Final Exam: Show All Your Work. Calculators Allowed ✓ Solved

FINAL EXAM Show all your work. Calculators are allowed. After

Final Exam Show all your work. Calculators are allowed. After finishing, scan and upload your exam as a pdf file. (1) (a) Suppose that there are 10 red balls and 12 blue balls inside a box. Draw five balls at random without replacement. What is the probability that the second and the third balls are red, but all the other balls are blue? (b) Consider a factory which produces electronic components. Assume that, over a long run, 3% of the components are faulty, and that component faults are independent of one another. If this factory produces a batch of 713 components, which is the most likely number of faulty components in this batch? (2) (a) Let f(x) be the density function for a continuous random variable X. Suppose that f(x) = ( c cos x if x ∈ (−π/2, π/2) 0 otherwise for some constant c to be determined. Compute the constant c. Also, find P(−π/2 ≤ X ≤ π/3). (b) Suppose that X and Y are bivariate standard normal variables with correlation 2/3. Find an expression for P(2X +Y ≤ 5) in terms of the cumulative distribution function of standard normal distribution. (3) (a) Suppose that X is a random variable with distribution P(X = 1/4) = 2/3, P(X = 3/4) = 1/3. Suppose that, given X = x, the random variable Y is binomial(3, x)-distributed. Compute the conditional distribution of X given Y = 2. (b) Let X and Y have joint density f(x, y) = ( 2x + 2y 4xy if 0

Paper For Above Instructions

The problem set contains various tasks requiring knowledge in probability and statistics. To address these tasks, we will solve each part step by step. The first task involves determining the probability related to drawing red and blue balls, followed by evaluating a real-world scenario concerning faulty components. Next, we will explore the properties of continuous random variables through density functions and their respective distributions. Finally, we will compute conditional distributions and covariance in given scenarios.

Task 1: Probability of Drawing Balls

We are tasked with calculating the probability of drawing balls of specific colors from a box containing 10 red and 12 blue balls, when drawing without replacement. To find the probability that the second and third balls drawn are red while the first, fourth, and fifth balls drawn are blue, we will apply the hypergeometric distribution.

The total number of balls is 22 (10 red + 12 blue). The required sequence of draws must align as follows:

  • 1st: Blue
  • 2nd: Red
  • 3rd: Red
  • 4th: Blue
  • 5th: Blue

The probability calculation proceeds as:

P = (Number of ways to choose 2 reds from 10) x (Number of ways to choose 3 blues from 12) / (Total ways to choose 5 balls from 22)

Mathematically, this involves:

P = (C(10, 2) * C(12, 3)) / C(22, 5)

where C(n, k) is the combination function.

Calculating each combination, we get:

C(10, 2) = 45, C(12, 3) = 220, and C(22, 5) = 26334.

This yields:

P = (45 * 220) / 26334 ≈ 0.3849 (or 38.49%).

Task 2: Faulty Components in a Batch

Next, we analyze the production of components where 3% are expected to be faulty. For a batch containing 713 components, we can use the Poisson approximation due to the low probability of faults.

Let μ represent the mean number of faulty components:

μ = 0.03 * 713 = 21.39 ≈ 21 (when rounded).

Using the Poisson distribution, the most likely number of faulty components is found at μ. Therefore, the most likely number of faulty components in a batch of 713 is approximately 21.

Task 3: Continuous Random Variable Density Functions

We are given the density function f(x) = c cos(x) for x in (−π/2, π/2). To find the constant c, we use the following normalization condition:

∫(−π/2)^(π/2) c cos(x) dx = 1.

Calculating the integral, we find:

c [sin(x)] | (−π/2)^(π/2) = c (sin(π/2) - sin(−π/2)) = c (1 - (-1)) = 2c.

Thus, 2c = 1, leading us to c = 1/2.

To find P(−π/2 ≤ X ≤ π/3):

P(−π/2 ≤ X ≤ π/3) = ∫(−π/2)^(π/3) (1/2) cos(x) dx = (1/2) [sin(x)] | (−π/2)^(π/3) = (1/2)(sin(π/3) - sin(−π/2)) = (1/2)(√3/2 + 1).

Task 4: Bivariate Standard Normal Variables

Next, we have bivariate standard normal variables X and Y with correlation coefficient 2/3. Hence, we need to find P(2X + Y ≤ 5).

This can be expressed as a linear combination of normal variables, thus simplifying to a standard normal distribution. This requires calculating the covariance of 2X + Y to express the desired probability in terms of the cumulative distribution function.

P(2X + Y ≤ 5) = Φ((5 - μ)/(σ)) where μ and σ must be computed based on the expectation and covariances between X and Y.

Task 5: Conditional Distributions

Under the specified joint distribution, we compute the conditional distribution of X given Y = 2 with initial distributions P(X = 1/4) and P(X = 3/4). The binomial property assists in refocusing on likelihood assessments.

Finally, the covariance Cov(X, Y) can be resolved using the identities tied to independence and the computed expectation of joint samples. Notably, pertaining to why X and Y do not possess a bivariate normal distribution requires examining their independence versus correlation delineations.

In conclusion, solving for the aforementioned tasks involves a series of fundamental probability principles and applications of statistical theories regarding distributions, expectations, and covariance estimations.

References

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