Statistics

Problem Set due Monday September 19 spell 1 According to results derived from Normal Distribution, about 1.09 percent of the students scored the equivalent or lower than Joe, which means Joe did better than only 1.09% of the test-takers. Since whiz was in the top 5% of the students, he scored 789 on the CEEB accord to the deliberation below. About 4.7% of the students scored within the range. The mean of the temperatures is 83.85714 and the normal is 84. The variation and ensample deviation for temperatures atomic moment 18 34.14286 and 5.843189, respectively. The mean of the numerate of chirrups/ fine is 51.14286 and the median is 50, with the variance and standard deviation creation 63.14286 and 7.946248, respectively. The covariance between temperature and chirrups is 45.3571. By standardizing covariance we can get correlation between the two inconstants, which is a number locating between -1 and 1. In this case, the correlation is 0.9769, which indicat es actually strong relations between temperatures and chirrups.

The domineering correlation shows they be positive related, meaning when one variable goes up (down), the some other variable also goes up (down). Number of chirrups = -60.25732 +1.328452 ×Temperature The intercept is -60.25732. Literally, when temperature locomote to 0, the number of chirrups would be -60.25732. However, since the number of chirrups will never be negative, the value of intercept is unreal in reality. The slope of the regression is 1.328452. This result reveals that the dependent variable and the independent variable are positi ve related. If temperature goes up (down) by! 1 degree, the number of chirrups/minute will go up (down) by 1.328452 unit accordingly.If you deprivation to get a liberal essay, order it on our website: BestEssayCheap.com

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