Q. 19 - Elementary Statistics

Question

Road Patrol. In the paper "Injuries and Risk Factors in a Vol.  K. Reynolds et al. reported on a study commissioned by the U.S. Army. The purpose of the study was to improve medical planning and identify risk factors during multiple-day road patrols by examining the acute effects of long-distance marches by light-infantry soldiers. Each soldier carried a standard U.S. Army rucksack. Meal-Ready-to-Eat packages. and other field equipment. A sample of $10$ participating soldiers revealed the following data on total load mass, in kilograms.

a.Obtain the sample mean of these 10 load masses.

b.Obtain the range of the load masses.

c.Obtain the sample standard deviation of the load masses

Step-by-Step Solution

Answer

Part(a) The sample mean is $45.7$

Part(b) The range of load masses is $17$

Part(c) The standard deviation is $4.09$

Part(a) Step 1: Given information

We are given a sample of participating soldiers with load mass.

Part(a) Step 2: Simplify

As we know mean of sample is sum of all data divided by total number of data. Therefore, the sample mean is 

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _{i=n}}{x}_i}{n} \\ \overline{x}=\frac{48+50+45+49+44+47+37+54+40+43}{10}=45.7 \end{gathered}$$

Part(b) Step 1: Given information

We are given a sample of participating soldiers with load mass.

Part(b) Step 2: Simplify

Range of sample is difference of highest value of data and lowest value of data.

Highest $=54$

Lowest $=37$

Range $=54-37=17$

Part(c) Step 1: Given information

We are given a sample of participating soldiers with load mass.

Part(c) Step 2: Simplify

Standard deviation of sample is given by 

$\sigma =\sum _{}\frac{x^2}{n}-{\left(\overline{x}\right)}^2$

Where,

$\overline{x}=$mean of sample

$n=$Total number of sample

$x=$data in sample

Now,

$$\begin{gathered} \frac{{\displaystyle \sum _{}}{x}^2}{n}=\frac{{48}^2+{50}^2+{45}^2+{49}^2+{44}^2+{47}^2+{37}^2+{54}^2+{40}^2+{43}^2}{10} \\ \frac{{\displaystyle \sum _{}{x}^2}}{n}=2110.9 \\ {\left(\overline{x}\right)}^2={\left(45.9\right)}^2=2106.81 \\ \sigma =2110.9-2106.81=4.09 \end{gathered}$$