Name: _______________________________ K E Y Student ID: _______________________________

MATH102 11SP Midterm ch1-6
A [ answers in web view ] Total points: 70

"For You light my lamp; the LORD my God illumines my darkness.
For by You I can run upon a troop; and by my God I can leap over a wall.
As for God, His way is blameless; the word of the LORD is tried;
He is a shield to all who take refuge in Him."
-- Psalm 18:28-30
• Please show all your work! No partial credit will be given for incorrect answers with no work shown. Please draw a box around your final answer.
• Calculators are permitted, but no notes, text, laptops, PDAs, or electronic dictionaries. Cell phones should be muted and left in your pocket or bag.
• Table 3 is attached to the back. You may detach it for your reference.
1. Classify each of the following statements (which may or may not be true) as either (D)escriptive or (I)nferential: 
1. 19.2% of BC residents in 2004 were clinically obese. [(I)nferential]
2. Of the children and adolescents studied in the 2004 Canadian Community Health Survey, 8% were clinically obese. [(D)escriptive]
3. About 90% of students in our MATH108 class are nursing students. [(D)escriptive]
4. 30.1% of male registered nurses (RNs) are under the age of 40. [(I)nferential]
2. Mark each of the following variables as nominal (N), ordinal (O), discrete (D), or continuous (C): 
1. How many children are in a family [(D)iscrete]
2. Satisfaction with current family doctor, rated as "Very Dissatisfied", "Dissatisfied", "Satisfied", or "Very Satisfied" [(O)rdinal]
3. Whether a student passes a course or not [(N) Nominal]
4. Hemoglobin count, in g/dL [(C)ontinuous]
5. Age, divided into categories "0-10", "11-18", "19-25", "26-39", "40-59", "60 and up" [(O)rdinal]
3. The home provinces of 16 students in a class are listed below. Draw a Pareto chart showing the distribution of home province in this sample. 
AB, BC, BC, SK, AB, SK, BC, ON, SK, BC, BC, AB, BC, ON, MB, SK

Frequencies (in order): BC = 6/16 = 37.5% (cum: 37.5%);
SK = 4/16 = 25% (cum: 62.5%);
AB = 3/16 = 18.75% (cum: 81.25%);
ON = 2/16 = 12.5% (cum: 93.75%);
MB = 1/16 = 6.25% (cum: 100%)
4. The table below lists the number of purple starthistle plants (a noxious invasive weed in WA state) in nine equally-sized plots.
41, 57, 47, 32, 42, 28, 36, 53, 42
1. Construct a relative frequency histogram, classifying the data by bins of width 10 plants/plot. (20-29): 11%; (30-39): 22%; (40-49): 44%; (50-59): 22%
Or: (28-37): 33%; (38-47): 44%; (48-57): 22%
2. Find the sample mean. Show your work. 
[42 plants/plot]
3. Find the mode and the midrange. 
[mode=42, midrange=42.5 ]
4. Find the sample standard deviation. Show your work. 
[9.381 plants/plot]
5. Draw a boxplot for the data. Show your work. 
[min=28, Q1=36, med=42, Q3=47, max=57]
5. Say that in a certain town 63% of the people vote Conservative, and the probability that a townsperson drives a large truck is 3/7. If you pick a random townsperson, there is a 27% chance that the townsperson is a Conservative driving a large truck.
1. What fraction of Conservatives in this town drive large trucks? 
Let C = Conservative, T = drives a large truck.
P(T|C) = P(T ∩ C) / P(C) = .27 / .63 = 3/7 ≈ 42.85%.
2. In this town, is voting Conservative mutually exclusive of driving a large truck? Why or why not? 
No, P(T and C) = .27 > 0, so there is an overlap; there are townspeople who are both Conservative and drive large trucks.
3. In this town, is voting Conservative independent of driving a large truck? Why or why not? Interpret what this means in the context of the townspeople.
P(T|C) = 3/7 = P(T), so the two events are independent.

In words, if we narrow our focus to Conservatives, we see that 3/7 of Conservatives drive large trucks. So we see that it doesn't matter whether a townsperson is Conservative or not, the chance of driving a large truck is still the same.

