The following data represents the average prices of gold (in dollars per fine ounce) for the years 1981 to 2000. Find the mean, variance, and standard deviation of the data. What percentage of the data lies within two standard deviations of the mean?
460, 376, 424, 361, 318,
368, 478, 438, 383, 385,
363, 345, 361, 385, 386,
389, 332, 295, 280, 280
(Source: U.S. Bureau of Mines and U.S. Geological
Survey)
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Solution:
ReplyDelete(a) Mean= (∑_(i=1)^n Xi)/n
= (460+368+363+389+376+478+345+332+424+438+361+295+361+383+385+280+318+385+386+280)/20
= 370.35
(b) Standard deviation = 53.99 (obtained from calculator)
(c) Variance = ∑(Xi - X (mean))/ (n - 1)
= 2915.08 (calculator obtained value)
(d) % = 1 - 1/k^2
= 1 - 1/2^2
= 75%