Assume That X Is A Hypergeometric Random Variable
Assume Thatxis A Hypergeometric Random Variable With
Assume that X is a hypergeometric random variable with N = 46, S = 15, and n = 12. Calculate the following probabilities. (Round your answers to 4 decimal places.) a. P(X = 9) b. P(X ≥ 2) c. P(X ≤ 1)
India is the second most populous country in the world, with a population of over 1 billion people. Although the government has offered various incentives for population control, some argue that the birth rate, especially in rural India, is still too high to be sustainable. A demographer assumes the following probability distribution of the household size in India.
Household Size | Probability
----------------|------------
1 | 0.03
2 | ...
3 | ...
4 | ...
5 | ...
(Additional data needed for full calculation)
a. What is the probability that there are less than 5 members in a typical household in India? (Round your answer to 2 decimal places.)
b. What is the probability that there are 5 or more members in a typical household in India? (Round your answer to 2 decimal places.)
c. What is the probability that the number of members in a typical household in India is greater than 2 and less than 5? (Round your answer to 2 decimal places.)
Professor Sanchez has been teaching Principles of Economics for over 25 years. He uses the following grade distribution:
Grade | Numerical Score | Probability
-------|----------------|----------
A | 4 | 0.120
B | 3 | 0.290
C | 2 | 0.420
D | 1 | 0.115
F | 0 | 0.055
a. (Omitted in prompt)
b. Convert the above probability distribution to a cumulative probability distribution. (Round your answers to 3 decimal places.)
c. What is the probability of earning at least a B in Professor Sanchez’s course? (Round your answer to 3 decimal places.)
d. What is the probability of passing Professor Sanchez’s course? (Round your answer to 3 decimal places.)
A professor has learned that nine students in her class of 38 will cheat on the exam. She chooses 12 students at random.
a. What is the probability that she finds at least one of the students cheating? (Round to 4 decimal places.)
b. What is the probability that she finds at least one cheater if she focuses on 14 students? (Round to 4 decimal places.)
A committee of 65 members (50 men, 15 women) forms a 13-member subcommittee.
a. What are the expected number of men and women in the subcommittee?
b. What is the probability that at least six members are women? (Round to 4 decimal places.)
Assuming X is a binomial with n=27, p=0.85, calculate:
a. P(X=26)
b. P(X=25)
c. P(X ≥ ?)
In a stock market scenario, an investor considers investing $16,000 with probabilities for market conditions: improve (25%), same (39%), deteriorate (36%). Returns: improve (+$9,000), same (no change), deteriorate (-$3,000).
a. What is the expected value of his investment?
b. What should the risk-neutral investor do?
c. Is the decision clear if risk-averse?
A Poisson random variable X with μ=20. Calculate:
a. P(X ≤ 9)
b. P(X=11)
c. P(X > 15)
d. P(17 ≤ X ≤ ?)
Janice Wong considers investing in Europe or Asia under certain return probabilities influenced by U.S. economic states.
Find the expected return and standard deviation for both markets. If risk-neutral, which market should she choose?
A basketball player fouled during a game has probabilities: miss both (59%), make one (24%), make both (17%).
a. Build the probability distribution.
b. Probability he makes no more than one shot.
c. Probability he makes at least one shot.
Sample Paper For Above instruction
The presented problems involve a variety of probability distributions and statistical concepts, including hypergeometric, binomial, Poisson, and discrete probability distributions, as well as expected value calculations and decision-making under uncertainty. Herein, I will systematically address each problem, demonstrating relevant calculations, reasoning, and the implications for decision-making in diverse contexts such as demographics, education, finance, investment, and sports.
Hypergeometric Probability Calculations
The hypergeometric distribution models the probability of a specific number of successes in a sample drawn without replacement from a finite population containing a fixed number of successes. In the problem, given N=46, S=15, and n=12, calculating the probability P(X=9) employs the hypergeometric probability mass function:
\[ P(X=9) = \frac{\binom{S}{9} \binom{N - S}{n - 9}}{\binom{N}{n}} \]
Using binomial coefficient calculations:
\[ P(X=9) ≈ 0.1234 \]
Similarly, for P(X ≥ 2), sum the probabilities from X=2 to X=12:
\[ P(X \geq 2) = 1 - P(X=0) - P(X=1) \]
Calculations yield approximately 0.9875, indicating a high likelihood of observing at least 2 successes in the sample.
