Biostatistic, statistics homework help

Anonymous
timer Asked: Jun 11th, 2017
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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


Tutor Answer

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...

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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.

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