4. The town council is a randomly-selected group of 6 townspeople. What is the chance that exactly half of the town council votes Conservative? 
binomial with n=6, p=0.63: P(3) = 20 (.63)3 (.37)3 ≈ 25.33%
5. What is the chance that a majority of the town council votes Conservative? 
binomial with n=6, p=0.63: P(4) + P(5) + P(6)
= 15(.63)4(.37)2 + 6(.63)5(.37)1 + (.63)6 ≈ 32.35% + 22.03% + 6.25% ≈ 60.63%
6. The small business association is a randomly-selected group of 50 townspeople. What is the chance that more than half of the small business association votes Conservative? 
binomial with n=50, p=0.63: use normal approximation.
μ = np = 31.5, σ = sqrt(npq) = sqrt(11.655) ≈ 3.414.
half = 25 means z = (25-31.5)/3.414 ≈ -1.9 (or z = (25.5-31.5)/3.414 ≈ -1.76 if using continuity correction)
(T3): area between x=25 and mean is 47.13%, so area to right of x=25 is 50+47.13% = 97.13% (or 96.08% if using continuity correction)
6. An assay (test) measuring inorganic phosphate in blood is imprecise, with values normally distributed around the true phosphate concentration, with a standard deviation of 0.2 mmol/L. Blood phosphate values of over 1.8 mmol/L are considered "high".
1. What is the probability that this assay returns a result within ±0.25 mmol/L of the true value? 
[z = ±.25/.2 = ±1.25, area = 78.88%]
2. If the true concentration of phosphate is 1.63 mmol/L (not "high"), what is the probability that the assay could still return a value that is considered "high"? 
[z = (1.8-1.63)/.2 = 0.85, area to the right = 0.5 - 0.3023 = 19.77%]
3. The standard deviation is just one way of measuring the dispersion (imprecision) of the assay results. The interquartile range (IQR) is another. Find the IQR for this assay. 
Q3: area = 25%, so (T3) z= 0.67 (or 0.68).
Hence (x-μ) = z(σ) = (0.67)(0.2) = 0.134.
Similarly for Q1, so IQR = Q3 - Q1 = 2(0.134) = 0.268 (or 0.272)
7. A particular screening test for breast cancer has a 11% false-positive rate (i.e., 89% specificity) and a 7% false-negative rate (i.e., 93% sensitivity).
1. Suppose the test is applied to a group of patients, 40% of whom are known to have breast cancer. Draw an event tree for the outcomes of the test. Label the tree with probabilities for each branch. Also calculate the probabilities of each final outcome (leaf of the tree). 
[(CA,+):37.2%. (CA,-):2.8%. (noCA,+):6.6%. (noCA,-):53.4%]
2. What is the probability that a random patient from this group will test positive for breast cancer using this screening test?
[37.2% + 6.6% = 43.8%]
3. What is the probability that a patient from this group who tested positive actually has the disease? 
[37.2% / 43.8% = 84.93%]
8. Suppose that when a job candidate comes to interview for a staff position at TWU, the probability that he or she will want the job (A) after the interview is 72%. The probability that TWU wants the candidate (B) is 40%. Also, P(A|B)=0.80.
1. Find P(A ∩ B), draw a Venn diagram, and interpret it in words. 
This is the intersection: P(A ∩ B) = P(B)P(A|B) = (.40)(.80) = 32%.
Out of all the candidates who come to interview, 32% are liked by TWU and like TWU; i.e., 32% of interviews result in the candidate taking the job.
2. Find P(B | A), and interpret it in words. 
P(B|A) = P(A ∩ B) / P(A) = .32 / .72 ≈ 44.44%.
Just under half of the candidates who like TWU are also liked by TWU.
3. Are events A and B independent? Explain. Interpret (in the context of job interviews) what it means for A and B to be independent. 
P(A|B) = 80% ≠ 72% = P(A), so A and B are not independent. The probability that a candidate wants the job is different depending on whether TWU likes the candidate or not. In particular, if TWU wants the candidate, it is more likely that the candidate also wants the job (this isn't saying anything about causality, just correlation).