Use The Following Context For Questions 1 Through 5
Use The Following Context For The Questions 1 Thru 5it Is Given That 6
Use the following context for questions 1 through 5: It is given that 60% of the employees at a large company are women. A simple random sample of 15 employees is selected from this company. We are applying the binomial distribution in this situation.
1. What is the mean of this binomial distribution?
2. What is the standard deviation of this binomial distribution?
3. Find the probability that the number of women in this sample will be exactly 8.
4. Find the probability that the number of women in this sample will be 8 or less.
5. Find the probability that the number of women in this sample will be more than 8.
Consider the following context for questions 6-8: A university has determined that the probability that a new student with a confirmed dorm assignment will actually take the dorm space is 0.87. The university has 800 spaces for new students.
6. How many students are expected to take the dorm space if the number of confirmed assignments is 800?
7. The university confirms the dorm assignments for 870 students. Use the binomial distribution to find the probability that the number of students actually taking the dorm space will be 800 or less.
8. Is the university taking too high a risk in sending confirmation to 870 students under the assumption that p=0.87 and that all students make decisions independently?
Consider the following statement for questions 9-15: Given that the lengths of human pregnancies are normally distributed with a mean of 266 days and standard deviation 16 days.
9. Find the probability that a human pregnancy will be less than 260 days long.
10. Find the probability that a human pregnancy will be less than 280 days long.
11. Find the probability that a human pregnancy will be between 260 days and 280 days long.
12. Find the 90th percentile of the length of human pregnancies.
13. Explain in one sentence, what does the answer for the 90th percentile mean in this context.
14. Find the probability that a human pregnancy will be longer than 280 days.
15. An insurance company has 10 cases of pregnancies among its insured people. Use the binomial distribution to find the probability that no more than 2 of these will be longer than 280 days.
Consider the following for questions 16-20: An insurance company has determined that the mean of the claim payments on collisions in a certain region is $1300 with a standard deviation of $2400. The company has 100 claims coming up and considers these as a simple random sample from the population with the same mean and standard deviation.
16. What is the mean of the means of simple random samples of size 100 from this population?
17. What is the standard deviation of the means of simple random samples of size 100 from this population?
18. Does the reference population in this situation have a normal distribution? Provide a reason.
19. Why can we still treat the distribution of sample means for simple random samples of size 100 as approximately normal?
20. Find the probability that the total claim of a simple random sample of 100 claims will be more than $200,000.
Paper For Above instruction
The application of probability distributions in various real-world scenarios provides crucial insights into understanding and predicting outcomes in fields such as business, healthcare, and social sciences. This comprehensive analysis explores the concepts of binomial and normal distributions and their practical implications based on given contexts involving employee demographics, student housing, pregnancy lengths, and insurance claim data.
Binomial Distribution Applications
The first scenario considers the employee demographic at a large company. With 60% of employees being women, we examine a random sample of 15 employees to understand the variability in the number of women present. The mean (expected value) of this binomial distribution can be calculated using the formula μ = np, where n is the sample size and p is the probability of success. Here, μ = 15 × 0.6 = 9 women. This indicates that, on average, 9 women are expected in any randomly selected group of 15 employees.
The standard deviation, representing the variability, is given by σ = √(np(1-p)), which equals √(15 × 0.6 × 0.4) ≈ 1.90. Probability calculations further elucidate specific likelihoods: the probability of exactly 8 women involves the binomial probability formula P(X = 8), calculated as C(15,8) × (0.6)^8 × (0.4)^7. The probability of 8 or fewer women, P(X ≤ 8), requires summing probabilities from 0 through 8, typically computed with binomial tables or statistical software. Conversely, the probability of more than 8 women, P(X > 8), equals 1 – P(X ≤ 8).
University Dorm Space Study
In the second context, the probability that a new student takes the dorm space is 0.87. For 800 assignments, the expected number of students to occupy dorms is directly obtained via the mean of the binomial distribution: E = np = 800 × 0.87 = 696 students. This expected value informs the university of typical occupancy levels, assisting in resource planning.
When 870 students are confirmed, applying the binomial distribution allows us to compute the probability that 800 or fewer will actually take the dorms. Using the normal approximation to the binomial (given the large n), the mean remains 696, and the standard deviation approximates to √(np(1-p)) ≈ √(870 × 0.87 × 0.13) ≈ 10.33. Standardizing and calculating the probability P(X ≤ 800) facilitates risk assessment. If this probability is very low, the university might be at risk of overconfirming students, leading to logistical issues or student dissatisfaction.
Normal Distribution in Pregnancy Lengths
The lengths of human pregnancies are modeled by a normal distribution with a mean of 266 days and standard deviation of 16 days. Probabilities such as less than 260 days or less than 280 days involve calculating z-scores { (X – μ) / σ } and consulting standard normal tables or software. For example, P(X
Similarly, P(X
Inverting the percentile, the probability that a pregnancy exceeds 280 days relates to the upper tail of the normal distribution, approximately 0.19. The binomial distribution becomes applicable when considering the probability that no more than 2 out of 10 pregnancies extend beyond 280 days; this involves summing binomial probabilities with p ≈ 0.19 or relevant transformation.
Insurance Claims and Statistical Inference
For the insurance claims, the mean total losses from 100 claims are estimated by multiplying the population mean ($1300) by the sample size, resulting in an expected total of $130,000. The variability in total claims can be assessed using the standard deviation of the sum, calculated as √n × σ, which yields approximately 2400 × √100 = $24,000. Thus, the distribution of total claims is centered around $130,000 with this variability.
Applying the Central Limit Theorem, the distribution of sample means is approximately normal, justifying the calculation of the probability that total claims exceed $200,000. Standardizing yields a z-score, and probability lookup indicates the likelihood of such an event, which informs risk management and reserve policies for the insurance company.
Conclusion
The application of binomial and normal distributions across these scenarios demonstrates their importance in decision-making processes, resource allocation, and risk assessment. Whether analyzing employee demographics, student housing logistics, pregnancy durations, or insurance losses, understanding these distributions allows organizations to make informed, data-driven choices that optimize outcomes and mitigate potential risks.
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