Question

In: Statistics and Probability

12. (15.30) It appears that people who are mildly obese are less active than leaner people....

12.

(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 372 minutes and standard deviation 67 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 524 minutes and standard deviation 108 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?______

13.

(15.34) The level of nitrogen oxides (NOX) in the exhaust of cars of a particular model varies Normally with mean 0.19 g/mi and standard deviation 0.053 g/mi. A company has 25 cars of this model in its fleet.

What is the level L (±0.0001) such that the probability that the average NOX level x¯¯¯x¯ for the fleet is greater than L is only 0.03?

L =______

14.

(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 point 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 point 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)? ______

Expert Solution

12)

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>1.02)=

1-P(Z<1.02)=

1-0.8461=

0.1539

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>-2.81)=

1-P(Z<-2.81)=

1-0.0025=

0.9975

13)

for 97th percentile critical value of z=		1.880
therefore corresponding value=mean+z*std deviation=			0.2099

14)

a) std deviation =1.3/2.97=0.4377

b)

for 99.7 % CI value of z=	2.97
standard deviation σ=	6.60
margin of error E =	1.3
required sample size n=(zσ/E)² =	228.0

orchestra answered 2 years ago

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