Question

about ages of people in apartodar regon A have te Rotowing ages , in years. Complete parts (a) tough (d)b a. Determine the range of these ages. The range is- Simplity your answer.) the sample standard deviation of these ages by using the defining formiula s b. Find sample size and x is the sample mean. The standard deviation is (Round to one decimal place as needed.) where n is the sze.le Standard deviation of these ages by usingthe computing formi..! -n-1h, ages by using the computing formula s the sample size. he standard deviation is Round to one decimal place as needed.) Compare your work in parts (b) and (c). answer to part (b) is (1)the answer to part (c). O the same as different from antitative data set under consideration has roughly a bell-shaped distribution. Apply the empirical ru

Answer 1

(a)

Range = Maximum - Minimum = 58 - 9 = 49

So,

Answer is:

49.

(b)

Sample mean ($\bar{x}$) is given by:

$\bar{x}=\frac{\sum_{i=1}^{5}x_{i}}{5}=\frac{160}{5}=32$

x	x - $\bar{x}$	(x -$\bar{x}$)2
21	-11	121
58	26	676
18	- 14	196
54	22	484
9	- 23	529
Total =		2006

So,

Standard Deviation is given by:

$s=\sqrt{\frac{\sum_{i=1}^{5}(x_{i}-\bar{x})^{2}}{4}}=\sqrt{\frac{2006}{4}}=\sqrt{501.5}=22.4$

So,

Answer is:

22.4

(c)

Sample mean ($\bar{x}$) is given by:

$\bar{x}=\frac{\sum_{i=1}^{5}x_{i}}{5}=\frac{160}{5}=32$

x	x2
21	441
58	3364
18	324
54	2916
9	81
Total = 160	7126

So,

Standard Deviation is given by:

$s=\sqrt{\frac{\sum x^{2}_{i}-\frac{(\sum x_{i})^{2}}{n}}{n-1}}$

$s=\sqrt{\frac{7126-\frac{160^{2}}{5}}{5-1}}=\sqrt{\frac{7126-5120}{4}}=\sqrt{\frac{2006}{4}}=\sqrt{501.5}=22.4$

So,

Answer is:

22.4

(d)

Answer to part (b) is same as the answer to part (c).