Exercise 18 Essay
6/29/2014
HLT362V
Kristy Osgood
Exercise 18
1. Assuming that the distribution is normal for weight relative to the ideal and 99% of the male participants scored between (53.68, 64.64), where did 95% of the values for weight relative to the ideal lie? Round you answer to two decimal places.
Answer: The mean of the weight is 5.48 and the standard deviation is 22.93
Mean ± 1.96(Deviation)
5.48 ± 1.96(22.93)
5.48  1.96(22.93) = 5.48  44.94 = (39.46)
5.48 + 1.96(22.93) = 5.48 + 44.94 = 50.42
Which is (39.46, 50.42)
2. Which of the following values from Table 1 tells us about variability of the scores in a distribution?
a) 60.22
b) 11.94
c) 22.57
d) 53.66
3. Assuming that the distribution for General Health …show more content…
Mean: 65.2
Deviation: 29.79
65.2 + 1.96(29.79) = 65.2 + 58.39= 123.59
65.2  1.96(29.79) = 65.2  58.39= 6.81
(6.81, 123.59)
8. Assuming that the distribution of scores is normal, 99% of HIVpositive body image score around the mean were between what two values? Round your answer to two decimal places.
9. Assuming that the distribution of scores for Role Functioning is normal, 99% of the men’s scores around the mean were between what values? Round your answer to two decimal places.
Mean: 50
Deviation: 46.29
50 + 1.96(46.29) = 50 + 90.73= 140.73
50  1.96(46.29) = 50  90.73= 40.73
(40.73, 140.73)
10. What are some of the limitations of this study that decrease the potential for generalizing the findings to the target population?
The number of individuals surveyed, different

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