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2_Problem_Set

# 2_Problem_Set - c s = 9.6 mean = 93.2 The calculation...

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Professor Owens PAM 210 Problem Set 2 2/1/08 Question #1 a. ; z-score 1= .8413; z-score -1= .1587 area between 9.7 and 14.3 = .6826 b. ; z-score 2= .9772; z-score -2= .0228 area between 7.4 and 16.6 = .9544 c. z-score 2= .9772; z-score -1= .1587 area between 9.7 and 16.6 = .8185 d. ;

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z-score -3= .0013; z-score 3= .9987 area less than 5.1 = .0013; area more than 18.9 = .0013 Question #2 The distribution for the amounts of chloroform presents in water source, with a mean of 34 μg/L and standard deviation of 53 μg/L, does not have a normal distribution. This is because the left side of the graph ends before one standard deviation while the right side continues on until zero. This is due to the mean being smaller than the standard deviation and yields a skewed graph. Question #3 a. min=78; max=112; median=94; estimated standard deviation: 9 b. for histogram: see attached; visual approximation of mean: 94; visual approximation of standard deviation: 9.5
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Unformatted text preview: c. ; s = 9.6; mean = 93.2 The calculation checks proved to be quite close to the actual calculated values. d. 1 st interval: 83.6 to 102.8; count: 13 2 nd interval: 74 to 112.4; count: 21 3 rd interval: 64.4 to 122; count: 21 The fractions falling in each interval are close to what would be expected following the empirical rule, the 1 st and 2 nd intervals being off by only one count each. Moore and McCabe 2.9 The graph shows a strong, positive, linear association between social distress and brain activity. The data suggest that there may exist a correlation between brain activity and social distress. Moore and McCabe 2.18 Mean growth for 0 nematodes: 10.65 1,000: 10.4 5,000: 5.6 10,000: 5.45 b. The data suggests that there is a negative association between nematode concentration and plant growth. In other words, generally as the number of nematodes increases, the amount of plant growth decreases....
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2_Problem_Set - c s = 9.6 mean = 93.2 The calculation...

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