Answer :
Certainly! Let's break down the steps to tackle each part of the question effectively.
### Step-by-Step Solution
#### Data:
The number of defective items rejected daily over the past 30 days is given as:
9, 13, 7, 5, 8, 12, 15, 5, 7, 3, 4, 8, 15, 17, 19, 6, 4, 10, 8, 22, 7, 5, 3, 9, 19, 14, 13, 18
For clarity, let's organize and sort these numbers:
`[3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 12, 13, 13, 14, 15, 15, 17, 18, 19, 19, 22]`
(Note: Ensure that only 28 days instead of the declared 30 days' data could be due to oversight. I will work with the given 28 numbers.)
### Part A: Construct a Histogram
1. Determine the Range and Bins:
- Minimum value = 3
- Maximum value = 22
- Range = 22 - 3 = 19
- If we choose to create 5 bins, a bin width could be calculated as:
[tex]\[ \text{Bin width} = \frac{\text{Range}}{\text{Number of bins}} = \frac{19}{5} \approx 3.8 \approx 4 \][/tex]
2. Define the Bins and Frequency Counts:
- Bin 1: [3, 6) → contains values 3, 3, 4, 4, 5, 5, 5 → Frequency = 7
- Bin 2: [7, 10) → contains values 7, 7, 7, 8, 8, 8, 9, 9, 10 → Frequency = 9
- Bin 3: [11, 14) → contains values 12, 13, 13, 14 → Frequency = 4
- Bin 4: [15, 18) → contains values 15, 15, 17, 18 → Frequency = 4
- Bin 5: [19, 22] → contains values 19, 19, 22 → Frequency = 3 (Note: considering 22 to be inclusive here)
3. Draw the Histogram:
- Each bin will be represented on the x-axis (Number of Defective Items).
- The frequency of the bins will be represented on the y-axis.
- Ensure to label the axes and include titles.
### Part B: Construct an Ogive
1. Sort the Data and Compute Cumulative Frequencies:
- Sorted data: `[3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 12, 13, 13, 14, 15, 15, 17, 18, 19, 19, 22]`
- Cumulative frequency:
[tex]\[ [2, 2+2, 4+2, 6+1, 7+3, 10+1, 11+3, 14+1, 15+3, 18+2, 20+1, 21+3, 24+3, 27+3, 30+4, 34+4, 38+4, 42+3, 45+2, 47+2, 49+1, 50+2, 52+2, 54+2, 56+2, 58+2, 60+1, 61+2, 63+1, 1+22] \][/tex]
In steps:
[tex]\[ [2, 4, 6, 7, 10, 11, 14, 15, 18, 20, 21, 24, 27, 30, 34, 38, 42, 45, 47, 49, 50, 52, 54, 56, 58, 60, 61, 63, 84 ] \][/tex]
2. Construct the Ogive Plot:
- The x-axis will be the number of defective items.
- The y-axis will be the cumulative frequency.
- Points will be plotted at the upper boundaries of each class, connected with a smooth line.
### Part C: Describe the Shape of the Histogram
- Analyze the distribution of frequencies.
- Shape:
- The histogram appears right-skewed (positively skewed), which means the right tail (side with higher values) is longer than the left.
- Most data points lie on the left side, indicating that lower numbers of defective items are more frequent days over the given period.
In summary, we have:
- Structured bins for the histogram,
- A cumulative frequency table for the ogive,
- Description provided based on the frequency distribution observed from the histogram.
Following the above steps should cover each part of the question comprehensively.
### Step-by-Step Solution
#### Data:
The number of defective items rejected daily over the past 30 days is given as:
9, 13, 7, 5, 8, 12, 15, 5, 7, 3, 4, 8, 15, 17, 19, 6, 4, 10, 8, 22, 7, 5, 3, 9, 19, 14, 13, 18
For clarity, let's organize and sort these numbers:
`[3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 12, 13, 13, 14, 15, 15, 17, 18, 19, 19, 22]`
(Note: Ensure that only 28 days instead of the declared 30 days' data could be due to oversight. I will work with the given 28 numbers.)
### Part A: Construct a Histogram
1. Determine the Range and Bins:
- Minimum value = 3
- Maximum value = 22
- Range = 22 - 3 = 19
- If we choose to create 5 bins, a bin width could be calculated as:
[tex]\[ \text{Bin width} = \frac{\text{Range}}{\text{Number of bins}} = \frac{19}{5} \approx 3.8 \approx 4 \][/tex]
2. Define the Bins and Frequency Counts:
- Bin 1: [3, 6) → contains values 3, 3, 4, 4, 5, 5, 5 → Frequency = 7
- Bin 2: [7, 10) → contains values 7, 7, 7, 8, 8, 8, 9, 9, 10 → Frequency = 9
- Bin 3: [11, 14) → contains values 12, 13, 13, 14 → Frequency = 4
- Bin 4: [15, 18) → contains values 15, 15, 17, 18 → Frequency = 4
- Bin 5: [19, 22] → contains values 19, 19, 22 → Frequency = 3 (Note: considering 22 to be inclusive here)
3. Draw the Histogram:
- Each bin will be represented on the x-axis (Number of Defective Items).
- The frequency of the bins will be represented on the y-axis.
- Ensure to label the axes and include titles.
### Part B: Construct an Ogive
1. Sort the Data and Compute Cumulative Frequencies:
- Sorted data: `[3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 12, 13, 13, 14, 15, 15, 17, 18, 19, 19, 22]`
- Cumulative frequency:
[tex]\[ [2, 2+2, 4+2, 6+1, 7+3, 10+1, 11+3, 14+1, 15+3, 18+2, 20+1, 21+3, 24+3, 27+3, 30+4, 34+4, 38+4, 42+3, 45+2, 47+2, 49+1, 50+2, 52+2, 54+2, 56+2, 58+2, 60+1, 61+2, 63+1, 1+22] \][/tex]
In steps:
[tex]\[ [2, 4, 6, 7, 10, 11, 14, 15, 18, 20, 21, 24, 27, 30, 34, 38, 42, 45, 47, 49, 50, 52, 54, 56, 58, 60, 61, 63, 84 ] \][/tex]
2. Construct the Ogive Plot:
- The x-axis will be the number of defective items.
- The y-axis will be the cumulative frequency.
- Points will be plotted at the upper boundaries of each class, connected with a smooth line.
### Part C: Describe the Shape of the Histogram
- Analyze the distribution of frequencies.
- Shape:
- The histogram appears right-skewed (positively skewed), which means the right tail (side with higher values) is longer than the left.
- Most data points lie on the left side, indicating that lower numbers of defective items are more frequent days over the given period.
In summary, we have:
- Structured bins for the histogram,
- A cumulative frequency table for the ogive,
- Description provided based on the frequency distribution observed from the histogram.
Following the above steps should cover each part of the question comprehensively.
