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

Question

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Answer 1

Question

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

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Answer 2

Question

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

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Answer 3

Question

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

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Answer 4

Question

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

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Answer 5

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

Answer 6

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

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)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=27.5 years

Therefore, the average age is 27.5 years_.

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=17312−1=15.727 (years)2

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

Substitute 173 (years)² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=17312−1=3.966 years

Therefore, the standard deviation of age is 3.966 years_.

Answer 7

Chapter 19, Problem 12P

To determine

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

Answer 8

Expert Solution & Answer

Answer to Problem 12P

The average height is 200.75 cm_.

The variance of height is 44.205 cm2_.

The standard deviation of height is 6.649 cm_.

The average mass is 99 kg_.

The variance of mass is 135.782 kg2_.

The standard deviation of mass is 11.653 kg_.

The average age is 27.5 years_.

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

The standard deviation of age is 3.966 years_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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 (x¯) using the equation.

x¯=1n∑t=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(xi−x¯)2n−1 (2)

Find the standard deviation (s) using the equation.

s=∑t=1n(xi−x¯)2n−1 (3)

Height:

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

S No	height (cm) (xi)	(xi−x¯), cm	(xi−x¯)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
	∑t=1nxi=2409		∑t=1n(xi−x¯)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=200.75 cm

Therefore, the average height is 200.75 cm_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=486.2512−1=44.205 cm2

Therefore, the variance of height is 44.205 cm2_.

Substitute 486.25 cm² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=486.2512−1=6.649 cm

Therefore, the standard deviation of height is 6.649 cm_.

Mass:

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

S No	mass (kg) (xi)	(xi−x¯), kg	(xi−x¯)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
	∑t=1nxi=1188		∑t=1n(xi−x¯)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=99 kg

Therefore, the average mass is 99 kg_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (2).

v=1493.612−1=135.782 kg2

Therefore, the variance of mass is 135.782 kg2_.

Substitute 1493.6 kg² for ∑t=1n(xi−x¯)2 and 12 for n in Equation (3).

s=1493.612−1=11.653 kg

Therefore, the standard deviation of mass is 11.653 kg_.

Age:

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

S No	Age (years) (xi)	(xi−x¯), years	(xi−x¯)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
	∑t=1nxi=330		∑t=1n(xi−x¯)2=173