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Name: Math 258: Practice Midterm Exam August 2020
- (10 points) Consider the following data set.
2.36 3.68 5.15 5.24 5.38 5.63 5.97 6.01 7.58 8. (a) Compute the mean, median, quartiles and the standard deviation.
x¯ =
(2.36 + 3.68 +... + 8.73) = 5. 573 , x˜ =
x 5 + x 6 2
Q 1 = x 3 = 5. 15 , Q 2 = ˜x = 5. 505 , Q 3 = x 8 = 6. 01 ,
V (X) =
[
]
≈ 3. 176 , s ≈
(b) Construct a boxplot of the data.
(c) Identify any outliers, and extreme outliers. First we define the fourth spread fs = Q 3 − Q 1 = 0.86. Next, we compute the boundaries of outliers Q 1 − 1. 5 fs = 3.86, Q 1 − 3. 0 fs = 2.57, Q 3 + 1. 5 fs = 7.3 and Q 3 + 3. 0 fs = 8.59. Then our rules are: x <︸ ︷︷ 2. (^57) ︸ extreme outliers
, (^2) ︸. 57 ≤ (^) ︷︷x < 3. (^86) ︸ outliers
, (^7) ︸. 3 < x︷︷ ≤ 8. (^59) ︸ outliers
, x <︸ ︷︷ 8. (^59) ︸ extreme outliers
We have outliers: 3. 68 , 7. 58 , extreme outliers: 2. 36 , 8. 73
(d) The data is: (circle one) symmetric
(e) The data is: (circle one) unimodal
- (10 points) There are 12 students in Chemistry Lab, and they are to be grouped into 4 teams of 3. How many ways groups be formed? First consider group 1. There are 12 students, and we must choose 3. The number of possibilities is
N 1 = C 3 , 12 =
Next, we have group 2. There are 9 remaining students, and so the number of ways we can form a group of 3 students is N 2 = C 3 , 9 =
Next, we have group 3. There are 6 remaining students, and so the number of ways we can form a group of 3 students is N 3 = C 3 , 6 =
Finally, there is only C 3 , 3 = 1 way to choose the fourth group, as there are only 3 remaining students. The total number of ways to form groups is N = N 1 · N 2 · N 3 · N 4 = 369, 600.
- (10 points) Suppose 25% of dogs are brown, and further that 20% of brown dogs have spots. What is the probability that a randomly selected dog is brown and has spots? Consider a randomly selected dog, and let A =”the dog is brown”, and B =”the dog has spots”. Then we are being asked P (A ∩ B). But we are given the following: P (A) = 0. 25 , P (B|A) = 0. 20. That is, we know 25% of dogs are brown, and of these, 20% have spots (a conditional probability). The law of conditional probabilty applies here, and we obtain P (A ∩ B) = P (A)P (B|A) = 0. 25 · 0 .20 = 0. 05. That is, 5% of dogs are brown with spots, which is 20% of 25% of all dogs.
- (10 points) Suppose 1 in 1,000 people have a virus. A test for detecting the virus has a 40% false-negative rate, and a 1% false-positive rate. If a test is positive, what is the probability that the person is actually infected? Which rate should be decreased to dramatically improve the reliability of the test? Consider a randomly selected person, and administer the test. Let A be the event that the person has the virus, and B be the event that the test is positive. Likewise, A′^ is when the person does not have the virus, and B′^ is when the test is negative. Then, we first observe that P (A) = 0. 001 , P (A′) = 0.999. Now, suppose that a test gives a false negative. That is, given that a person has the virus, what is the probability that the test incorrectly indicates a negative result? We have P (B′|A) = 0. 40. As for false positives, in which a person does not have the virus, but the test incorrectly indicates they do, we have P (B|A′) = 0. 01. Our goal is to determine the efficacy of the test, which is the probability P (A|B) that a person has the virus, given that the test is positive. We will need to use Baye’s theorem
P (A|B) =
P (A ∩ B)
P (B)
along with the laws of total probability and conditional probabilty, P (B) = P (B ∩ A) + P (B ∩ A′) = P (A)P (B|A) + P (A′)P (B|A′). Based on what is known, we can infer that the probability of a properly detected case of the virus is P (B|A) = 1 − P (B|A′) = 0.6. Then, we have
P (A|B) =
P (A)P (B|A)
P (A)P (B|A) + P (A′)P (B|A′)
Note: The test is only effective approximately 5% of the time. While it may seem like the false negative rate is the problem, it is in fact the false positives. This is because 1% of 999 people is much larger than 60% of 1 person. So, if we have a randomly selecte positive test, it is highly likely it came from the larger population of those not infected by the virus. In order to make the test more effective, reduce the rate of false positives to a miniscule number. Then, the effectiveness is
P (A|B) =
P (A)P (B|A)
P (A)P (B|A) +
P (A)P (B|A)
P (A)P (B|A)
- (15 points) A customer can choose to purchase items from any (or all, or none) of brands A, B and C. Based on shopping data, the following probabilities are known: P (A) = 0. 4 , P (B) = 0. 75 , P (C) = 0. 3 , P (A ∩ B ∩ C) = 0. 01 , P (A ∩ B) = 0. 3 , P (B ∩ C) = 0. 18 , P (A ∩ C) = 0. 03.
(c) Suppose balls are selected at random, and placed back in the jar, until 4 green balls are observed. What is the probability that exactly 2 red balls are drawn?
X ∼ negbin
=⇒ P (X = 2) = nb
(d) Suppose that 4 balls are randomly removed from the jar, without replacement. What is the probability
that 2 red balls have been selected? X ∼ hyper (4, 5 , 11) =⇒ P (X = 2) =
2
2
4
- (5 points) The gray fox population is stable in Michigan, but a catch-and-release campaign is conducted to get a population estimate. In the first phase, 250 gray foxes are caught and tagged. In the second phase, 40 are caught, and 3 are found with tags. Estimate the gray fox population. We would like to estimate N = totalpopulation. From phase one, M = 250 gray foxes are tagged, so the proportion is p =
N
In the second phase, the proportion of tagged gray foxes in our sample is
ˆp =
m n
We find N by equating the sample and population proportions, 250 N
=⇒ N =
