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Sagot :
Let's go through the solution step by step to complete the table and calculate the mean time delay.
(a) Completing the Table:
1. Frequency for the [tex]\( 5 < t \leqslant 10 \)[/tex] Interval:
To find the frequency for the interval [tex]\( 5 < t \leqslant 10 \)[/tex], we need to consider the proportion of frequencies to angles for the sectors provided. After calculations, we determine that the frequency for this interval is 23.
[tex]\[ \text{Frequency for } 5 < t \leqslant 10 = 23 \][/tex]
2. Angle for the [tex]\( 10 < t \leqslant 20 \)[/tex] Interval:
The total angle in a pie chart is 360 degrees. Subtracting the angles we are given:
[tex]\[ \text{Angle Sum} = 360 - (168 + 92 + 8) = 360 - 268 = 92 \text{ degrees} \][/tex]
Therefore, the angle for the [tex]\( 10 < t \leqslant 20 \)[/tex] interval is 92 degrees.
[tex]\[ \text{Angle for } 10 < t \leqslant 20 = 92 \text{ degrees} \][/tex]
3. Frequency for the [tex]\( 20 < t \leqslant 25 \)[/tex] Interval:
We must ensure the total frequency matches. The total frequencies provided and found so far sum to a value where no more frequency remains to be allocated to the [tex]\( 20 < t \leqslant 25 \)[/tex] interval. Thus, the frequency for this interval ends up being 0.
[tex]\[ \text{Frequency for } 20 < t \leqslant 25 = 0 \][/tex]
The completed table is:
\begin{tabular}{|c|c|c|}
\hline
Time delay (t minutes) & Frequency & Angle (in degrees) \\
\hline
[tex]$1 < t \leqslant 5$[/tex] & 42 & 168 \\
\hline
[tex]$5 < t \leqslant 10$[/tex] & 23 & 92 \\
\hline
[tex]$10 < t \leqslant 20$[/tex] & 15 & 92 \\
\hline
[tex]$20 < t \leqslant 25$[/tex] & 0 & 0 \\
\hline
[tex]$25 < t \leqslant 40$[/tex] & 2 & 8 \\
\hline
\end{tabular}
(b) Estimating the Mean Time Delay:
We use the midpoints of each interval to estimate the mean time delay. The midpoint for each interval is:
[tex]\[ \text{Midpoint for } 1 < t \leqslant 5 = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ \text{Midpoint for } 5 < t \leqslant 10 = \frac{5 + 10}{2} = 7.5 \][/tex]
[tex]\[ \text{Midpoint for } 10 < t \leqslant 20 = \frac{10 + 20}{2} = 15 \][/tex]
[tex]\[ \text{Midpoint for } 20 < t \leqslant 25 = \frac{20 + 25}{2} = 22.5 \][/tex]
[tex]\[ \text{Midpoint for } 25 < t \leqslant 40 = \frac{25 + 40}{2} = 32.5 \][/tex]
The estimated mean time delay is calculated using the weighted average:
[tex]\[ \text{Mean Delay} = \frac{(3 \times 42) + (7.5 \times 23) + (15 \times 15) + (22.5 \times 0) + (32.5 \times 2)}{42 + 23 + 15 + 0 + 2} \][/tex]
[tex]\[ \text{Mean Delay} = \frac{(126) + (172.5) + (225) + (0) + (65)}{82} \][/tex]
[tex]\[ \text{Mean Delay} = \frac{588.5}{82} \approx 7.177 \text{ minutes} \][/tex]
Rounded to three significant figures, the estimated mean delay is:
[tex]\[ \text{Estimated Mean Delay} = 7.177 \text{ minutes} \][/tex]
Therefore, the final mean time delay to 3 significant figures is:
[tex]\[ 7.18 \text{ minutes} \][/tex]
(a) Completing the Table:
1. Frequency for the [tex]\( 5 < t \leqslant 10 \)[/tex] Interval:
To find the frequency for the interval [tex]\( 5 < t \leqslant 10 \)[/tex], we need to consider the proportion of frequencies to angles for the sectors provided. After calculations, we determine that the frequency for this interval is 23.
[tex]\[ \text{Frequency for } 5 < t \leqslant 10 = 23 \][/tex]
2. Angle for the [tex]\( 10 < t \leqslant 20 \)[/tex] Interval:
The total angle in a pie chart is 360 degrees. Subtracting the angles we are given:
[tex]\[ \text{Angle Sum} = 360 - (168 + 92 + 8) = 360 - 268 = 92 \text{ degrees} \][/tex]
Therefore, the angle for the [tex]\( 10 < t \leqslant 20 \)[/tex] interval is 92 degrees.
[tex]\[ \text{Angle for } 10 < t \leqslant 20 = 92 \text{ degrees} \][/tex]
3. Frequency for the [tex]\( 20 < t \leqslant 25 \)[/tex] Interval:
We must ensure the total frequency matches. The total frequencies provided and found so far sum to a value where no more frequency remains to be allocated to the [tex]\( 20 < t \leqslant 25 \)[/tex] interval. Thus, the frequency for this interval ends up being 0.
[tex]\[ \text{Frequency for } 20 < t \leqslant 25 = 0 \][/tex]
The completed table is:
\begin{tabular}{|c|c|c|}
\hline
Time delay (t minutes) & Frequency & Angle (in degrees) \\
\hline
[tex]$1 < t \leqslant 5$[/tex] & 42 & 168 \\
\hline
[tex]$5 < t \leqslant 10$[/tex] & 23 & 92 \\
\hline
[tex]$10 < t \leqslant 20$[/tex] & 15 & 92 \\
\hline
[tex]$20 < t \leqslant 25$[/tex] & 0 & 0 \\
\hline
[tex]$25 < t \leqslant 40$[/tex] & 2 & 8 \\
\hline
\end{tabular}
(b) Estimating the Mean Time Delay:
We use the midpoints of each interval to estimate the mean time delay. The midpoint for each interval is:
[tex]\[ \text{Midpoint for } 1 < t \leqslant 5 = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ \text{Midpoint for } 5 < t \leqslant 10 = \frac{5 + 10}{2} = 7.5 \][/tex]
[tex]\[ \text{Midpoint for } 10 < t \leqslant 20 = \frac{10 + 20}{2} = 15 \][/tex]
[tex]\[ \text{Midpoint for } 20 < t \leqslant 25 = \frac{20 + 25}{2} = 22.5 \][/tex]
[tex]\[ \text{Midpoint for } 25 < t \leqslant 40 = \frac{25 + 40}{2} = 32.5 \][/tex]
The estimated mean time delay is calculated using the weighted average:
[tex]\[ \text{Mean Delay} = \frac{(3 \times 42) + (7.5 \times 23) + (15 \times 15) + (22.5 \times 0) + (32.5 \times 2)}{42 + 23 + 15 + 0 + 2} \][/tex]
[tex]\[ \text{Mean Delay} = \frac{(126) + (172.5) + (225) + (0) + (65)}{82} \][/tex]
[tex]\[ \text{Mean Delay} = \frac{588.5}{82} \approx 7.177 \text{ minutes} \][/tex]
Rounded to three significant figures, the estimated mean delay is:
[tex]\[ \text{Estimated Mean Delay} = 7.177 \text{ minutes} \][/tex]
Therefore, the final mean time delay to 3 significant figures is:
[tex]\[ 7.18 \text{ minutes} \][/tex]
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