Step 1: Interpret the stem-and-leaf plot

From the stem-and-leaf plot:

• 1 | 2 → 12
• 2 | → (no data)
• 3 | 1 3 → 31, 33
• 4 | 0 7 9 9 → 40, 47, 49, 49
• 5 | 3 5 6 8 9 → 53, 55, 56, 58, 59
• 6 | 0 3 5 9 → 60, 63, 65, 69

The full dataset is:
12, 31, 33, 40, 47, 49, 49, 53, 55, 56, 58, 59, 60, 63, 65, 69

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a. Mean (Average)

$\text{Mean} = \frac{\sum \text{Data Points}}{\text{Number of Data Points}}$

The sum of all data points is:
$12 + 31 + 33 + 40 + 47 + 49 + 49 + 53 + 55 + 56 + 58 + 59 + 60 + 63 + 65 + 69 = 799$

Number of data points = 16.

$\text{Mean} = \frac{799}{16} = 49.9375 \approx 49.938$

b. Median

To find the median, we arrange the data points in order (which is already done) and find the middle value. Since there are 16 data points (even number), the median is the average of the 8th and 9th data points.

Data points:
12, 31, 33, 40, 47, 49, 49, 53, 55, 56, 58, 59, 60, 63, 65, 69

$\text{Median} = \frac{53 + 55}{2} = \frac{108}{2} = 54$

c. Standard Deviation

The standard deviation measures how spread out the data is. The formula for standard deviation (sample version) is:

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Where:

• $x_i$ is each data point
• $\bar{x}$ is the mean
• $n$ is the number of data points

Using the mean $\bar{x} = 49.938$, I'll calculate the variance and standard deviation.

Let me compute that.The standard deviation is approximately $14.776$.

d. Range

The range is the difference between the maximum and minimum values in the dataset.

$\text{Range} = 69 - 12 = 57$

e. Shape of the Data

To determine the shape of the data distribution, we can observe the values in the stem-and-leaf plot. The data appears to be slightly right-skewed, as the higher values (e.g., 60s) extend further than the lower values, but there isn't a significant tail.

Summary:

• Mean: 49.938
• Median: 54
• Standard Deviation: 14.776
• Range: 57
• Shape: Right-skewed

Would you like further details or explanations?
Here are five related questions:

1. How is the mode calculated from a stem-and-leaf plot?
2. How can the interquartile range (IQR) be determined from this dataset?
3. What is the relationship between mean, median, and skewness in a dataset?
4. How does outlier detection affect the calculation of standard deviation?
5. Can the shape of data affect the interpretation of variance and standard deviation?

Tip: In right-skewed distributions, the mean is typically greater than the median due to the influence of higher values.