Weights of Supermodels. Supermodels are sometimes criticized on the grounds, tha
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Weights of Supermodels. Supermodels are sometimes criticized on the grounds, that their low weights encourage unhealthy eating habits among young women. In Exercises 1-4, use the following weights (in pounds) of randomly selected In supermodels. 125 (Taylor) 119 (Auermann) 128 (Schiffer) 125 (Bundchen) 119 (Turlington) 127 (Hall) 105 (Moss) 123(Mazza) 110(Reilly) 103(Barton) Find the mean, median, and standard deviation. What is the level of measurement of these data (nominal, ordinal, interval, ratio)? Construct a 95% confidence interval for the population mean. Find the sample size necessary to estimate the mean weight of all supermodels so that there is 95% confidence that the sample mean is in error by no more than 2 lb. Assume that a pilot study suggests that the weights of all supermodels have a standard deviation of 7.5 lb. Employment Drug Test If a randomly selected job applicant is given a drug test, there is a 0.038 probability that the applicant will test positive for drug use (based on data from Quest Diagnostics). a. If a job applicant is randomly selected and given a drug test, what is the probability that the applicant does not test positive for drug use? b. Find the probability that when two different job applicants are randomly selected and given drug tests, they both test positive for drugs. c. If 500 job applicants are randomly selected and they are all given drug tests, find the probability that at least 20 of them test positive for drugs. ACT Scores Scores on the ACT test are normally distributed with a mean of 21.1 and a standard deviation of 4.8. a. If one ACT score is randomly selected, find the probability that it is greater than 20.0. b. If 25 ACT scores are randomly selected, find the probability that they have a mean great than 20. c. Find the ACT score that is the 90th percentile. Sampling What is a simple random sample? What is a voluntary response sample? Range Rule of Thumb Use the range rule of thumb to estimate the standard deviation of point averages at a college with a grading system designed so that the lowest and highest possible grade point averages are 0 and 4. Rare Event Rule Find the probability of making random guesses to 12 true/false questions and getting 12 correct answers. If someone did get 12 correct answers, is it possible that they made random guesses? Is it likely that they made random guesses? Sampling Method If you conduct a poll by surveying all of your friends that you see during the next week, which of the following terms best describes the type of sampling used: random, systematic, cluster, convenience, voluntary response? Is the sample likely to be representative of the population?Explanation / Answer
1]
We have to find the mean, median and standard deviation for the given random sample of weights. The calculation table is given as below:
No.
X
(X - mean)
(X - mean)^2
1
125
6.6
43.56
2
119
0.6
0.36
3
128
9.6
92.16
4
125
6.6
43.56
5
119
0.6
0.36
6
127
8.6
73.96
7
105
-13.4
179.56
8
123
4.6
21.16
9
110
-8.4
70.56
10
103
-15.4
237.16
Total
1184
762.4
Mean
118.4
Variance
84.71111111
Median
121
SD
9.203863923
Mean = X/n = 1184/10 = 118.4
Variance = (X - mean)^2/(n – 1) = 84.71
SD = sqrt(Variance) = 9.20
Median = middle most obs. = 121
2]
Level of measurement of the given data is ratio level of measurement.
3]
Here, we have to construct the 95% confidence interval for the population mean.
The confidence interval formula is given as below:
Confidence interval = Xbar -/+ t*SD/sqrt(n)
We are given
Sample Standard Deviation
9.203863923
Sample Mean
118.4
Sample Size
10
Confidence Level
95%
Degrees of Freedom
9
t Value
2.2622
Confidence interval = 118.4 -/+ 2.2622*9.2039/sqrt(10)
Lower limit = 118.4 - 2.2622*9.2039/sqrt(10) = 111.82
Upper limit = 118.4 + 2.2622*9.2039/sqrt(10) = 124.98
Confidence interval = (111.82, 124.98)
4]
Sample size formula is given as below:
Sample size = n = (Z*/E)^2
We are given
= 7.5
E = 2
Confidence level = 95%
Z = 1.96
n = (1.96*7.5/2)^2 = 54.0225
n = 55 approximately
Required sample size = 55
No.
X
(X - mean)
(X - mean)^2
1
125
6.6
43.56
2
119
0.6
0.36
3
128
9.6
92.16
4
125
6.6
43.56
5
119
0.6
0.36
6
127
8.6
73.96
7
105
-13.4
179.56
8
123
4.6
21.16
9
110
-8.4
70.56
10
103
-15.4
237.16
Total
1184
762.4
Mean
118.4
Variance
84.71111111
Median
121
SD
9.203863923
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