Question

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.
Age: 25-34. 35-44. 45-54. 55-64
Number 22.1. 31.5. 37.7. 25.3
(Millions)

Answer 1

Answer:

The mean is 45.178

The standard deviation is 10.3325

Step-by-step explanation:

The mean is 45.178

The standard deviation is 10.3325

Answer 2

Answer:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{N}= \frac{5267.7}{116.6}= 45.178$

And the sample variance can be calculated with this formula:

$s^2 = \frac{\sum fx^2 -\frac{(\sum x*f)^2}{n}}{n-1}= \frac{250323.15 - \frac{(5267.7)^2}{116.6}}{116.6-1} = 106.7601$

And the deviation would be:

$s = \sqrt{106.7601}= 10.332$

Step-by-step explanation:

For this case we can calculate the mean with the following table

Age        Number (f)    Midpoint    x*f                 x^2 *f

25-34         22.1            29.5         651.95         19232.525

35-44         31.5            39.5         1244.25       49147.875

45-54         37.7           49.5         1866.15         92374.425

55-64         25.3           59.5        1505.35        89568.325

Total          116.6                             5267.7        250323.15

And the mean can be calculatd with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{N}= \frac{5267.7}{116.6}= 45.178$

And the sample variance can be calculated with this formula:

$s^2 = \frac{\sum fx^2 -\frac{(\sum x*f)^2}{n}}{n-1}= \frac{250323.15 - \frac{(5267.7)^2}{116.6}}{116.6-1} = 106.7601$

And the deviation would be:

$s = \sqrt{106.7601}= 10.332$