EBK ENGINEERING FUNDAMENTALS: AN INTROD
EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
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Chapter 19, Problem 12P
To determine

Find the average, variance, and standard deviation for the height, age, and mass of a basketball team.

Expert Solution & Answer
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Answer to Problem 12P

The average height is 200.75cm_.

The variance of height is 44.205cm2_.

The standard deviation of height is 6.649cm_.

The average mass is 99kg_.

The variance of mass is 135.782kg2_.

The standard deviation of mass is 11.653kg_.

The average age is 27.5years_.

The variance of age is 15.727(years)2_.

The standard deviation of age is 3.966years_.

Explanation of Solution

Consider Golden state warriors (GSW) basketball team for the analysis.

Tabulate the details collected from the GSW team as in Table 1.

PlayersAge (years)height (cm)mass (kg)
12820197.5
230206108.9
33019086.2
428211122.5
524206101.6
63519897.5
72518881.2
82220699.8
93320187.1
102620397.5
1123201112.9
122619895.3

Find the average (x¯) using the equation.

x¯=1nt=1nxi (1)

Here, the data points are xi and the number of data points is n.

The number of observations is 12.

Find the variance (v) using the equation.

v=t=1n(xix¯)2n1 (2)

Find the standard deviation (s) using the equation.

s=t=1n(xix¯)2n1 (3)

Height:

Calculate the values to find the average, variance, and standard deviation as in Table 2.

S Noheight (cm) (xi)(xix¯), cm(xix¯)2, cm2
12010.250.0625
22065.2527.5625
3190–10.75115.563
421110.25105.063
52065.2527.5625
6198–2.757.5625
7188–12.75162.563
82065.2527.5625
92010.250.0625
102032.255.0625
112010.250.0625
12198–2.757.5625
t=1nxi=2409t=1n(xix¯)2=486.25

Refer the Table 2 for the calculation values.

Substitute 2409 cm for t=1nxi and 12 for n in Equation (1).

x¯=240912=200.75cm

Therefore, the average height is 200.75cm_.

Substitute 486.25 cm2 for t=1n(xix¯)2 and 12 for n in Equation (2).

v=486.25121=44.205cm2

Therefore, the variance of height is 44.205cm2_.

Substitute 486.25 cm2 for t=1n(xix¯)2 and 12 for n in Equation (3).

s=486.25121=6.649cm

Therefore, the standard deviation of height is 6.649cm_.

Mass:

Calculate the values to find the average, variance, and standard deviation as in Table 3.

S Nomass (kg) (xi)(xix¯), kg(xix¯)2, kg2
197.5–1.52.25
2108.99.998.01
386.2–12.8163.84
4122.523.5552.25
5101.62.66.76
697.5–1.52.25
781.2–17.8316.84
899.80.80.64
987.1–11.9141.61
1097.5–1.52.25
11112.913.9193.21
1295.3–3.713.69
t=1nxi=1188t=1n(xix¯)2=1493.6

Refer to Table 2 for the calculation values.

Substitute 1188 kg for t=1nxi and 12 for n in Equation (1).

x¯=118812=99kg

Therefore, the average mass is 99kg_.

Substitute 1493.6 kg2 for t=1n(xix¯)2 and 12 for n in Equation (2).

v=1493.6121=135.782kg2

Therefore, the variance of mass is 135.782kg2_.

Substitute 1493.6 kg2 for t=1n(xix¯)2 and 12 for n in Equation (3).

s=1493.6121=11.653kg

Therefore, the standard deviation of mass is 11.653kg_.

Age:

Calculate the values to find the average, variance, and standard deviation as in Table 4.

S NoAge (years) (xi)(xix¯), years(xix¯)2, (years)2
1280.50.25
2302.56.25
3302.56.25
4280.50.25
524–3.512.25
6357.556.25
725–2.56.25
822–5.530.25
9335.530.25
1026–1.52.25
1123–4.520.25
1226–1.52.25
t=1nxi=330t=1n(xix¯)2=173

Refer the Table 2 for the calculation values.

Substitute 330 years for t=1nxi and 12 for n in Equation (1).

x¯=33012=27.5years

Therefore, the average age is 27.5years_.

Substitute 173 (years)2 for t=1n(xix¯)2 and 12 for n in Equation (2).

v=173121=15.727(years)2

Therefore, the variance of age is 15.727(years)2_.

Substitute 173 (years)2 for t=1n(xix¯)2 and 12 for n in Equation (3).

s=173121=3.966years

Therefore, the standard deviation of age is 3.966years_.

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