Standard Deviation and Decimal Places Essay

Submitted By roxb74
Words: 489
Pages: 2

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 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
(-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 Perceptions is normal, 95% of the females’ scores around the mean were between what values? Round your answer to two decimal places.
Mean (39.71)
Deviation (25.46)
39.71 + 1.96(25.46) = 39.71 + 49.9 = 89.61
39.71 - 1.96(25.46) = 39.71 - 49.9 = (-10.19)
89.61, -10.19

4. Assuming that the distribution of scores for Pain is normal, 95% of the men’s scores around the mean were between what two values? Round your answer to two decimal places.
Mean (52.53)
Deviation (30.90)
52.53 + 1.96(30.90) = 52.53 + 60.56 = 113.09
52.53 - 1.96(30.90) = 52.53 - 60.56 = (-8.03)
(-8.03, 113.09)

5. Were the body image scores significantly different for women versus men? Provide rationale for your answer.
Male: (33.28, 93.5)= 93.5-33.28= 60.22
Female: (39.89, 106.25)= 106.25-39.89= 66.36
No, the information shows that there is not a big difference in the two numbers. The difference is 6.06. So there is a small difference.

6. Assuming that the distribution of Mental Health scores for men is normal, where are 99% of the men’s mental health scores around the mean in this distribution? Round your answer to two decimal places.
Mean: 57.09
Deviation: 23.72
57.09 + 1.96(23.72) = 57.09 + 46.49 = 103.58
57.09 - 1.96(23.72) = 57.09 - 46.49 = 10.6
(10.6, 103.58)

7. Assuming that the distribution of…