Question

​Recently, a random sample of 2534 year olds was​ asked, "How much do you currently have in​ savings, not including retirement​ savings?" The data in the table represent the responses to the survey. Approximate the mean and standard deviation amount of savings.

Savings Lower Limit Upper Limit Frequency
0-199 0 199 345
200-399 200 399 97
400-599 400 599 52
600-799 600 799 21
800-999 800 999 9
1000-1199 1000 1199 8
1200-1399 1200 1399 3

Answer 1

The approximations of the mean and the standard deviation are 233.3 and 229.82, respectively

How to determine the mean?

The table of values is given as:

Savings Lower Limit Upper Limit     Frequency

0-199         0         199                       345

200-399   200   399                         97

400-599   400   599                         52

600-799   600   799                         21

800-999   800   999                         9

1000-1199 1000  1199                         8

1200-1399 1200  1399                      3

Rewrite the table to include the class midpoint and the frequency

x               f

99.5         345

299.5        97

499.5        52

699.5        21

899.5        9

1099.5        8

1299.5        3

The mean is calculated as:

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{99.5* 345 + 299.5* 97 + 499.5* 52 + 699.5 * 21 + 899.5 * 9 + 1099.5 * 8 + 1299.5 * 3}{345 + 97 + 52 + 21 + 9 + 8 +3}$

Evaluate

$\bar x = 233.331775701$

Approximate

$\bar x = 233.3$

Hence, the approximation of the mean is 233.3

How to determine the standard deviation?

The standard deviation is calculated as:

$\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}$

So, we have:

$\sigma= \sqrt{\frac{(99.5-233.3)^2* 345 + (299.5-233.3)^2* 97 +...... + (1299.5 -233.3)^2* 3}{345 + 97 + 52 + 21 + 9 + 8 +3}$

Evaluate

$\sigma = 229.82$

Hence, the approximation of the standard deviation is 229.82

Read more about mean and standard deviation at:

https://brainly.com/question/475676

#SPJ1