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Sagot :
### Solution
First, we need to find an estimate for the mean height of the students given the height intervals and frequencies.
#### Step-by-Step:
1. Find the midpoints of each height interval:
- For [tex]\( 120 < t < 124 \)[/tex], the midpoint is [tex]\( \frac{120 + 124}{2} = 122 \)[/tex] cm
- For [tex]\( 124 < t \leqslant 128 \)[/tex], the midpoint is [tex]\( \frac{124 + 128}{2} = 126 \)[/tex] cm
- For [tex]\( 128 < t < 132 \)[/tex], the midpoint is [tex]\( \frac{128 + 132}{2} = 130 \)[/tex] cm
- For [tex]\( 132 < t \leqslant 136 \)[/tex], the midpoint is [tex]\( \frac{132 + 136}{2} = 134 \)[/tex] cm
- For [tex]\( 136 < t \leqslant 140 \)[/tex], the midpoint is [tex]\( \frac{136 + 140}{2} = 138 \)[/tex] cm
2. List the heights (midpoints) and their corresponding frequencies:
[tex]\[ \begin{array}{ccc} \text{Height Interval} & \text{Midpoint (cm)} & \text{Frequency (f)} \\ \hline 120 < t < 124 & 122 & 7 \\ 124 < t \leqslant 128 & 126 & 8 \\ 128 < t < 132 & 130 & 13 \\ 132 < t \leqslant 136 & 134 & 9 \\ 136 < t \leqslant 140 & 138 & 3 \\ \end{array} \][/tex]
3. Calculate the sum of the product of each midpoint and its frequency:
[tex]\[ \begin{aligned} & (122 \times 7) + (126 \times 8) + (130 \times 13) + (134 \times 9) + (138 \times 3) \\ & = 854 + 1008 + 1690 + 1206 + 414 \\ & = 5172 \, \text{cm} \end{aligned} \][/tex]
4. Calculate the total number of students:
[tex]\[ 7 + 8 + 13 + 9 + 3 = 40 \][/tex]
5. Calculate the mean height:
[tex]\[ \text{Mean height} = \frac{\text{Sum of (midpoints} \times \text{frequencies)}}{\text{Total number of students}} = \frac{5172}{40} = 129.3 \, \text{cm} \][/tex]
### Answer to part (a)
The estimated mean height of the students is [tex]\( 129.3 \)[/tex] cm.
### Answer to part (b)
The answer to part (a) is an estimate because the data provided is grouped into intervals. By using the midpoints of these intervals to represent the data, we approximate the actual heights of students within each range. This method doesn't account for the specific distribution of data within each interval, leading to an estimation rather than a precise calculation.
First, we need to find an estimate for the mean height of the students given the height intervals and frequencies.
#### Step-by-Step:
1. Find the midpoints of each height interval:
- For [tex]\( 120 < t < 124 \)[/tex], the midpoint is [tex]\( \frac{120 + 124}{2} = 122 \)[/tex] cm
- For [tex]\( 124 < t \leqslant 128 \)[/tex], the midpoint is [tex]\( \frac{124 + 128}{2} = 126 \)[/tex] cm
- For [tex]\( 128 < t < 132 \)[/tex], the midpoint is [tex]\( \frac{128 + 132}{2} = 130 \)[/tex] cm
- For [tex]\( 132 < t \leqslant 136 \)[/tex], the midpoint is [tex]\( \frac{132 + 136}{2} = 134 \)[/tex] cm
- For [tex]\( 136 < t \leqslant 140 \)[/tex], the midpoint is [tex]\( \frac{136 + 140}{2} = 138 \)[/tex] cm
2. List the heights (midpoints) and their corresponding frequencies:
[tex]\[ \begin{array}{ccc} \text{Height Interval} & \text{Midpoint (cm)} & \text{Frequency (f)} \\ \hline 120 < t < 124 & 122 & 7 \\ 124 < t \leqslant 128 & 126 & 8 \\ 128 < t < 132 & 130 & 13 \\ 132 < t \leqslant 136 & 134 & 9 \\ 136 < t \leqslant 140 & 138 & 3 \\ \end{array} \][/tex]
3. Calculate the sum of the product of each midpoint and its frequency:
[tex]\[ \begin{aligned} & (122 \times 7) + (126 \times 8) + (130 \times 13) + (134 \times 9) + (138 \times 3) \\ & = 854 + 1008 + 1690 + 1206 + 414 \\ & = 5172 \, \text{cm} \end{aligned} \][/tex]
4. Calculate the total number of students:
[tex]\[ 7 + 8 + 13 + 9 + 3 = 40 \][/tex]
5. Calculate the mean height:
[tex]\[ \text{Mean height} = \frac{\text{Sum of (midpoints} \times \text{frequencies)}}{\text{Total number of students}} = \frac{5172}{40} = 129.3 \, \text{cm} \][/tex]
### Answer to part (a)
The estimated mean height of the students is [tex]\( 129.3 \)[/tex] cm.
### Answer to part (b)
The answer to part (a) is an estimate because the data provided is grouped into intervals. By using the midpoints of these intervals to represent the data, we approximate the actual heights of students within each range. This method doesn't account for the specific distribution of data within each interval, leading to an estimation rather than a precise calculation.
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