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    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. x=5.48, SD=22.93 5.48+1.96(22.93) = 170.5992 5.48-1.96(22.93)=80.7136 (80.71,170.60) 2. Which of the following values from Table 1 tells us about variability of the scores in a distribution? c. 22.57 3. Assuming that the distribution for General

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    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. a. X =5.48, SD = 22.93 b. 5.48 – 1.96(22.93) = AND 5.48 + 1.96(22.93) = c. 5.48 – 44.9428 = AND 5.48 + 44.9428 = d. -39.4628 AND 50.422 i. (-39.46, 50.42) 2. Which of the following values from Table 1 tells us

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  • Exercise 18

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    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. Relative mean weight = (-53.68 + 64.64)/2 = 5.48. (Already provided) S.D = 22.93 Calculation: X bar 1.96 (SD) at 95% Confidence : 5.48 + 1.96(22.93) = 50.42 : 5.48 - 1.96(22.93) = -39.46 Expressed as: (-39.46, 50.42)

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    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. X =5.48, SD = 22.93 5.48 – 1.96(22.93) = AND 5.48 + 1.96(22.93) = 5.48 – 44.9428 = AND 5.48 + 44.9428 = -39.4628 AND 50.422 (-39.46, 50.42) 2.Which of the following values from Table 1 tells us about variability of the scores in a distribution

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    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. Calculation X+1.96(SD) X=5.48 SD=22.93 5.48+1.96(22.93) 5.48-1.96(22.93) =50.42 = -39.46 Assuming that the distribution is normal for weight relative to the ideal, 95% of the male participants scores fall between (-39.46, 50.42).

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    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. x=5.48, SD=22.93 5.48+1.96(22.93) = 170.5992 5.48-1.96(22.93)=80.7136 (80.71,170.60) 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 Ans. C 22

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  • Hlt-362v Exercise 18

    • 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 answer to two decimal places. • Which of the following values from Table 1 tells us about variability of the scores in a distribution? • Assuming that the distribution for General Health Perceptions is normal, 95% of the females’ scores around the mean were between what

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    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. 2. Which of the following values from Table 1 tells us about variability of the scores in a distribution? 3. Assuming that the distribution for General Health Perceptions is normal, 95% of the females’ scores around the mean were between what

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