Age	Lower Limit	Upper Limit	Number (millions)

25-34	25	34	25600000

35-44	35	44	30200000

45-54	45	54	38400000

55-64	55	64	24500000
The following data represent the number of people aged 25 to 64 years covered by health insurance​ (private or​ government) in 2018. Approximate the mean and standard deviation for age.


To approximate the mean and standard deviation of the age distribution, we'll follow these steps:

### Step 1: Calculate the midpoint of each age group
The midpoint (\(x_i\)) of each age group is found by averaging the lower and upper limits of the age group.

\[
\text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}
\]

| Age Group | Lower Limit | Upper Limit | Midpoint (\(x_i\)) | Number of People (Frequency, \(f_i\)) |
|-----------|-------------|-------------|-------------------|--------------------------------------|
| 25-34     | 25          | 34          | \( \frac{25 + 34}{2} = 29.5 \) | 25,600,000 |
| 35-44     | 35          | 44          | \( \frac{35 + 44}{2} = 39.5 \) | 30,200,000 |
| 45-54     | 45          | 54          | \( \frac{45 + 54}{2} = 49.5 \) | 38,400,000 |
| 55-64     | 55          | 64          | \( \frac{55 + 64}{2} = 59.5 \) | 24,500,000 |

### Step 2: Compute the weighted mean
The mean (\(\mu\)) is computed as the weighted average of the midpoints using the formula:

\[
\mu = \frac{\sum (f_i \cdot x_i)}{\sum f_i}
\]

Where:
- \(f_i\) is the frequency (number of people) in each group,
- \(x_i\) is the midpoint of each age group.

Let's calculate:

\[
\mu = \frac{(25,600,000 \cdot 29.5) + (30,200,000 \cdot 39.5) + (38,400,000 \cdot 49.5) + (24,500,000 \cdot 59.5)}{25,600,000 + 30,200,000 + 38,400,000 + 24,500,000}
\]

Summing the products:

\[
\mu = \frac{(756,800,000) + (1,191,900,000) + (1,900,800,000) + (1,459,750,000)}{118,700,000}
\]

\[
\mu = \frac{5,309,250,000}{118,700,000} \approx 44.73
\]

So, the **mean age** is approximately **44.73 years**.

### Step 3: Compute the weighted standard deviation
The standard deviation (\(\sigma\)) is calculated using:

\[
\sigma = \sqrt{\frac{\sum f_i (x_i - \mu)^2}{\sum f_i}}
\]

Where:
- \(f_i\) is the frequency,
- \(x_i\) is the midpoint,
- \(\mu\) is the mean.

Now, calculate each \((x_i - \mu)^2\) and multiply by the corresponding \(f_i\):

| Midpoint (\(x_i\)) | \(x_i - \mu\) | \((x_i - \mu)^2\) | \(f_i \cdot (x_i - \mu)^2\) |
|-------------------|---------------|------------------|------------------------------|
| 29.5              | 29.5 - 44.73 = -15.23  | 231.94           | 25,600,000 \(\cdot\) 231.94 = 5,935,664,000   |
| 39.5              | 39.5 - 44.73 = -5.23   | 27.34            | 30,200,000 \(\cdot\) 27.34  = 824,468,000    |
| 49.5              | 49.5 - 44.73 = 4.77    | 22.75            | 38,400,000 \(\cdot\) 22.75  = 873,600,000    |
| 59.5              | 59.5 - 44.73 = 14.77   | 218.21           | 24,500,000 \(\cdot\) 218.21 = 5,343,145,000   |

Sum of all the \(f_i \cdot (x_i - \mu)^2\):

\[
\sum f_i (x_i - \mu)^2 = 5,935,664,000 + 824,468,000 + 873,600,000 + 5,343,145,000 = 12,976,877,000
\]

Finally, the standard deviation:

\[
\sigma = \sqrt{\frac{12,976,877,000}{118,700,000}} \approx \sqrt{109.33} \approx 10.46
\]

Thus, the **standard deviation** is approximately **10.46 years**.

---

### Summary:
- **Mean age**: 44.73 years
- **Standard deviation**: 10.46 years

Do you want more details on any step or any other questions?

---

Here are 5 relative questions based on this information:
1. How does a weighted mean differ from a simple mean?
2. Why is the midpoint used to approximate the mean in grouped data?
3. How can this method be extended to handle other types of grouped data?
4. What does the standard deviation tell us about the age distribution?
5. How would the mean and standard deviation change if a new age group was added?

**Tip:** Standard deviation measures the dispersion of data around the mean—low values mean less spread, high values mean more spread!

The following data represent the number of people aged 25 to 64 years covered by health insurance in 2018. Approximate the mean and standard deviation for age:

Age Lower Limit Upper Limit Number (millions)

25-34 25 34 25600000

35-44 35 44 30200000

45-54 45 54 38400000

55-64 55 64 24500000

Explore how to approximate the mean and standard deviation for the age distribution of people aged 25 to 64 years covered by health insurance in 2018. Understand key statistical concepts like weighted mean and standard deviation, with step-by-step calculations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation