Calculate the mean, standard deviation, and variance for the given measured values.
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,
ˉx=x1+x2+x3+............+xn−1+xnn=1nn∑i=1xi (1)
Here,
ˉx 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=n∑i=1(xi−ˉx)2n−1 (2)
From equation 19.6 in the textbook, the formula to find standard deviation is,
s=√n∑i=1(xi−ˉx)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 (ˉx),
ˉx=3.50+3.55+3.45+3.60+3.55+3.40+3.40+3.65+3.35+3.6010=35.0510ˉx=3.505
Substitute all the value of lumber width for xi up to the range n, 3.505 for ˉx, 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 ˉx, 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 (ˉx),
ˉx=1.00+0.95+1.05+1.10+1.00+0.90+0.85+1.05+0.95+0.9010=9.7510ˉx=0.975
Substitute all the value of spherical balls for xi up to the range n, 0.975 for ˉx, 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 ˉx, 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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