To address your questions, I'll provide the answers based on the provided information.
Question 1.1
The box-and-whiskers plots are not visible, but I will explain how to determine the answers based on typical attributes of box-and-whiskers plots.
1.1.1 What data set has the smaller range?
To determine the range of each data set, look at the maximum and minimum values of each plot. The range is calculated as:
\[ \text{Range} = \text{Maximum} - \text{Minimum} \]
The dataset with the smaller range is the one where this calculation yields a smaller result.
1.1.2 Which data set shows a greater spread in the middle 50% of the data?
The spread of the middle 50% of the data can be determined by looking at the interquartile range (IQR), which is the distance between the first quartile (Q1) and the third quartile (Q3). The dataset with a larger box in the plot has a greater spread in the middle 50%.
1.1.3 Is data set A positively skewed, negatively skewed, or symmetrical?
To determine skewness:
- Positively skewed: The tail on the right side is longer or fatter than the left side.
- Negatively skewed: The tail on the left side is longer or fatter than the right side.
- Symmetrical: Both sides are balanced. Examine the whiskers and the position of the median within the box of dataset A.
1.1.4 Which data set has the larger standard deviation? Explain your answer.
The standard deviation measures how much individual data points deviate from the mean. Generally, a larger spread indicated in a box-and-whiskers plot (wider boxes and whiskers) suggests a larger standard deviation. Compare the entire stretch and shape of the plots. The data set with the wider variation will have the larger standard deviation.
Question 1.2
Given the dataset of weekly late submissions:
15, 18, 22, 17, 25, 19, 21, 17, 13, 23
1.2.1 Calculate the average weekly number of late submissions. \[ \text{Average} = \frac{\text{Sum of all values}}{\text{Number of values}} = \frac{15 + 18 + 22 + 17 + 25 + 19 + 21 + 17 + 13 + 23}{10} = \frac{200}{10} = 20 \] Average = 20 (3 marks)
1.2.2 Calculate the standard deviation of the weekly late submissions. First, find the variance:
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Calculate the mean: (as above, 20)
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Calculate the deviations from the mean and square them:
- (15-20)² = 25
- (18-20)² = 4
- (22-20)² = 4
- (17-20)² = 9
- (25-20)² = 25
- (19-20)² = 1
- (21-20)² = 1
- (17-20)² = 9
- (13-20)² = 49
- (23-20)² = 9
Now sum these squared deviations: \[ 25 + 4 + 4 + 9 + 25 + 1 + 1 + 9 + 49 + 9 = 136 \]
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Find the variance (for the sample, divide by (n-1)): \[ \text{Variance} = \frac{136}{10 - 1} = \frac{136}{9} \approx 15.11 \]
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Finally, take the square root to find the standard deviation: \[ \text{Standard deviation} \approx \sqrt{15.11} \approx 3.89 \] Standard Deviation ≈ 3.89 (3 marks)
1.2.3 Determine the number of weeks where the weekly late submissions were outside one standard deviation from the average.
Average = 20, Standard Deviation ≈ 3.89
\[ \text{Outside one standard deviation} = (20 - 3.89) \text{ to } (20 + 3.89) \]
\[ \approx 16.11 \text{ to } 23.89 \]
Now we can see which values fall outside this range:
- Below 16.11: 15, 18, 17, 17, 13 (5 values)
- Above 23.89: 25, 23 (2 values)
Total weeks outside one standard deviation: 5 + 2 = 7 weeks. (3 marks)
If you have access to the box-and-whiskers plots, apply this reasoning to obtain specific values.