Download PDF by Murray Spiegel, Larry Stephens: Schaum's outline of theory and problems of statistics

By Murray Spiegel, Larry Stephens

ISBN-10: 0071485848

ISBN-13: 9780071485845

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Example text

120 0:0001980 ð416; 000Þð0:000187Þ pffiffiffiffiffiffiffiffiffiffiffi 73:84 Evaluate each of the following, given that U ¼ À2, V ¼ 12, W ¼ 3, X ¼ À4, Y ¼ 9, and Z ¼ 16, where all numbers are assumed to be exact. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðW À 2Þ2 ðY À 5Þ2 (a) 4U þ 6V À 2W (g) þ V Z (b) XYZ UVW (h) X À3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðY À 4Þ2 þ ðU þ 5Þ2 (c) 2X À 3Y UW þ XV (i) X 3 þ 5X 2 À 6X À 8 32 VARIABLES AND GRAPHS (d ) 3ðU À XÞ2 þ Y ( j) (e) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U 2 À 2UV þ W (f) 3Xð4Y þ 3ZÞ À 2Yð6X À 5ZÞ À 25 [CHAP.

Graphic representation of relative-frequency distributions can be obtained from the histogram or frequency polygon simply by changing the vertical scale from frequency to relative frequency, keeping exactly the same diagram. The resulting graphs are called relative-frequency histograms (or percentage histograms) and relative-frequency polygons (or percentage polygons), respectively. 40 FREQUENCY DISTRIBUTIONS [CHAP. 2 CUMULATIVE-FREQUENCY DISTRIBUTIONS AND OGIVES The total frequency of all values less than the upper class boundary of a given class interval is called the cumulative frequency up to and including that class interval.

6667 into the original equation, we have 0:8239 À ðÀ0:1761Þ ¼ 1. 6667. CHAP. 41 VARIABLES AND GRAPHS 29 Solve the logarithmic equation lnðxÞ2 À 1 ¼ 0. SOLUTION The equation may be factored as ½lnðxÞ þ 1Š½lnðxÞ À 1Š ¼ 0. Setting the factor lnðxÞ þ 1 ¼ 0, we have lnðxÞ ¼ À1 or x ¼ eÀ1 ¼ 0:3678. Setting the factor lnðxÞ À 1 ¼ 0, we have lnðxÞ ¼ 1 or x ¼ e1 ¼ 2:7183. Both values are solutions to the equation. 42 Solve the following equation for x: 2 logðx þ 1Þ À 3 logðx þ 1Þ ¼ 2. SOLUTION The equation may be rewritten as log½ðx þ 1Þ2 =ðx þ 1Þ3 Š ¼ 2 or log½1=ðx þ 1ފ ¼ 2 or logð1Þ À logðx þ 1Þ ¼ 2 or 0 À logðx þ 1Þ ¼ 2 or logðx þ 1Þ ¼ À2 or x þ 1 ¼ 10À2 or x ¼ À0:99.

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Schaum's outline of theory and problems of statistics by Murray Spiegel, Larry Stephens

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