Answer :
To address the problem thoroughly, we will provide a detailed solution for each part.
(a) Construct a cumulative frequency distribution.
First, we organize the data into bins specified as intervals:
- \( 1000-1999 \)
- \( 2000-2999 \)
- \( 3000-3999 \)
- \( 4000-4999 \)
Using the given data, we count the number of observations that fall into each interval and then create the cumulative frequency for each interval.
[tex]\[ \begin{aligned} & \text{Bin } 1000-1999 \text{ has 4 observations (1260, 1913, 1050, 1753).} \\ & \text{Cumulative Frequency: } 4 \\ & \text{Bin } 2000-2999 \text{ has 6 observations (2400, 2500, 2200, 2968, 2798, 2651).} \\ & \text{Cumulative Frequency: } 4 + 6 = 10 \\ & \text{Bin } 3000-3999 \text{ has 2 observations (3110, 3640).} \\ & \text{Cumulative Frequency: } 10 + 2 = 12 \\ & \text{Bin } 4000-4999 \text{ has 4 observations (4247, 4800, 4752, 4400).} \\ & \text{Cumulative Frequency: } 12 + 4 = 16 \\ \end{aligned} \][/tex]
Completing the table we get:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Students in a District} & \text{Cumulative Frequency} \\ \hline 1000-1999 & 4 \\ \hline 2000-2999 & 10 \\ \hline 3000-3999 & 12 \\ \hline 4000-4999 & 16 \\ \hline \end{tabular} \][/tex]
(b) Construct a cumulative relative frequency distribution.
The total number of observations is 16. We calculate the relative frequency by dividing each cumulative frequency by the total number of observations.
[tex]\[ \begin{aligned} & \text{For bin } 1000-1999: \frac{4}{16} = 0.25 \\ & \text{For bin } 2000-2999: \frac{10}{16} = 0.625 \\ & \text{For bin } 3000-3999: \frac{12}{16} = 0.75 \\ & \text{For bin } 4000-4999: \frac{16}{16} = 1.0 \\ \end{aligned} \][/tex]
So the table of cumulative relative frequency distribution is:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Students in a District} & \text{Cumulative Relative Frequency} \\ \hline 1000-1999 & 0.25 \\ \hline 2000-2999 & 0.625 \\ \hline 3000-3999 & 0.75 \\ \hline 4000-4999 & 1.0 \\ \hline \end{tabular} \][/tex]
(c) Draw a frequency polygon.
The frequency polygon is drawn using the midpoints of each bin. The midpoints are calculated as follows:
[tex]\[ \begin{aligned} & \text{Midpoint of } 1000-1999: \frac{1000 + 1999}{2} = 1500 \\ & \text{Midpoint of } 2000-2999: \frac{2000 + 2999}{2} = 2500 \\ & \text{Midpoint of } 3000-3999: \frac{3000 + 3999}{2} = 3500 \\ & \text{Midpoint of } 4000-4999: \frac{4000 + 4999}{2} = 4500 \\ \end{aligned} \][/tex]
The x-coordinates for the frequency polygon are \([1000, 1500, 2500, 3500, 4500, 5000]\) and the corresponding y-coordinates (frequencies at each midpoint and 0 at each end) are \([0, 4, 6, 2, 4, 0]\).
(d) Draw a relative frequency ogive.
An ogive is a plot of cumulative relative frequencies against the upper boundary of each bin.
The x-coordinates for the relative frequency ogive are \([1500, 2500, 3500, 4500]\) and the y-coordinates (cumulative relative frequencies) are \([0.25, 0.625, 0.75, 1.0]\).
By following these steps, we achieve each part of the problem systematically and accurately.
(a) Construct a cumulative frequency distribution.
First, we organize the data into bins specified as intervals:
- \( 1000-1999 \)
- \( 2000-2999 \)
- \( 3000-3999 \)
- \( 4000-4999 \)
Using the given data, we count the number of observations that fall into each interval and then create the cumulative frequency for each interval.
[tex]\[ \begin{aligned} & \text{Bin } 1000-1999 \text{ has 4 observations (1260, 1913, 1050, 1753).} \\ & \text{Cumulative Frequency: } 4 \\ & \text{Bin } 2000-2999 \text{ has 6 observations (2400, 2500, 2200, 2968, 2798, 2651).} \\ & \text{Cumulative Frequency: } 4 + 6 = 10 \\ & \text{Bin } 3000-3999 \text{ has 2 observations (3110, 3640).} \\ & \text{Cumulative Frequency: } 10 + 2 = 12 \\ & \text{Bin } 4000-4999 \text{ has 4 observations (4247, 4800, 4752, 4400).} \\ & \text{Cumulative Frequency: } 12 + 4 = 16 \\ \end{aligned} \][/tex]
Completing the table we get:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Students in a District} & \text{Cumulative Frequency} \\ \hline 1000-1999 & 4 \\ \hline 2000-2999 & 10 \\ \hline 3000-3999 & 12 \\ \hline 4000-4999 & 16 \\ \hline \end{tabular} \][/tex]
(b) Construct a cumulative relative frequency distribution.
The total number of observations is 16. We calculate the relative frequency by dividing each cumulative frequency by the total number of observations.
[tex]\[ \begin{aligned} & \text{For bin } 1000-1999: \frac{4}{16} = 0.25 \\ & \text{For bin } 2000-2999: \frac{10}{16} = 0.625 \\ & \text{For bin } 3000-3999: \frac{12}{16} = 0.75 \\ & \text{For bin } 4000-4999: \frac{16}{16} = 1.0 \\ \end{aligned} \][/tex]
So the table of cumulative relative frequency distribution is:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Students in a District} & \text{Cumulative Relative Frequency} \\ \hline 1000-1999 & 0.25 \\ \hline 2000-2999 & 0.625 \\ \hline 3000-3999 & 0.75 \\ \hline 4000-4999 & 1.0 \\ \hline \end{tabular} \][/tex]
(c) Draw a frequency polygon.
The frequency polygon is drawn using the midpoints of each bin. The midpoints are calculated as follows:
[tex]\[ \begin{aligned} & \text{Midpoint of } 1000-1999: \frac{1000 + 1999}{2} = 1500 \\ & \text{Midpoint of } 2000-2999: \frac{2000 + 2999}{2} = 2500 \\ & \text{Midpoint of } 3000-3999: \frac{3000 + 3999}{2} = 3500 \\ & \text{Midpoint of } 4000-4999: \frac{4000 + 4999}{2} = 4500 \\ \end{aligned} \][/tex]
The x-coordinates for the frequency polygon are \([1000, 1500, 2500, 3500, 4500, 5000]\) and the corresponding y-coordinates (frequencies at each midpoint and 0 at each end) are \([0, 4, 6, 2, 4, 0]\).
(d) Draw a relative frequency ogive.
An ogive is a plot of cumulative relative frequencies against the upper boundary of each bin.
The x-coordinates for the relative frequency ogive are \([1500, 2500, 3500, 4500]\) and the y-coordinates (cumulative relative frequencies) are \([0.25, 0.625, 0.75, 1.0]\).
By following these steps, we achieve each part of the problem systematically and accurately.
