# EXERCISE 18 Mean Standard Deviation And 95 And 99 Of The Normal Curve

1003 words 5 pages
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
To find the 99% of the women’s physical functioning scores you would use the formula:. This formula was found on page 132 of Exercise 18 in the second paragraph. The Mean=57.09 and the SD=23.72. These scores were found on pg. 134 in table 2 column labeled Female in the Physical Functioning category.

Formula: 65.20±2.58(29.79)
65.20-2.58(29.79) = 65.20-76.86
65.20-76.86= -11.66
65.20+2.58(29.79) = 65.20+76.86
65.20+76.86= 142.06

8. Assuming that the distribution of scores is normal, 99% of HIV-positive body image scores around the mean were between what two values? Round your answer to two decimal places. To find the 99% of HIV-positive body image scores you would use the formula:. This formula was found on page 132 of Exercise 18 in the second paragraph. The Mean=68.0 and the SD=17.0. These scores were found on pg. 133 in the second paragraph.

Formula: 68.00±2.58(17.0)
68.00-2.58(17.00) = 68.00-43.86
68.00-43.86= 24.14
68.00+2.58(17.00) = 68.00+43.86
68.00+43.86= 111.86