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Sagot :
Let's go through the solution methodically for each part of the question.
### Part (a): Finding the Class Interval that Contains the Median
Given the data:
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Number of \\ hours $(h)$ \end{tabular} & Frequency \\ \hline $0 \leqslant h<1$ & 4 \\ \hline $1 \leqslant h<2$ & 8 \\ \hline $2 \leqslant h<3$ & 11 \\ \hline $3 \leqslant h<4$ & 7 \\ \hline \end{tabular} \][/tex]
The total number of children is 30. The median position is located at [tex]\( \left( \frac{30}{2} \right) \)[/tex] which is the 15th child.
To find the interval that contains the 15th child, we calculate the cumulative frequency:
- Cumulative frequency for [tex]\(0 \leqslant h<1\)[/tex]: [tex]\(4\)[/tex]
- Cumulative frequency for [tex]\(1 \leqslant h<2\)[/tex]: [tex]\(4 + 8 = 12\)[/tex]
- Cumulative frequency for [tex]\(2 \leqslant h<3\)[/tex]: [tex]\(12 + 11 = 23\)[/tex]
- Cumulative frequency for [tex]\(3 \leqslant h<4\)[/tex]: [tex]\(23 + 7 = 30\)[/tex]
From the cumulative frequencies, we see that the 15th child falls within the interval [tex]\(2 \leqslant h<3\)[/tex] since the cumulative frequency just before it is 12 and the next cumulative frequency is 23, which includes the 15th position.
Thus, the class interval that contains the median is:
[tex]\[ 2 < h < 3 \][/tex]
### Part (b): Estimating the Mean Number of Hours
To estimate the mean number of hours, we need to calculate the midpoint of each class interval and then use the frequency of each class:
- Midpoint of [tex]\(0 \leqslant h<1\)[/tex] is [tex]\((0 + 1) / 2 = 0.5\)[/tex]
- Midpoint of [tex]\(1 \leqslant h<2\)[/tex] is [tex]\((1 + 2) / 2 = 1.5\)[/tex]
- Midpoint of [tex]\(2 \leqslant h<3\)[/tex] is [tex]\((2 + 3) / 2 = 2.5\)[/tex]
- Midpoint of [tex]\(3 \leqslant h<4\)[/tex] is [tex]\((3 + 4) / 2 = 3.5\)[/tex]
Next, we calculate the total sum of the products of midpoints and frequencies:
[tex]\[ 0.5 \times 4 = 2.0 \][/tex]
[tex]\[ 1.5 \times 8 = 12.0 \][/tex]
[tex]\[ 2.5 \times 11 = 27.5 \][/tex]
[tex]\[ 3.5 \times 7 = 24.5 \][/tex]
Adding these products gives the total sum:
[tex]\[ 2.0 + 12.0 + 27.5 + 24.5 = 66.0 \][/tex]
The estimated mean number of hours is then calculated by dividing this total sum by the number of children (30):
[tex]\[ \text{Estimated mean} = \frac{66.0}{30} = 2.2 \][/tex]
### Final Summary:
a) The class interval that contains the median:
[tex]\[ 2 < h < 3 \][/tex]
b) The estimated mean number of hours:
[tex]\[ 2.2 \][/tex]
### Part (a): Finding the Class Interval that Contains the Median
Given the data:
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Number of \\ hours $(h)$ \end{tabular} & Frequency \\ \hline $0 \leqslant h<1$ & 4 \\ \hline $1 \leqslant h<2$ & 8 \\ \hline $2 \leqslant h<3$ & 11 \\ \hline $3 \leqslant h<4$ & 7 \\ \hline \end{tabular} \][/tex]
The total number of children is 30. The median position is located at [tex]\( \left( \frac{30}{2} \right) \)[/tex] which is the 15th child.
To find the interval that contains the 15th child, we calculate the cumulative frequency:
- Cumulative frequency for [tex]\(0 \leqslant h<1\)[/tex]: [tex]\(4\)[/tex]
- Cumulative frequency for [tex]\(1 \leqslant h<2\)[/tex]: [tex]\(4 + 8 = 12\)[/tex]
- Cumulative frequency for [tex]\(2 \leqslant h<3\)[/tex]: [tex]\(12 + 11 = 23\)[/tex]
- Cumulative frequency for [tex]\(3 \leqslant h<4\)[/tex]: [tex]\(23 + 7 = 30\)[/tex]
From the cumulative frequencies, we see that the 15th child falls within the interval [tex]\(2 \leqslant h<3\)[/tex] since the cumulative frequency just before it is 12 and the next cumulative frequency is 23, which includes the 15th position.
Thus, the class interval that contains the median is:
[tex]\[ 2 < h < 3 \][/tex]
### Part (b): Estimating the Mean Number of Hours
To estimate the mean number of hours, we need to calculate the midpoint of each class interval and then use the frequency of each class:
- Midpoint of [tex]\(0 \leqslant h<1\)[/tex] is [tex]\((0 + 1) / 2 = 0.5\)[/tex]
- Midpoint of [tex]\(1 \leqslant h<2\)[/tex] is [tex]\((1 + 2) / 2 = 1.5\)[/tex]
- Midpoint of [tex]\(2 \leqslant h<3\)[/tex] is [tex]\((2 + 3) / 2 = 2.5\)[/tex]
- Midpoint of [tex]\(3 \leqslant h<4\)[/tex] is [tex]\((3 + 4) / 2 = 3.5\)[/tex]
Next, we calculate the total sum of the products of midpoints and frequencies:
[tex]\[ 0.5 \times 4 = 2.0 \][/tex]
[tex]\[ 1.5 \times 8 = 12.0 \][/tex]
[tex]\[ 2.5 \times 11 = 27.5 \][/tex]
[tex]\[ 3.5 \times 7 = 24.5 \][/tex]
Adding these products gives the total sum:
[tex]\[ 2.0 + 12.0 + 27.5 + 24.5 = 66.0 \][/tex]
The estimated mean number of hours is then calculated by dividing this total sum by the number of children (30):
[tex]\[ \text{Estimated mean} = \frac{66.0}{30} = 2.2 \][/tex]
### Final Summary:
a) The class interval that contains the median:
[tex]\[ 2 < h < 3 \][/tex]
b) The estimated mean number of hours:
[tex]\[ 2.2 \][/tex]
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