Question

(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 366 minutes and standard deviation 68 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 525 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. Usez-scores rounded to two decimal places or your calculator 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 400 minutes? _____ What is the probability (±±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 400 minutes? ____

(15.38) To estimate the mean score μμ of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information you know that the scores are approximately Normal with standard deviation about 6.6. You want your sample mean x¯¯¯x¯ to estimate μμ with an error of no more than 1.3 points in either direction.

(a) What standard deviation (±±0.0001) must x¯¯¯x¯ have so that 99.7% of all samples give an x¯¯¯x¯ within 1.3 points of μμ? ___

(b) How large an SRS do you need in order to reduce the standard deviation of x¯¯¯x¯ to the value you found in part (a)? ___

could you show how to get on a calculator also??

Answer 1

1)

a) probability (±±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 400 minutes

probability =P(X>400)=P(Z>(400-366)/(68/sqrt(6))=P(Z>1.22)=1-P(Z<1.22)=1-0.8888=0.1112

b)

probability (±±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 400 minutes :

probability =P(X>400)=P(Z>(400-525)/(106/sqrt(6))=P(Z>-2.89)=1-P(Z<-2.89)=1-0.0019=0.9981

(15.38)

a)

standard deviation =margin of error /3 =1.3/3=0.4333

b)

sample size =(population standard deviation/sample standard deviation)^2 =(6.6*3/1.3)^2=232