Question

The article ``Oxygen Consumption During Fire Suppression: Error of Heart Rate Estimation" reported the following data on oxygen consumption (ml/kg/min) for a sample of ten firefighters performing a fire suppression simulation:

29.5 49.3 30.6 28.2 28.0 26.3 33.9 29.4 23.5 31.6
Compute the following:
a. The sample range.
b. The sample variance s2 from the definition (i.e. first computing deviations, then squaring them, etc.).
c. The sample standard deviation.
d. s2 using the shortcut method.

Answer 1

C

Step-by-step explanation:Hope this helps!:>

Answer 2

Kindly check explanation

Explanation:

Given the data:

29.5 49.3 30.6 28.2 28.0 26.3 33.9 29.4 23.5 31.6

Sort in ascending order of magnitude :

23.5,26.3,28.0,28.2,29.4,29.5,30.6,31.6,33.9,49.3

A) Range of the sample data:

(maximum - minimum)

49.3 - 23.5 = 25.8

B) sample variance :

S² = ∑ (x − m)²/ n − 1.

m = mean ; n = number of observations

m = (23.5 + 26.3 + 28.0 + 28.2 + 29.4 + 29.5 + 30.6 + 31.6 + 33.9 + 49.3) = 310.3 / 10 = 31.03

∑ (x − m)² = 443.801

n - 1 = 10 - 1 = 9

S² = ∑ (x − m)²/ n − 1

S² = 443.801 / 9

S² = 49.311222 = 49.31

C) The standard deviation (s):

S = √[∑ (x − m)²/ n − 1]

√variance = √49.311222

S = 7.0221949

S = 7.022

D) using statistical calculator,

S² = 49.31