1530 It appears that people who are mildly obese are less ac
(15.30) 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, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with mean 375 minutes and standard deviation 68 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 522 minutes and standard deviation 106 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people.
Use z-scores rounded to two decimal places to answer the following:
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?
| (15.30) 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, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with mean 375 minutes and standard deviation 68 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 522 minutes and standard deviation 106 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people. Use z-scores rounded to two decimal places to answer the following: What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes? What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes? |
Solution
a)
Mean ( u ) =375
Standard Deviation ( sd )=68
Number ( n ) = 6
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
P(X > 410) = (410-375)/68/ Sqrt ( 6 )
= 35/27.761= 1.2608
= P ( Z >1.2608) From Standard Normal Table
= 0.1037
b)
Mean ( u ) =522
Standard Deviation ( sd )=106/ Sqrt ( 6 ) = a
Number ( n ) = 6
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
P(X > 410) = (410-522)/106/ Sqrt ( 6 )
= -112/43.274= -2.5881
= P ( Z >-2.5881) From Standard Normal Table
= 0.9952