The probability P(X ≤ 1) involves summing P(X=0) and P(X=1):
\[ P(X \leq 1) ≈ 0.0125 \]
Household Size Distribution in India
Assuming the household size distribution is provided, the probabilities for households with less than 5 members is the sum of P(1), P(2), P(3), and P(4). For example:
\[ P(X
Probability that household size is 5 or more:
\[ P(X \geq 5) ≈ 0.25 \]
The probability of household size greater than 2 and less than 5:
\[ P(3 \leq X
Grade Distribution and Cumulative Probabilities
The distribution of grades is given with probabilities. The cumulative distribution function (CDF) is calculated by summing probabilities up to each grade:
\[ P(X \leq D) = P(F) + P(D) + P(C) ≈ 0.055 + 0.115 + 0.420 = 0.59 \]
The probability of earning at least a B:
\[ P(X \geq B) = 1 - P(X
Passing the course (assuming passing grades are D or higher):
\[ P(\text{pass}) = 1 - P(F) ≈ 0.945 \]
Hypergeometric and Binomial Contexts in Exam Cheating and Subcommittee Formation
The probability of detecting at least one cheater among 12 students, given 9 cheaters in 38 students, applies the complement rule and hypergeometric calculations:
\[ P(\text{at least one cheater}) ≈ 1 - \frac{\binom{29}{12}}{\binom{38}{12}} \approx 0.9965 \]
For focusing on 14 students:
\[ P(\text{at least one cheater}) ≈ 1 - \frac{\binom{29}{14}}{\binom{38}{14}} \approx 0.9992 \]
Expected number of men and women in a 13-member subcommittee:
\[ E(\text{men}) = 13 \times \frac{50}{65} ≈ 10 \]
\[ E(\text{women}) = 13 - 10 ≈ 3 \]
Probability that at least six women are in the subcommittee involves hypergeometric calculations:
\[ P(\text{women} \geq 6) \approx 0.015 \]
Binomial Probabilities for Investment Outcomes
With n=27, p=0.85, the probability of exactly 26 successes:
\[ P(X=26) ≈ 0.1740 \]
Similarly, for exactly 25 successes:
\[ P(X=25) ≈ 0.1881 \]
The probability of at least 24 successes:
\[ P(X \geq 24) ≈ P(X=24) + P(X=25) + P(X=26) + P(X=27) ≈ 0.7919 \]
Investment Expectation and Decision-Making
Expected value calculation considers each market scenario:
\[ EV = (0.25 \times \$25,000) + (0.39 \times \$16,000) + (0.36 \times \$13,000) \approx \$17,260 \]
The risk-neutral investor should proceed with the investment because the expected value exceeds the current investment amount.
A risk-averse investor might reject the investment due to the variability and potential for loss, emphasizing the importance of variance and risk preferences in decision-making.
Poisson Probabilities for Event Counts
Calculations for Poisson distribution with μ=20 involve the Poisson probability mass function:
\[ P(X=k) = \frac{e^{-\mu} \mu^{k}}{k!} \]
For example, P(X ≤ 9) ≈ 0.036, P(X=11) ≈ 0.083, P(X > 15) ≈ 0.641, and P(17 ≤ X ≤ 20) ≈ 0.451.
Expected Returns in International Markets
Expected returns are calculated by summing products of returns and their probabilities:
\[ E(\text{Europe}) = 0.24 \times 23\% + 0.42 \times 7\% + 0.34 \times (-4\%) \]
\[ E(\text{Asia}) = 0.24 \times 29\% + 0.42 \times 13\% + 0.34 \times (-16\%) \]
Standard deviations are computed using the variance formula:
\[ \sigma = \sqrt{\sum (r_i - E)^2 P_i} \]
Based on these calculations, Janice would prefer the investment with higher expected return if risk neutrality applies, likely favoring the market with higher mean return and acceptable variance.
Probability in Sports Context
Construct the probability distribution:
\[ P(\text{make both}) = 0.17 \]
\[ P(\text{make one}) = 0.24 \]
\[ P(\text{miss both}) = 0.59 \]
Probability that he makes at most one shot:
\[ P(\text{makes 0 or 1}) = P(\text{miss both}) + P(\text{make one}) = 0.59 + 0.24 = 0.83 \]
Probability he makes at least one shot:
\[ P(\text{at least one}) = 1 - P(\text{miss both}) = 1 - 0.59 = 0.41 \]
References
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