Question

An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation:

28.2 49.2 30.9 28.8 28.0 25.9 34.0 29.0 23.8 30.1

Compute the following. (Round your answers to four decimal places.)

a. The sample range.
b. The sample variance s2 from the definition
c. The sample standard deviation
d. s2 using the shortcut method.

Answer 1

a)Range= 25.9

b)Variance =49.344

c)Standard deviation=7.0245

Explanation:

Data : 28.2 49.2 30.9 28.8 28.0 25.9 34.0 29.0 23.8 30.1

Maximum Value : 49.4

Minimum Value : 23.5

Range= Maximum - Minimum

Range= 49.4 -23.5 = 25.9

(b)

`Mean=\bar{x}=(\sum x)/(n)=(309.2)/(10)=30.92\\\\Variance=s^(2)=\frac{\sum \left ( x-\bar{x} \right )^(2)}{n-1}=49.344`

(c) Sample standard deviation:

`s=\sqrt{\frac{\sum \left ( x-\bar{x} \right )^(2)}{n-1}}=7.0245`

d)s2 using the shortcut method.

X
`X^2`

28.6 817.96

49.4 2440.36

30.3 918.09

28.2 795.24

28.9 835.21

26.4 696.96

33.8 1142.44

29.9 894.01

23.5 552.25

30.2 912.04

Total 309.2 10004.56

`s=\sqrt{(\sum x^(2)_(i)-(\left ( \sum x_(i) \right )^(2))/(n))/(n-1)}\\s=(10004.56-(309.2^(2))/(10))/(9)\\s=7.0245`