This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
To construct a continuous series from mid-values, we first need to determine the class interval width and then calculate the lower and upper limits for each class. Here is the given data: |c|c|c|c|c|c|c| Mid values & 15 & 25 & 35 & 45 & 55 & 65 \\ Frequency & 20 & 19 & 17 & 62 & 41 & 73 \\ Step 1: Calculate the class interval width (i). The class interval width is the difference between any two consecutive mid-values. i = 25 - 15 = 10 So, the class interval width is 10. Step 2: Calculate the lower limit (L) and upper limit (U) for each class. The lower limit (L) and upper limit (U) for a class can be found using the mid-value (M) and half of the class interval width ((i)/(2)). Here, (i)/(2) = (10)/(2) = 5. The formulas are: L = M - (i)/(2) U = M + (i)/(2) For each mid-value: • For M = 15: L = 15 - 5 = 10 U = 15 + 5 = 20 Class interval: 10 - 20 • For M = 25: L = 25 - 5 = 20 U = 25 + 5 = 30 Class interval: 20 - 30 • For M = 35: L = 35 - 5 = 30 U = 35 + 5 = 40 Class interval: 30 - 40 • For M = 45: L = 45 - 5 = 40 U = 45 + 5 = 50 Class interval: 40 - 50 • For M = 55: L = 55 - 5 = 50 U = 55 + 5 = 60 Class interval: 50 - 60 • For M = 65: L = 65 - 5 = 60 U = 65 + 5 = 70 Class interval: 60 - 70 Step 3: Construct the continuous series. Combine the calculated class intervals with their corresponding frequencies. |c|c| Class Interval & Frequency \\ 10 - 20 & 20 \\ 20 - 30 & 19 \\ 30 - 40 & 17 \\ 40 - 50 & 62 \\ 50 - 60 & 41 \\ 60 - 70 & 73 \\ That's 2 down. 3 left today — send the next one.