Sure, let's break down the solution step-by-step for each part of the question.
### (a) Construct a histogram to represent the data and estimate the mode from it. Justify the answer of the mode by calculation.
1. Data Representation in Histogram:
- The histogram will have weight intervals on the x-axis and the number of students on the y-axis.
- The intervals are: `40-<45`, `45-<50`, `50-<55`, `55-<60`, `60-<70`, and `70-<80`.
- The corresponding frequencies (number of students) are: `4`, `13`, `17`, `40`, `39`, `7`.
2. Histogram Construction:
- Draw a bar over each interval with a height corresponding to the number of students in that interval.
3. Estimating the Mode:
- The mode is the class interval with the highest frequency.
- From the given data:
- `40-<45`: 4 students
- `45-<50`: 13 students
- `50-<55`: 17 students
- `55-<60`: 40 students
- `60-<70`: 39 students
- `70-<80`: 7 students
- The class interval `55-<60` has the highest number of students, which is `40`.
Conclusion:
- The mode of the distribution is the class interval `55-<60`.
### (b) Construct a '<' cumulative frequency polygon:
To construct a '<' cumulative frequency polygon, we need to compute the cumulative frequency for each class interval.
1. Cumulative Frequency Table:
- Cumulative frequency for `40-<45` is `4`.
- Cumulative frequency for `45-<50` is `4 + 13 = 17`.
- Cumulative frequency for `50-<55` is `17 + 17 = 34`.
- Cumulative frequency for `55-<60` is `34 + 40 = 74`.
- Cumulative frequency for `60-<70` is `74 + 39 = 113`.
- Cumulative frequency for `70-<80` is `113 + 7 = 120`.
2. Constructing the Polygon:
- Plot the cumulative frequencies against the upper boundary of each weight interval and join these points with a line.
### (i) Estimate the median from the cumulative frequency polygon:
1. Sample Size:
- The total sample size is `120`.
2. Median Position:
- The median position is at the `(n / 2)`-th value, where `n` is the sample size:
- Median position = `120 / 2 = 60`.
3. Estimating Median from Cumulative Frequency:
- Identify the cumulative frequency immediately greater than or equal to `60`, which is `74` in the cumulative frequency table.
- The corresponding interval for this cumulative frequency is `55-<60`.
4. Linear Interpolation to Find Median:
- The lower boundary of the median class interval (`55-<60`) is `55`.
- Frequency of the median class interval is `40`.
- Cumulative frequency just before the median class interval is `34`.
- Class width of the median interval is `5`.
The median [tex]\( M \)[/tex] is calculated using the formula:
[tex]\[
M = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times w
\][/tex]
where:
- [tex]\( L \)[/tex] = lower boundary of the median class interval = `55`,
- [tex]\( n \)[/tex] = sample size = `120`,
- [tex]\( CF \)[/tex] = cumulative frequency before the median class = `34`,
- [tex]\( f \)[/tex] = frequency of the median class = `40`,
- [tex]\( w \)[/tex] = class width = `5`.
Substituting the values:
[tex]\[
M = 55 + \left(\frac{60 - 34}{40}\right) \times 5 = 55 + \left(\frac{26}{40}\right) \times 5 = 55 + 3.25 = 58.25
\][/tex]
Conclusion:
- The estimated median is `58.25`.
### (ii) Estimate the percentage of overweight students (weight ≥ 65 kg):
1. Identifying Overweight Students:
- Students classified as overweight are those in the weight range `65-<70` and `70-<80`.
- There are `39` students in the `60-<70` interval. Some of them (assumed to be half in the middle of 60 to 70) can be considered as 65 or above.
- We can assume about half of `39` be in the range 65-<70 kg which is around 19.5 students.
- Additionally, there are `7` students in the `70-<80` interval. So total overweight students would be `19.5 + 7 = 26.5` approximately 26 out of 120.
2. Calculating Percentage:
- Total number of students = `120`.
- Number of overweight students = `46`.
[tex]\[
\text{Percentage of overweight students} = \left(\frac{46}{120}\right) \times 100 \approx 38.33
\][/tex]
Conclusion:
- The estimated percentage of overweight students is approximately `38.33%`.
