Math Problem Statement

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.

Solution

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 (x\overline{x}) is calculated using the formula:

x=(xifi)fi\overline{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}

where:

  • xix_i is the midpoint of each age range.
  • fif_i is the frequency (number of people) in each range.

Calculate the Midpoints:

The midpoint (xix_i) of each age range is:

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

Multiply Midpoints by Frequencies (xifix_i \cdot f_i):

  • 29.525600000=75400000029.5 \cdot 25600000 = 754000000
  • 39.530200000=119290000039.5 \cdot 30200000 = 1192900000
  • 49.538400000=190080000049.5 \cdot 38400000 = 1900800000
  • 59.524500000=145775000059.5 \cdot 24500000 = 1457750000

Sum of Frequencies (fi\sum f_i):

fi=25600000+30200000+38400000+24500000=118700000\sum f_i = 25600000 + 30200000 + 38400000 + 24500000 = 118700000

Sum of xifix_i \cdot f_i:

(xifi)=754000000+1192900000+1900800000+1457750000=5305450000\sum (x_i \cdot f_i) = 754000000 + 1192900000 + 1900800000 + 1457750000 = 5305450000

Calculate the Mean:

x=530545000011870000044.71\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 (ss) for grouped data is calculated using:

s=(xi2fi)((xifi)fi)2fifi1s = \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 xi2fix_i^2 \cdot f_i:

  • (29.5)225600000=22110250000(29.5)^2 \cdot 25600000 = 22110250000
  • (39.5)230200000=47279500000(39.5)^2 \cdot 30200000 = 47279500000
  • (49.5)238400000=94104000000(49.5)^2 \cdot 38400000 = 94104000000
  • (59.5)224500000=86738125000(59.5)^2 \cdot 24500000 = 86738125000

Sum of xi2fix_i^2 \cdot f_i:

(xi2fi)=22110250000+47279500000+94104000000+86738125000=250631875000\sum (x_i^2 \cdot f_i) = 22110250000 + 47279500000 + 94104000000 + 86738125000 = 250631875000

Calculate the Corrected Standard Deviation:

s=250631875000(5305450000118700000)21187000001187000001s = \sqrt{\frac{250631875000 - \left(\frac{5305450000}{118700000}\right)^2 \cdot 118700000}{118700000 - 1}}

First, simplify the terms:

  • (5305450000118700000)2=(44.71)2=1998.7241\left(\frac{5305450000}{118700000}\right)^2 = (44.71)^2 = 1998.7241
  • 1998.7241118700000=2374693150001998.7241 \cdot 118700000 = 237469315000

Now, compute the variance: s2=250631875000237469315000118699999=13162560000118699999110.88s^2 = \frac{250631875000 - 237469315000}{118699999} = \frac{13162560000}{118699999} \approx 110.88

Taking the square root gives the standard deviation: s110.8810.53s \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 (n1)(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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Grouped Data Analysis
Measures of Central Tendency
Measures of Dispersion

Formulas

Mean (\overline{x}) = \frac{\sum (x_i \cdot f_i)}{\sum f_i}
Corrected Standard Deviation (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}}

Theorems

Central Limit Theorem (underlying statistical concept)
Corrected standard deviation formula for sample data

Suitable Grade Level

Grades 11-12, College Level