Let's solve the problem step-by-step by first identifying and computing the measures of central tendency and then moving on to the measures of dispersion for the cholesterol levels of the 7 patients.

Given Cholesterol Levels:

155, 220, 134, 216, 200, 144, 191

Measures of Central Tendency:

1. Mean (Average): $\text{Mean} = \frac{\sum \text{Cholesterol Levels}}{n}$ Where $n = 7$ (the number of patients).

   $\text{Mean} = \frac{155 + 220 + 134 + 216 + 200 + 144 + 191}{7}$ $\text{Mean} = \frac{1260}{7} = 180$

   So, the mean cholesterol level is 180.

2. Median (Middle value when data is ordered): First, arrange the values in ascending order:
   134, 144, 155, 191, 200, 216, 220

   Since there are 7 values (odd number of values), the median is the middle value, which is the 4th value in the ordered list:
   $\text{Median} = 191$

3. Mode (Most frequent value): Since all values are unique, there is no mode for this dataset.

Measures of Dispersion:

1. Range: The range is the difference between the highest and lowest values in the dataset. $\text{Range} = \text{Max} - \text{Min} = 220 - 134 = 86$

2. Variance: Variance is the average of the squared differences from the mean. We first calculate the differences of each value from the mean, square them, and then find the average.

   $\text{Variance} = \frac{\sum (\text{Cholesterol Level} - \text{Mean})^2}{n}$ $\text{Variance} = \frac{(155-180)^2 + (220-180)^2 + (134-180)^2 + (216-180)^2 + (200-180)^2 + (144-180)^2 + (191-180)^2}{7}$ $= \frac{(-25)^2 + 40^2 + (-46)^2 + 36^2 + 20^2 + (-36)^2 + 11^2}{7}$ $= \frac{625 + 1600 + 2116 + 1296 + 400 + 1296 + 121}{7} = \frac{7454}{7} = 1064.86$

   So, the variance is approximately 1064.86.

3. Standard Deviation: The standard deviation is the square root of the variance: $\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{1064.86} \approx 32.63$

   So, the standard deviation is approximately 32.63.

Summary:

• Mean (Average): 180
• Median: 191
• Mode: None
• Range: 86
• Variance: 1064.86
• Standard Deviation: 32.63

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Would you like further clarification on any part of this solution, or do you need more details?

Here are 5 related questions:

1. How would you calculate the mean for a larger dataset?
2. What is the significance of the range in data analysis?
3. Can you explain how variance and standard deviation help understand data variability?
4. What happens to the variance and standard deviation if we increase the sample size?
5. Why is the median sometimes preferred over the mean in measuring central tendency?

Tip: When working with small datasets, the mean can be sensitive to outliers. The median is often more robust in such cases.