Answer to Question #330241 in Statistics and Probability for zia

Question

The following data are the measures of the diameters of 36 rivet heads in 1/100 of an inch.

6.72 6.77 6.82 6.70 6.78 6.70 6.62 6.75

6.66 6.66 6.64 6.76 6.73 6.80 6.72 6.76

6.76 6.68 6.66 6.62 6.72 6.76 6.70 6.78

6.76 6.67 6.70 6.72 6.74 6.81 6.79 6.78

6.66 6.76 6.76 6.72

Compute the Arithmetic Mean, variance, standard deviation, Coecient of Variation, Coe-

cient of Skewness and Coecient of Kurtosis..

Answer 1

First let's sort the data, write appearing frequencies of each value and corresponding probabilities of appearing (the number of different values will be 17):

6.62 - 2, p = 0.117647

6.64 - 1, p = 0.0588235

6.66 - 4, p = 0.235294

6.67 - 1, p = 0.0588235

6.68 - 1, p = 0.0588235

6.7 - 4, p = 0.235294

6.72 - 5, p = 0.294118

6.73 - 1, p = 0.0588235

6.74 - 1, p = 0.0588235

6.75 - 1, p = 0.0588235

6.76 - 7, p = 0.411765

6.77 - 1, p = 0.0588235

6.78 - 3, p = 0.176471

6.79 - 1, p = 0.0588235

6.8 - 1, p = 0.0588235

6.81 - 1, p = 0.0588235

6.82 - 1, p = 0.0588235

Let n = 36

Mean:

$\mu=\frac{1}{n}\sum_{k=1}^nx_k=\\=\frac{1}{36}(6.62\cdot2+6.64+6.66\cdot4+6.67+\\ +6.68+6.7\cdot4+6.72\cdot5+6.73+6.74+\\ +6.75+6.76\cdot7+6.77+6.78\cdot3+6.79+\\ 6.8+6.81+6.82)\approx6.73$

Variance:

$\sigma^2=\frac{1}{n-1}\sum_{k=1}^n(x_k-\mu)^2=\\ =\frac{1}{35}(0.11^2\cdot2+0.09^2+0.07^2\cdot4+0.06^2+\\ +0.05^2+0.03^2\cdot4+0.01^2\cdot5+0.01^2+0.02^2+\\ +0.03^2\cdot7+0.04^2+0.05^2\cdot3+0.06^2+\\ +0.07^2+0.08^2+0.09^2)\approx0.002885714\approx0.00289$

Standard deviation:

$\sigma=\sqrt{\sigma^2}\approx0.0537$

Coefficient of variation:

$CV=\frac{\sigma}{\mu}\approx0.008$

Coefficient of skewness:

$b=\frac{1}{n}\sum_{k=1}^n(x_k-\mu)^3/\sigma^3=\\ \frac{1}{36\cdot0.0537^3}((-0.11)^3\cdot2+(-0.09)^3+\\ +(-0.07)^3\cdot4+(-0.06)^3+\\ +(-0.05)^3+(-0.03)^3\cdot4+(-0.01)^3\cdot5+\\ +0.01^3+0.02^3+\\ +0.03^3\cdot7+0.04^3+0.05^3\cdot3+0.06^3+\\ +0.07^3+0.08^3+0.09^3)\approx−18$

Coefficient of Kurtosis:

$K=\frac{n(n+1)}{(n-1)(n-2)(n-3)}\frac{1}{\sigma^4}\sum_{k=1}^n(x_k-\mu)^4=\\ =\frac{36\cdot37}{35\cdot34\cdot33}\frac{1}{0.0537^4}(0.11^4\cdot2+0.09^4+\\ +0.07^4\cdot4+0.06^4+\\ +0.05^4+0.03^4\cdot4+0.01^4\cdot5+0.01^4+0.02^4+\\ +0.03^4\cdot7+0.04^4+0.05^4\cdot3+0.06^4+\\ +0.07^2+0.08^2+0.09^2)\approx2.64$