lab_exercise_2

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Belmont University *

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6007

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Statistics

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Jan 9, 2024

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docx

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2

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PubH 6002 Lab Exercise 2 1. One of the variables collected in the North Carolina Birth Registry is pounds gained during pregnancy. According to data from the entire registry in 2001, the number of pounds gained during pregnancy was approximately normally distributed with a mean of 30.23 pounds and a standard deviation of 13.84 pounds. Calculate the probability that a randomly selected mother in North Carolina gained: a. Less than 15 pounds Z = (15 – 30.23) / 13.84 = -1.10 P(Z < -1.10) = 0.1357 = 14% b. Between 20 and 30 pounds Z = (20-30.23) / 13.84 = -0.74 Z = (30-30.23) / 13.84 = -0.02 c. More than 40 pounds 2. Serum cholesterol is an important risk factor for coronary disease. We can show that serum cholesterol is approximately normally distributed with mean 219 mg/dL and variance 2500 mg/dL. a. The clinically desirable range for cholesterol is less than 200 mg/dL. What is the probability that a randomly selected person has a clinically desirable cholesterol level? Z = (200-219) / (2500)^1/2 = -0.38 P(Z < -0.38) = 0.35 = 35% b. Most doctors believe that cholesterol levels over 240 mg/dL indicate a high enough risk for heart disease to warrant treatment. What is the probability that a randomly selected person has high cholesterol, i.e., a cholesterol level greater than 240 mg/dL? Z = (240-219) / 50 = 0.42 P (Z > 0.42) = 1 - 0.66 = 34% c. What is the probability that a randomly selected person has borderline high cholesterol, i.e., a cholesterol level greater than 200 mg/dL but less than 240 mg/dL? P(200 < X < 240) = 1 – (0.35+0.34) = 31% 3. The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If we stipulate that a baby is premature if the length of pregnancy is in the lowest 4%, find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care. - 1.75 = (X – 268) / 15 4. A professor informs her biostatistics class that a test is very difficult, but the grades will be curved. Scores for the test are normally distributed with a mean of 25 and a standard deviation of 5. 1
a. If she curves the grades by adding 50 to each grade, what is the new mean? What is the new standard deviation? b. If the grades are instead curved according to the following scheme, find the numerical limits for each letter grade. A = Top 10% Z = 1.28 31.4 B = Scores above the bottom 70% and below the top 10% 0.52 and 1.28 27.6 to 31.4 C = Scores above the bottom 30% and below the top 30% -0.52 and 0.52 22.4 to 27.6 D = Scores above the bottom 10% and below the top 70% F = Bottom 10% c. Which of the two abovementioned methods of curving the grades is fairer? Please explain why. Method (b) because variation is accounted for in addition to the mean. 2
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