Chapter 19, Problem 12P

To determine

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

Answer to Problem 12P

The average height is $\underset{\_}{200.75\text{cm}}$.

The variance of height is $\underset{\_}{44.205{\text{cm}}^2}$.

The standard deviation of height is $\underset{\_}{6.649\text{cm}}$.

The average mass is $\underset{\_}{99\text{kg}}$.

The variance of mass is $\underset{\_}{135.782{\text{kg}}^2}$.

The standard deviation of mass is $\underset{\_}{11.653\text{kg}}$.

The average age is $\underset{\_}{27.5\text{years}}$.

The variance of age is $\underset{\_}{15.727{\left(\text{years}\right)}^2}$.

The standard deviation of age is $\underset{\_}{3.966\text{years}}$.

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.

Players	Age (years)	height (cm)	mass (kg)
1	28	201	97.5
2	30	206	108.9
3	30	190	86.2
4	28	211	122.5
5	24	206	101.6
6	35	198	97.5
7	25	188	81.2
8	22	206	99.8
9	33	201	87.1
10	26	203	97.5
11	23	201	112.9
12	26	198	95.3

Find the average $\left(\overline{x}\right)$ using the equation.

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{t=1}^n{x}_i}$ (1)

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

The number of observations is 12.

Find the variance (v) using the equation.

$v=\frac{{\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}$ (2)

Find the standard deviation (s) using the equation.

$s=\sqrt{\frac{{\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}}$ (3)

Height:

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

S No	height (cm) $\left(x_i\right)$	$\left(x_i-\overline{x}\right)$, cm	${\left(x_i-\overline{x}\right)}^2$, cm2
1	201	0.25	0.0625
2	206	5.25	27.5625
3	190	–10.75	115.563
4	211	10.25	105.063
5	206	5.25	27.5625
6	198	–2.75	7.5625
7	188	–12.75	162.563
8	206	5.25	27.5625
9	201	0.25	0.0625
10	203	2.25	5.0625
11	201	0.25	0.0625
12	198	–2.75	7.5625
	${\displaystyle \sum _{t=1}^n{x}_i}=2409$		${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}=486.25$

Refer the Table 2 for the calculation values.

Substitute 2409 cm for ${\displaystyle \sum _{t=1}^n{x}_i}$ and 12 for n in Equation (1).

$\begin{array}{c}\overline{x}=\frac{2409}{12}\\ =200.75\text{cm}\end{array}$

Therefore, the average height is $\underset{\_}{200.75\text{cm}}$.

Substitute 486.25 cm² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (2).

$\begin{array}{c}v=\frac{486.25}{12-1}\\ =44.205{\text{cm}}^2\end{array}$

Therefore, the variance of height is $\underset{\_}{44.205{\text{cm}}^2}$.

Substitute 486.25 cm² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (3).

$\begin{array}{c}s=\sqrt{\frac{486.25}{12-1}}\\ =6.649\text{cm}\end{array}$

Therefore, the standard deviation of height is $\underset{\_}{6.649\text{cm}}$.

Mass:

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

S No	mass (kg) $\left(x_i\right)$	$\left(x_i-\overline{x}\right)$, kg	${\left(x_i-\overline{x}\right)}^2$, kg2
1	97.5	–1.5	2.25
2	108.9	9.9	98.01
3	86.2	–12.8	163.84
4	122.5	23.5	552.25
5	101.6	2.6	6.76
6	97.5	–1.5	2.25
7	81.2	–17.8	316.84
8	99.8	0.8	0.64
9	87.1	–11.9	141.61
10	97.5	–1.5	2.25
11	112.9	13.9	193.21
12	95.3	–3.7	13.69
	${\displaystyle \sum _{t=1}^n{x}_i}=1188$		${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}=1493.6$

Refer to Table 2 for the calculation values.

Substitute 1188 kg for ${\displaystyle \sum _{t=1}^n{x}_i}$ and 12 for n in Equation (1).

$\begin{array}{c}\overline{x}=\frac{1188}{12}\\ =99\text{kg}\end{array}$

Therefore, the average mass is $\underset{\_}{99\text{kg}}$.

Substitute 1493.6 kg² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (2).

$\begin{array}{c}v=\frac{1493.6}{12-1}\\ =135.782{\text{kg}}^2\end{array}$

Therefore, the variance of mass is $\underset{\_}{135.782{\text{kg}}^2}$.

Substitute 1493.6 kg² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (3).

$\begin{array}{c}s=\sqrt{\frac{1493.6}{12-1}}\\ =11.653\text{kg}\end{array}$

Therefore, the standard deviation of mass is $\underset{\_}{11.653\text{kg}}$.

Age:

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

S No	Age (years) $\left(x_i\right)$	$\left(x_i-\overline{x}\right)$, years	${\left(x_i-\overline{x}\right)}^2$, (years)2
1	28	0.5	0.25
2	30	2.5	6.25
3	30	2.5	6.25
4	28	0.5	0.25
5	24	–3.5	12.25
6	35	7.5	56.25
7	25	–2.5	6.25
8	22	–5.5	30.25
9	33	5.5	30.25
10	26	–1.5	2.25
11	23	–4.5	20.25
12	26	–1.5	2.25
	${\displaystyle \sum _{t=1}^n{x}_i}=330$		${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}=173$

Refer the Table 2 for the calculation values.

Substitute 330 years for ${\displaystyle \sum _{t=1}^n{x}_i}$ and 12 for n in Equation (1).

$\begin{array}{c}\overline{x}=\frac{330}{12}\\ =27.5\text{years}\end{array}$

Therefore, the average age is $\underset{\_}{27.5\text{years}}$.

Substitute 173 (years)² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (2).

$\begin{array}{c}v=\frac{173}{12-1}\\ =15.727{\left(\text{years}\right)}^2\end{array}$

Therefore, the variance of age is $\underset{\_}{15.727{\left(\text{years}\right)}^2}$.

Substitute 173 (years)² for ${\displaystyle \sum _{t=1}^n{\left(x_i-\overline{x}\right)}^2}$ and 12 for n in Equation (3).

$\begin{array}{c}s=\sqrt{\frac{173}{12-1}}\\ =3.966\text{years}\end{array}$

Therefore, the standard deviation of age is $\underset{\_}{3.966\text{years}}$.