Biostatistic, statistics homework help

Anonymous
timer Asked: Jun 11th, 2017
account_balance_wallet \$10

Question Description

1.  Glucose levels in patients free of diabetes are assumed to follow a normal distribution with a mean of120 and a standard deviation of 16. What proportion of patients has glucose levels exceeding 115?

2.  The mean total cholesterol level in women free of cardiovascular disease is 205 with a standard deviation of 19.2. What proportion of women has total serum cholesterol levels exceeding 200?

3.  The gestation period for human births can be taken as normally distributed with a mean of 266 days and a standard deviation of 16 days. If a gestation period of 276 days, what percentile among human births is this?

4.  Suppose that the probability of a child living in urban area in the US is obese is 20%. A social worker sees 15 children living in urban areas. What is the probability that none are obese?

5.  Approximately 30% of obese patients develop diabetes. If a physician sees 10 patients who are obese, what is the probability that half of them will develop diabetes?

6.  The following data was collected in survey of 8th graders and summarize their cell phone status.

 No of cell phone Conventional Cell phone user Smart Phone user Total Boys 55 65 35 155 Girls 31 78 27 136 Total 86 143 62 291

What proportion of the 8th graders has conventional cell phones?

7.  BMI in children is approximately normally distributed within a mean 24.5 and standard deviation 6.2. A BMI between 25 and 30 is considered overweight. What proportion of children is overweight?

8.  The following table shows a distribution of BMI in children living US and European urban neighborhoods. (The data is in millions.)

 Neighborhood Normal weight Overweight Obese Total US 125 50 40 215 Europe 101 42 21 164 Total 226 92 61 379

If a child is selected at random,

a.  What is the probability they are overweight?

b.  What is the probability that a child living in the US urban neighborhood is overweight?

9.  The risk of hepatoma among alcoholics without cirrhosis of the liver is 24%. Suppose we observe 7 alcoholics without cirrhosis. What is the probability that exactly 1 of these 7 people have hepatoma?

10.

 Screening test Impaired Glucose Tolerance Not Impaired Positive 17 13 Negative 8 37

A new non-invasive screening test is proposed that is claimed to be able to identify patients with impaired glucose tolerance based on a battery of questions related to health behaviors. The new test is given 75 patients. Based on each patient’s responses to the questions they are classified as positive, or negative for impaired glucose tolerance. Each patient also submits a blood sample and their glucose tolerance status is determined. The results are tabulated below.

a.  What is the sensitivity of the screening test?

b.  What is the false positive fraction of the screening test?

11.  The following table shows the results of a screening test hypothesized to detect persons at risk of side effects of a cosmetic surgery.

 Side Effects Positive Side Effects Absent Total Screen Positive 12 6 18 Screen Negative 85 204 289 Total 97 210 307

a.  Compute the sensitivity of the test

b.  Compute the specificity of the test

c.  Compute the false positive fraction

d.  Compute the false negative fraction

12.   The following table shows the results of the a screening test hypothesized to identify persons at risk for rare blood disease

 No Disease Disease Total Screen Positive 1274 28 1302 Screen Negative 51 45 96 Total 1325 73 1398

a.  Compute the sensitivity of the test

b.  Compute the specificity of the test

c.  Compute the false positive fraction

d.  Compute the false negative fraction

jesusale932
School: Rice University

Here is my answer :)

1. Glucose levels in patients free of diabetes are assumed to follow a normal distribution
with a mean of 120 and a standard deviation of 16. What proportion of patients has
glucose levels exceeding 115?

Mean of μ = 120
Standard Deviation of σ = 16
x−μ
σ

115−μ
)
σ

P(x>115) = P(

>

P(x>115) = P(z>

115−120
)
16

P(x>115) = P(z>-0.31)
P(x>115) = 1 - P(z115) = 0.6227

2. The mean total cholesterol level in women free of cardiovascular disease is 205 with
a standard deviation of 19.2. What proportion of women has total serum cholesterol
levels exceeding 200?

Mean of μ = 205
Standard Deviation of σ = 19.2
x−μ
σ

200−μ
)
σ

P(x>200) = P(

>

P(x>200) = P(z>

200−205
)
19.2

P(x>200) = P(z>-0.2604)
P(x>200) = 1 - P(z200) = 0.6027

3. The gestation period for human births can be taken as normally distributed with a
mean of 266...

flag Report DMCA
Review

Anonymous
Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors