diff --git a/source/exams/exam-01.ptx b/source/exams/exam-01.ptx new file mode 100644 index 0000000..99e8f93 --- /dev/null +++ b/source/exams/exam-01.ptx @@ -0,0 +1,490 @@ + + + + + Exam 1 + + + +

+ Show all relevant work. +

+ + + + + + + +

+ Write either True or False for each of the following statements. + No justification is required. +

+ + + + + +

+ Let A, B be any sets. + Then |A \cap B| = |A| \cdot |B|. +

+ + + +

+ False. +

+ + + + + + +

+ Let A, B be any events. + Then \Pr(A \mid B) = \Pr(B \mid A). +

+ + + +

+ False. +

+ + + + + + +

+ Given a joint distribution for random variables X and Y, then it must be the case that \Pr(X = 0, Y = 0) = \Pr(X = 0)\Pr(Y = 0). +

+ + + +

+ False. +

+ + + + + + +

+ Suppose we flip a coin repeatedly until we first see heads and let T be the number of flips. + Then \Pr(T = 2) \geq \Pr(T = 4). +

+ + + +

+ True. +

+ + + + + + +

+ If molecules are observed leaving a cell after 1.5 minutes, 1.8 minutes, 2.1 minutes, 2.2 minutes, and 2.4 minutes, then the maximum likelihood estimation for the rate at which molecules leave the cell is 1/2 per minute. +

+ + + +

+ True. +

+ + + + + + +

+ Suppose \L(\theta) is a likelihood function, and \L(2) = 0.04. + Then the probability that \theta = 2 is 0.04. +

+ + + +

+ False. +

+ + + + + + + + + +

+ Consider the sample space S = \{1, 2, 3, 4, 5\} with probability distribution given in the table below. + Let A = \{1, 2, 3\} and B = \{1, 3, 5\}. + Calculate the following probabilities. +

+ + + + x + 1 + 2 + 3 + 4 + 5 + + + + \Pr(x) + 0.1 + 0.2 + 0.3 + 0.3 + 0.1 + + + + + + + +

+ \Pr(A \cup B) +

+ + + +

+ A\cup B = \{ 1, 2, 3, 5 \}, so \Pr(A\cup B) = 0.1 + 0.2 + 0.3 + 0.1 = \boxed{0.7} +

+ + + + + + +

+ \Pr(A \cap B) +

+ + + +

+ A\cap B = \{ 1, 3 \}, so \Pr(A\cap B) = 0.1 + 0.3 = \boxed{0.4} +

+ + + + + + +

+ \Pr(A \mid B) +

+ + + +

+ \Pr(A\mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{0.4}{0.1 + 0.3 + 0.1} = \frac{0.4}{0.5} = \boxed{0.8} +

+ + + + + + + + + +

+ Consider the random variable X with probability distribution below. +

+ + + + x + 1 + 2 + 3 + 4 + 5 + + + + \Pr(x) + 0.2 + 0.3 + 0.1 + 0.25 + 0.15 + + + + + + + +

+ Find \E(X). +

+ + + +

+ + \E(X) \amp = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.25) + (5)(0.15) = \boxed{2.85} + +

+ + + + + + +

+ Find \Var(X). +

+ + + +

+ + \E(X^2) \amp = (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1) + (4)^2(0.25) + (5)^2(0.15) = 10.05 + \Var(X) \amp = \E(X^2) - \left(\E(X)\right)^2 = 10.05 - (2.85)^2 = \boxed{1.9275} + +

+ + + + + + +

+ A diagnostic test is developed to detect a disease present in 2.1% of the population. + For a patient who has the disease, the test will accurately give a positive result 76% of the time. + When the patient does not have the disease, the test will accurately give a negative result 98.2% of the time. +

+ +

+ For a patient who receives a negative test, what is the probability they do not have the disease? +

+ + + +

+ Let P be the event of receiving a positive test result and D be the event of having the disease. + The given information is: \Pr(D) = 0.021, \Pr(P\mid D) = 0.76, and \Pr(P^c \mid D^c) = 0.982. + Then, using Bayes' Theorem: + + \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c \mid D^c)\Pr(D^c)}{\Pr(P^c \mid D^c)\Pr(D^c) + \Pr(P^c \mid D)\Pr(D)} + \amp = \frac{\Pr(P^c \mid D^c)(1 - \Pr(D))}{\Pr(P^c \mid D^c)(1 - \Pr(D)) + (1 - \Pr(P \mid D))\Pr(D)} + \amp = \frac{(0.982)(1 - 0.021)}{(0.982)(1 - 0.021) + (1 - 0.76)(0.021)} + \amp \approx \boxed{0.995} + +

+ + + + + + + + +

+ In each of the following scenarios, state the maximum likelihood estimation for the unknown parameter indicated. +

+ + + + + +

+ A radioactive material is observed for 5 hours. + 120 particle emissions are seen. + \lambda is the hourly rate of particle emissions. +

+ + + +

+ 120 emissions per 5 hours is an hourly rate of \boxed{\widehat{\lambda} = 120/5 = 24 \text{ per hour}} +

+ + + + + + +

+ In each of 5 trials, a coin is flipped until heads is seen. + The number of flips in each trial is 3, 4, 3, 5, and 6. + p is the probability of the coin coming up heads on a flip. +

+ + + +

+ There are 3 + 4 + 3 + 5 + 6 = 21 total flips, 5 of which are heads, so the MLE is \boxed{\widehat{p} = \frac{5}{21}} +

+ + + + + + +

+ Let X be a continuous random variable with c.d.f. + \displaystyle{F(x) = \frac{1}{8}(x^3 + 3x + 4)}, where x \in [-1, 1]. +

+ + + + + +

+ Find the p.d.f. + f(x) for X. +

+ + + +

+ f(x) = F'(x) = \frac{1}{8}\left(3x^2 + 3\right) = \frac{3}{8}\left(x^2 + 1\right) +

+ + + + + + +

+ Find \E(X). +

+ + + +

+ + \E(X) \amp = \int_{-1}^1 x \cdot f(x)\ dx = \int_{-1}^1 x\cdot \frac{3}{8}\left(x^2 + 1\right)\ dx + \amp = \frac{3}{8} \int_{-1}^1 x^3 + x\ dx = \frac{3}{8} \left(\frac{x^4}{4} + \frac{x^2}{2}\right)\bigg|_{-1}^1 + \amp = \frac{3}{8} \left[\left(\frac{1}{4} + \frac{1}{2}\right) - \left( \frac{1}{4} + \frac{1}{2}\right)\right] = \boxed{0} + +

+ + + + + + +

+ Find \Var(X). +

+ + + +

+ + \E(X^2) \amp = \int_{-1}^1 x^2 \cdot f(x)\ dx = \int_{-1}^1 x^2\cdot \frac{3}{8}\left(x^2 + 1\right)\ dx + \amp = \frac{3}{8} \int_{-1}^1 x^4 + x^2\ dx = \frac{3}{8} \left(\frac{x^5}{5} + \frac{x^3}{3}\right)\bigg|_{-1}^1 + \amp = \frac{3}{8} \left[\left(\frac{1}{5} + \frac{1}{3}\right) - \left( \frac{-1}{5} - \frac{1}{3}\right)\right] = \frac{3}{8} \cdot \frac{16}{15} = \frac{2}{5} + \Var(X) \amp = \E(X^2) - (\E(X))^2 = \frac{2}{5} - 0^2 = \boxed{\frac{2}{5}} + +

+ + + + + + + + + +

+ Consider the joint distribution for X and Y below. + Are X and Y independent? +

+ + + + + X = 0 + X = 1 + X = 2 + + + + Y = 1 + 0.1 + 0.25 + 0.2 + + + + Y = 2 + 0.2 + 0.1 + 0.15 + + + + + +

+ + \Pr(X = 0, Y = 1) \amp = 0.1 + \Pr(X = 0) \amp = 0.1 + 0.2 = 0.3 + \Pr(Y = 1) \amp = 0.1 + 0.25 + 0.2 = 0.55 + \Pr(X = 0)\Pr(Y = 1) \amp = (0.3)(0.55) = 0.165 \neq 0.1, + + so X, Y are not independent. +

+ + + + + +

+ Suppose a parameter -1/2 \leq \theta \leq 1 has likelihood function \L(\theta) = \theta^2 - \theta^3. + Find the maximum likelihood estimation of \theta. +

+ + + +

+ \L is a continuous function and [-1/2, 1] is a closed interval, so we use the CIM. + + \L'(\theta) \amp = 2\theta - 3\theta^2 + \L'(\theta) = 0 \text{ when } 0 \amp = 2\theta - 3\theta^2 + 0 \amp =\theta (2 - 3\theta) + \theta \amp = 0, \frac{2}{3} + + Then: +

+ + + + \theta + -1/2 + 2/3 + 1 + + + + \L(\theta) + 3/8 + 4/27 + 0 + + + +

+ 3/8 is the largest value, so the MLE is \boxed{\widehat{\theta} = -1/2}. +

+ + + + \ No newline at end of file diff --git a/source/main.ptx b/source/main.ptx index 4c1af32..0fa5d9a 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -41,6 +41,13 @@ + + Exams + + + + + Recitations