Answer :
To estimate the mean height of the students, we need to follow these steps:
1. Identify the height intervals and their corresponding frequencies:
- Height [tex]\( 120 < x \leq 130 \)[/tex]: Frequency = 5
- Height [tex]\( 130 < x \leq 140 \)[/tex]: Frequency = 12
- Height [tex]\( 140 < x \leq 150 \)[/tex]: Frequency = 3
2. Calculate the midpoint for each height interval:
- The midpoint for the interval [tex]\( 120 < x \leq 130 \)[/tex] is [tex]\( \frac{120 + 130}{2} = 125.0 \)[/tex].
- The midpoint for the interval [tex]\( 130 < x \leq 140 \)[/tex] is [tex]\( \frac{130 + 140}{2} = 135.0 \)[/tex].
- The midpoint for the interval [tex]\( 140 < x \leq 150 \)[/tex] is [tex]\( \frac{140 + 150}{2} = 145.0 \)[/tex].
3. Multiply each midpoint by its corresponding frequency to get the sum of products [tex]\( ( \text{midpoint} \times \text{frequency} ) \)[/tex]:
- For the interval [tex]\( 120 < x \leq 130 \)[/tex]: [tex]\( 125.0 \times 5 = 625.0 \)[/tex]
- For the interval [tex]\( 130 < x \leq 140 \)[/tex]: [tex]\( 135.0 \times 12 = 1620.0 \)[/tex]
- For the interval [tex]\( 140 < x \leq 150 \)[/tex]: [tex]\( 145.0 \times 3 = 435.0 \)[/tex]
4. Calculate the sum of these products:
Sum = [tex]\( 625.0 + 1620.0 + 435.0 = 2680.0 \)[/tex].
5. Calculate the total number of students (total frequency):
Total frequency = [tex]\( 5 + 12 + 3 = 20 \)[/tex] students.
6. Estimate the mean height by dividing the sum of the products by the total frequency:
Mean height = [tex]\( \frac{2680.0}{20} = 134.0 \)[/tex].
Therefore, the estimated mean height of the students is [tex]\( 134.0 \)[/tex] cm.
1. Identify the height intervals and their corresponding frequencies:
- Height [tex]\( 120 < x \leq 130 \)[/tex]: Frequency = 5
- Height [tex]\( 130 < x \leq 140 \)[/tex]: Frequency = 12
- Height [tex]\( 140 < x \leq 150 \)[/tex]: Frequency = 3
2. Calculate the midpoint for each height interval:
- The midpoint for the interval [tex]\( 120 < x \leq 130 \)[/tex] is [tex]\( \frac{120 + 130}{2} = 125.0 \)[/tex].
- The midpoint for the interval [tex]\( 130 < x \leq 140 \)[/tex] is [tex]\( \frac{130 + 140}{2} = 135.0 \)[/tex].
- The midpoint for the interval [tex]\( 140 < x \leq 150 \)[/tex] is [tex]\( \frac{140 + 150}{2} = 145.0 \)[/tex].
3. Multiply each midpoint by its corresponding frequency to get the sum of products [tex]\( ( \text{midpoint} \times \text{frequency} ) \)[/tex]:
- For the interval [tex]\( 120 < x \leq 130 \)[/tex]: [tex]\( 125.0 \times 5 = 625.0 \)[/tex]
- For the interval [tex]\( 130 < x \leq 140 \)[/tex]: [tex]\( 135.0 \times 12 = 1620.0 \)[/tex]
- For the interval [tex]\( 140 < x \leq 150 \)[/tex]: [tex]\( 145.0 \times 3 = 435.0 \)[/tex]
4. Calculate the sum of these products:
Sum = [tex]\( 625.0 + 1620.0 + 435.0 = 2680.0 \)[/tex].
5. Calculate the total number of students (total frequency):
Total frequency = [tex]\( 5 + 12 + 3 = 20 \)[/tex] students.
6. Estimate the mean height by dividing the sum of the products by the total frequency:
Mean height = [tex]\( \frac{2680.0}{20} = 134.0 \)[/tex].
Therefore, the estimated mean height of the students is [tex]\( 134.0 \)[/tex] cm.