It appears that people who are mildly obese are less active than leaner people.
One study looked at the average number of minutes per day that people spend standing or walking.
Among mildly obese people, minutes of activity varied according to the N(373, 67) distribution.
Minutes of activity for lean people had the N(526, 107) distribution.
Within what limits do the active minutes for 95% of the people in each group fall?
Use the 68–95–99.7 rule.
Within what limits do the active minutes for 95% of the people in the mildly obese group fall?
A. Within 306 to 440 minutes.
B. Within 134 to 612 minutes.
C. Less than 239 or more than 507 minutes.
D. Within 239 to 507 minutes.
Please help, thank you! I don't know how to set it up or where to even begin.
Z = (score-mean)/SD
95% = Z = ±1.96
Values for A-D are not 95%.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the percentages related to the Z scores.
D. Within 239 to 507 minutes.
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, minutes of activity is normally distributed with mean 373 minutes and standard deviation of 67 minutes. The least active 20% of these individuals spend at most 317 minutes walking or standing because 317 = _____*67 + 373. Enter your answer rounded to two decimal places.
To solve this problem, you can use the 68–95–99.7 rule, also known as the empirical rule, which states that for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Let's calculate the limits of active minutes for 95% of the people in each group.
For the mildly obese group:
Mean = 373 minutes
Standard deviation = 67 minutes
Mean ± 2 * standard deviation will give us the range within which 95% of the data falls.
Lower limit = Mean - 2 * standard deviation
Upper limit = Mean + 2 * standard deviation
Lower limit = 373 - 2 * 67 = 239
Upper limit = 373 + 2 * 67 = 507
So, the active minutes for 95% of people in the mildly obese group fall within 239 to 507 minutes, which is option D.
Please note that this solution assumes that the distribution of minutes of activity for both groups follows a normal distribution.