be consistent across all classes. In your example:\r
\n" ); document.write( "\n" ); document.write( "* 40 - 35 = 5
\n" ); document.write( "* 45 - 40 = 5
\n" ); document.write( "* And so on...\r
\n" ); document.write( "\n" ); document.write( "Therefore, the class width is **5**.\r
\n" ); document.write( "\n" ); document.write( "**2. Class Midpoints:**\r
\n" ); document.write( "\n" ); document.write( "The class midpoint is the average of the upper and lower limits of each class interval.\r
\n" ); document.write( "\n" ); document.write( "* 35-39: (35 + 39) / 2 = **37**
\n" ); document.write( "* 40-44: (40 + 44) / 2 = **42**
\n" ); document.write( "* 45-49: (45 + 49) / 2 = **47**
\n" ); document.write( "* 50-54: (50 + 54) / 2 = **52**
\n" ); document.write( "* 55-59: (55 + 59) / 2 = **57**
\n" ); document.write( "* 60-64: (60 + 64) / 2 = **62**
\n" ); document.write( "* 65-69: (65 + 69) / 2 = **67**\r
\n" ); document.write( "\n" ); document.write( "**3. Class Boundaries:**\r
\n" ); document.write( "\n" ); document.write( "Class boundaries are the values that separate the classes. They are halfway between the upper limit of one class and the lower limit of the next class.\r
\n" ); document.write( "\n" ); document.write( "* The lower boundary of the first class is typically found by subtracting half the class width from the lowest limit of the first class: 35 - (5/2) = 32.5. However, since your data starts at 35, we can use 34.5 to ensure all data points are included.
\n" ); document.write( "* The upper boundary of the last class is found by adding half the class width to the highest limit of the last class: 69 + (5/2) = 71.5. However, since your data ends at 69, we can use 69.5 to ensure all data points are included.\r
\n" ); document.write( "\n" ); document.write( "Here's a complete table summarizing the results:\r
\n" ); document.write( "\n" ); document.write( "| Temperature | Frequency | Class Midpoint | Class Boundaries |
\n" ); document.write( "|---|---|---|---|
\n" ); document.write( "| 35-39 | 1 | 37 | 34.5 - 39.5 |
\n" ); document.write( "| 40-44 | 3 | 42 | 39.5 - 44.5 |
\n" ); document.write( "| 45-49 | 5 | 47 | 44.5 - 49.5 |
\n" ); document.write( "| 50-54 | 11 | 52 | 49.5 - 54.5 |
\n" ); document.write( "| 55-59 | 7 | 57 | 54.5 - 59.5 |
\n" ); document.write( "| 60-64 | 7 | 62 | 59.5 - 64.5 |
\n" ); document.write( "| 65-69 | 7 | 67 | 64.5 - 69.5 |
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