• Question 1
Needs Grading
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. (20 pts)
What proportion of patients has glucose levels exceeding 115?
Selected
Answer: z = (x - mean)/std
z = (115 - 120)/16
z = -0.3125
To determine the proportion of people whose level is below
115 from table in back of book the Z value is 0.3783
Therefore we have the probability that Z< 0.378 we need to
subtract it from 1
1-.03783= 0.6217
P(X>115) = P(Z>115-120/16)=P(Z>-0.31)=1-0.3783=0.6217
Correct
Answer: 0.62
Response [None Given]
Feedback:
• Question 2
Needs Grading
A ferritin test is a popular test to measure a person’s current iron stores. In women, ferritin is
approximately normally distributed with a mean of 89 ng/mL and a standard deviation of 23
ng/mL.
What is the probability that a woman has a ferritin value of 100 or less?
Selected
Answer: P ( x < 100) μ = 89 ng/ml
σ = 23 ng/ml
P ( x < 100) = area to the left of 100 therefore we
convert : z = x - µ / σ and so
Z = 100 - ( 89) / 23 = 0.478261≈ 0.48 = P( Z < 0.48 )
= Area to the left of 0.48 so
The probability of a woman having a ferritin value of
100 or less is 68.44%
Correct
Answer: 0/68)(15 pts)
Response [None Given]
Feedback:
, • Question 3
Needs Grading
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?
Selected
Answer: The probability that half of them will develop
diabetes is 0.1029
Correct
Answer: (0.10) (15 pts)
Response [None Given]
Feedback:
• Question 4
Needs Grading
The following table shows the distribution of BMI in children living in US and European urban
neighborhoods. (The data are in millions.)
Neighborhood Normal Weight
US 125
Europe 101
Total 226
If a child is selected at random,
a. What is the probability they are overweight? (7.5 pts)
b. What is the probability that a child living in a US urban neighborhood is overweight? (7.5
pts)
Selected
Answer: a. 50 +42/379 =92/379 =0.24 It is a 24% probability
that they are overweight
b. 50/379=0.13 it is a 13% probability that a child
living in a US urban neighborhood is overweight
Correct
Answer: a. 0.24 (7.5 pts)
b. 0.23 (7.5 pts)
Response [None Given]
Feedback:
• Question 5
Needs Grading
Needs Grading
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. (20 pts)
What proportion of patients has glucose levels exceeding 115?
Selected
Answer: z = (x - mean)/std
z = (115 - 120)/16
z = -0.3125
To determine the proportion of people whose level is below
115 from table in back of book the Z value is 0.3783
Therefore we have the probability that Z< 0.378 we need to
subtract it from 1
1-.03783= 0.6217
P(X>115) = P(Z>115-120/16)=P(Z>-0.31)=1-0.3783=0.6217
Correct
Answer: 0.62
Response [None Given]
Feedback:
• Question 2
Needs Grading
A ferritin test is a popular test to measure a person’s current iron stores. In women, ferritin is
approximately normally distributed with a mean of 89 ng/mL and a standard deviation of 23
ng/mL.
What is the probability that a woman has a ferritin value of 100 or less?
Selected
Answer: P ( x < 100) μ = 89 ng/ml
σ = 23 ng/ml
P ( x < 100) = area to the left of 100 therefore we
convert : z = x - µ / σ and so
Z = 100 - ( 89) / 23 = 0.478261≈ 0.48 = P( Z < 0.48 )
= Area to the left of 0.48 so
The probability of a woman having a ferritin value of
100 or less is 68.44%
Correct
Answer: 0/68)(15 pts)
Response [None Given]
Feedback:
, • Question 3
Needs Grading
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?
Selected
Answer: The probability that half of them will develop
diabetes is 0.1029
Correct
Answer: (0.10) (15 pts)
Response [None Given]
Feedback:
• Question 4
Needs Grading
The following table shows the distribution of BMI in children living in US and European urban
neighborhoods. (The data are in millions.)
Neighborhood Normal Weight
US 125
Europe 101
Total 226
If a child is selected at random,
a. What is the probability they are overweight? (7.5 pts)
b. What is the probability that a child living in a US urban neighborhood is overweight? (7.5
pts)
Selected
Answer: a. 50 +42/379 =92/379 =0.24 It is a 24% probability
that they are overweight
b. 50/379=0.13 it is a 13% probability that a child
living in a US urban neighborhood is overweight
Correct
Answer: a. 0.24 (7.5 pts)
b. 0.23 (7.5 pts)
Response [None Given]
Feedback:
• Question 5
Needs Grading
