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 corrected standard deviation for age.

To approximate the mean and corrected standard deviation for the given age data, we'll treat the age ranges as grouped data. Here’s how we can calculate both:

### Step 1: Calculate the Mean
For grouped data, the mean (\(\overline{x}\)) is calculated using the formula:

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

where:
- \(x_i\) is the midpoint of each age range.
- \(f_i\) is the frequency (number of people) in each range.

#### Calculate the Midpoints:
The midpoint (\(x_i\)) of each age range is:

- For 25-34: \((25 + 34) / 2 = 29.5\)
- For 35-44: \((35 + 44) / 2 = 39.5\)
- For 45-54: \((45 + 54) / 2 = 49.5\)
- For 55-64: \((55 + 64) / 2 = 59.5\)

#### Multiply Midpoints by Frequencies (\(x_i \cdot f_i\)):
- \(29.5 \cdot 25600000 = 754000000\)
- \(39.5 \cdot 30200000 = 1192900000\)
- \(49.5 \cdot 38400000 = 1900800000\)
- \(59.5 \cdot 24500000 = 1457750000\)

#### Sum of Frequencies (\(\sum f_i\)):
\[
\sum f_i = 25600000 + 30200000 + 38400000 + 24500000 = 118700000
\]

#### Sum of \(x_i \cdot f_i\):
\[
\sum (x_i \cdot f_i) = 754000000 + 1192900000 + 1900800000 + 1457750000 = 5305450000
\]

#### Calculate the Mean:
\[
\overline{x} = \frac{5305450000}{118700000} \approx 44.71
\]

So, the approximate mean age is **44.71 years**.

---

### Step 2: Calculate the Corrected Standard Deviation
The corrected standard deviation (\(s\)) for grouped data is calculated using:

\[
s = \sqrt{\frac{\sum (x_i^2 \cdot f_i) - \left(\frac{\sum (x_i \cdot f_i)}{\sum f_i}\right)^2 \cdot \sum f_i}{\sum f_i - 1}}
\]

#### Calculate \(x_i^2 \cdot f_i\):
- \( (29.5)^2 \cdot 25600000 = 22110250000 \)
- \( (39.5)^2 \cdot 30200000 = 47279500000 \)
- \( (49.5)^2 \cdot 38400000 = 94104000000 \)
- \( (59.5)^2 \cdot 24500000 = 86738125000 \)

#### Sum of \(x_i^2 \cdot f_i\):
\[
\sum (x_i^2 \cdot f_i) = 22110250000 + 47279500000 + 94104000000 + 86738125000 = 250631875000
\]

#### Calculate the Corrected Standard Deviation:
\[
s = \sqrt{\frac{250631875000 - \left(\frac{5305450000}{118700000}\right)^2 \cdot 118700000}{118700000 - 1}}
\]

First, simplify the terms:
- \( \left(\frac{5305450000}{118700000}\right)^2 = (44.71)^2 = 1998.7241 \)
- \( 1998.7241 \cdot 118700000 = 237469315000 \)

Now, compute the variance:
\[
s^2 = \frac{250631875000 - 237469315000}{118699999} = \frac{13162560000}{118699999} \approx 110.88
\]

Taking the square root gives the standard deviation:
\[
s \approx \sqrt{110.88} \approx 10.53
\]

So, the approximate corrected standard deviation is **10.53 years**.

---

### Final Answer:
- **Mean age:** 44.71 years
- **Corrected standard deviation:** 10.53 years

Would you like further details on any of these calculations or have other questions? 

Here are 5 related questions that you might find useful:
1. How is grouped data different from ungrouped data in statistical calculations?
2. Why do we use the corrected formula for standard deviation with \((n-1)\)?
3. What is the significance of calculating the midpoint in grouped data?
4. How does the choice of age ranges affect the precision of the mean?
5. Can these methods be used for non-continuous data?

**Tip:** When working with large numbers, keeping intermediate results ensures accuracy and helps in tracking errors during calculations.

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 corrected standard deviation for age.

Explore how to approximate the mean and corrected standard deviation for grouped data based on the number of people aged 25 to 64 covered by health insurance in 2018. This problem covers core statistical concepts such as central tendency and dispersion, providing a step-by-step solution suitable for Grades 11-12 and college-level students.

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