Activity C.1.1 Identify and compute the appropriate measures of central tendency using the step-by-step process of the following. Please include to answer the measures of dispersion (Range, Variance and Standard Deviation)

2.Find out the average cholesterol level of 7 sample patients (155, 220, 134, 216, 200, 144 and 191) Cholesterol is related to heart health. Understanding this average helps doctors know how to keep their patients healthy.

### Step-by-Step Solution for Measures of Central Tendency and Dispersion:

#### 1. Data Set:
Cholesterol levels of 7 patients:  
\[ 155, 220, 134, 216, 200, 144, 191 \]

---

### **Measures of Central Tendency:**

#### **Step 1: Compute the Mean (Average)**

The mean is calculated by summing up all the data points and dividing by the number of data points.

\[
\text{Mean} = \frac{155 + 220 + 134 + 216 + 200 + 144 + 191}{7}
\]

\[
\text{Mean} = \frac{1260}{7} = 180
\]

Thus, the mean cholesterol level is **180 mg/dL**.

---

#### **Step 2: Compute the Median**

To calculate the median, we first arrange the data in ascending order:

\[ 134, 144, 155, 191, 200, 216, 220 \]

Since there are 7 data points, the median is the 4th value in the ordered list.

\[
\text{Median} = 191
\]

Thus, the median cholesterol level is **191 mg/dL**.

---

#### **Step 3: Compute the Mode**

The mode is the most frequently occurring value in the dataset. In this case, each value appears only once, so there is no mode.

\[
\text{Mode} = \text{None}
\]

---

### **Measures of Dispersion:**

#### **Step 4: Compute the Range**

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

\[
\text{Range} = \text{Max} - \text{Min} = 220 - 134 = 86
\]

Thus, the range of cholesterol levels is **86 mg/dL**.

---

#### **Step 5: Compute the Variance**

Variance is calculated by finding the squared differences between each data point and the mean, then taking the average of those squared differences.

\[
\text{Variance} = \frac{(155 - 180)^2 + (220 - 180)^2 + (134 - 180)^2 + (216 - 180)^2 + (200 - 180)^2 + (144 - 180)^2 + (191 - 180)^2}{7}
\]

Let's calculate each term:

\[
(155 - 180)^2 = (-25)^2 = 625
\]
\[
(220 - 180)^2 = 40^2 = 1600
\]
\[
(134 - 180)^2 = (-46)^2 = 2116
\]
\[
(216 - 180)^2 = 36^2 = 1296
\]
\[
(200 - 180)^2 = 20^2 = 400
\]
\[
(144 - 180)^2 = (-36)^2 = 1296
\]
\[
(191 - 180)^2 = 11^2 = 121
\]

Sum of squared differences:

\[
625 + 1600 + 2116 + 1296 + 400 + 1296 + 121 = 7454
\]

\[
\text{Variance} = \frac{7454}{7} = 1064.86
\]

Thus, the variance of cholesterol levels is approximately **1064.86**.

---

#### **Step 6: Compute the Standard Deviation**

The standard deviation is the square root of the variance:

\[
\text{Standard Deviation} = \sqrt{1064.86} \approx 32.63
\]

Thus, the standard deviation of cholesterol levels is approximately **32.63 mg/dL**.

---

### **Summary of Results:**

- **Mean (Average):** 180 mg/dL
- **Median:** 191 mg/dL
- **Mode:** None
- **Range:** 86 mg/dL
- **Variance:** 1064.86
- **Standard Deviation:** 32.63 mg/dL

---

Would you like more details or clarification on any part? Here are some related questions to consider:

1. What does a high standard deviation indicate about cholesterol levels?
2. How do the mean and median differ in terms of usefulness?
3. Why is understanding the range important for medical professionals?
4. How would the variance change if there were an outlier in the data set?
5. Can the standard deviation ever be larger than the mean?

**Tip:** The standard deviation helps assess the variability in patient data—higher values mean more diverse cholesterol levels.

Activity C.1.1 Identify and compute the appropriate measures of central tendency using the step-by-step process of the following. Please include to answer the measures of dispersion (Range, Variance and Standard Deviation). 2. Find out the average cholesterol level of 7 sample patients (155, 220, 134, 216, 200, 144, and 191). Cholesterol is related to heart health. Understanding this average helps doctors know how to keep their patients healthy.

Learn how to calculate the mean, median, range, variance, and standard deviation of cholesterol levels for 7 patients (155, 220, 134, 216, 200, 144, and 191). This example helps understand heart health and statistical analysis, suitable for students in Grades 10-12.

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