Calculate the mean, standard deviation, and variance for the given measured values.

Question

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Answer 1

Question

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

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Answer 2

Question

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

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Answer 3

Question

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

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Answer 4

Question

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

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Answer 5

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Answer 6

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Answer 7

Chapter 19, Problem 9P

To determine

Calculate the mean, standard deviation, and variance for the given measured values.

Answer 8

Expert Solution & Answer

Answer to Problem 9P

The mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively.

The mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The given measured values of lumber width and steel spherical balls are shown below.

Lumber Width (in.)	Steel spherical balls (cm)
3.50	1.00
3.55	0.95
2.55	1.05
3.60	1.10
3.55	1.00
3.40	0.90
3.40	0.85
3.65	1.05
3.35	0.95
3.60	0.90

The total number of measured values, $n=10$ .

Formula used:

From equation 19.1 in the textbook, the formula to find mean for any sample is,

$x¯=x1+x2+x3+............+xn−1+xnn=1n∑i=1nxi$ (1)

Here,

$x¯$ is the mean,

$xi$ is the data points,

$n$ is the number of data points.

From equation 19.5 in the textbook, the formula to find the variance is,

$v=∑i=1n(xi−x¯)2n−1$ (2)

From equation 19.6 in the textbook, the formula to find standard deviation is,

$s=∑i=1n(xi−x¯)2n−1$ (3)

Calculation:

Calculation for Lumber width:

Substitute all the value of lumber width for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510x¯=3.505$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1=0.092259v=0.010$

Substitute all the value of lumber width for $xi$ up to the range $n$ , $3.505$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(3.50−3.505)2+(3.55−3.505)2+(3.45−3.505)2+(3.60−3.505)2+(3.55−3.505)2+(3.40−3.505)2+(3.40−3.505)2+(3.65−3.505)2+(3.35−3.505)2+(3.60−3.505)210−1$

$s=0.092259s=0.010s=0.101$

Calculation for steel spherical balls:

Substitute all the value of spherical balls for $xi$ up to the range $n$ , and $10$ for $n$ in equation (1) to calculate mean $(x¯)$ ,

$x¯=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510x¯=0.975$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (2) to find variance $(v)$ ,

$v=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629v=0.006$

Substitute all the value of spherical balls for $xi$ up to the range $n$ , $0.975$ for $x¯$ , and $10$ for $n$ in equation (3) to find standard deviation $(s)$ ,

$s=(1.00−0.975)2+(0.95−0.975)2+(1.05−0.975)2+(1.10−0.975)2+(1.00−0.975)2+(0.90−0.975)2+(0.85−0.975)2+(1.05−0.975)2+(0.95−0.975)2+(0.90−0.975)210−1=0.05629=0.006s=0.079$

Therefore, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Conclusion:

Thus, the mean, variance, and standard deviation for the given values of lumber width are $3.505$ , $0.010$ , and $0.101$ respectively, and the mean, variance, and standard deviation for the given values of steel spherical balls are $0.975$ , $0.006$ , and $0.079$ respectively.

Answer 12

Ch. 19.3 - Prob. 1BYG Ch. 19.3 - Prob. 2BYG Ch. 19.3 - Prob. 3BYG Ch. 19.3 - VocabularyState the meaning of the following...Ch. 19.5 - Prob. 1BYG Ch. 19.5 - Prob. 2BYG Ch. 19.5 - Prob. 3BYG Ch. 19.5 - Prob. 4BYG Ch. 19.5 - Prob. 5BYG Ch. 19.5 - Prob. BYGV

Calculate the mean, standard deviation, and variance for the given measured values.

Calculate the mean, standard deviation, and variance for the given measured values.

Answer to Problem 9P

Explanation of Solution

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