Name: Ashley Lee
Class: HLT-362 Applied Statistics for Healthcare Professionals Date: 04/01/2015
EXERCISE 18 • Mean, Standard Deviation, and 95% and 99% of the Normal Curve
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 your answer to two decimal places.
In order to find where 95% of the values for the weight of relative to the ideal lies you would use the formula that is presented in the text on page 132 of Exercise 18. This formula is:. The = MEAN (5.48) and the (SD) =Standard Deviation (22.93). These numbers were derived from table 1 on pg.133 under the column labeled Male. The problem is worked out as such:
5.48-1.96(22.93) = 5.48-44.94 5.48-44.94= -39.46
5.48+1.96(22.93) = 5.48+44.94
2. Which of the following values from Table 1 tells us about variability of the scores in a distribution? a. 60.22
c. 22.57 ←Answer
The answer for question number 2 is (C). The SD indicates the variability. In the answer set the only choice that was SD was choice (C) the other options were Mean scores listed in table 1.
3. Assuming that the distribution for General Health Perceptions is normal, 95% of the females’ scores around the mean were between what values? Round your answer to two decimal places. Again you would use the formula information provided in the text on page 132 of Exercise 18. On pg. 134 of the workbook Exercise 18 table 2 you can find the Mean=39.71 and SD= 25.46 for the Females in the study.
39.71-1.96(25.46) = 39.71-49.90 39.71-49.90= -10.19
39.71+1.96(25.46) = 39.71+49.90
4. Assuming that the...
References: Grove, S. (2007, February 26). Grand Canyon University - Digital Resources. Retrieved March 23, 2015, from http://gcumedia.com/digital-resources/elsevier/2007/statistics-for-health-care-research_a-practical-workbook-custom_ebook_1e.php
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