Expected Number Of Admissions, Variance, And Standard Deviat

Expected number of admissions, variance, and standard deviation of a discrete probability distribution

The assignment involves calculating the expected number of admissions at Kinzua University based on a given probability distribution, as well as determining the variance and standard deviation of these admissions. Additionally, the assignment includes analyzing hypergeometric probabilities related to charitable contributions, applying the properties of the hypergeometric distribution, and understanding the assumptions behind it. Moreover, it explores the probability of chance occurrences in the context of stock market trends based on historical data. Further, the task involves working with uniform and normal distributions to calculate probabilities and percentiles related to operational times, population proportions, and product longevity, followed by interpretations and providing valid statistical guidelines supported by credible references.

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Introduction

Statistical analysis forms the backbone of decision-making in various fields, including education management, finance, market research, and quality control. The analysis of probability distributions—discrete, uniform, and normal—provides essential tools for understanding variability, estimating likelihoods, and making predictions. This paper explores multiple statistical concepts and applications, focusing on expected values, variances, standard deviations, hypergeometric and binomial distributions, and the normal distribution. These concepts are illustrated through real-world scenarios such as university admissions, charitable contributions, stock market predictions, and manufacturing quality controls.

Expected Admissions and Variability in Student Enrollment

To estimate the expected number of student admissions for the fall semester at Kinzua University, we consider the probabilities associated with different admission levels. The given data suggest three potential admission figures: 1,060, 1,320, and 1,620 students, with respective probabilities of 0.3, 0.2, and 0.5. The expected value (mean) for admissions is computed as:

E(X) = Σ [x P(x)] = (1060 0.3) + (1320 0.2) + (1620 0.5) = 318 + 264 + 810 = 1,392

Thus, the expected number of admissions is approximately 1,392 students.

To assess the variability around this expectation, we compute the variance and standard deviation. The variance of a discrete distribution is calculated as:

Var(X) = Σ [ (x - μ)^2 * P(x) ]

where μ = 1,392. Computing each squared deviation multiplied by the probability:

  • (1060 - 1392)^2 0.3 = (332)^2 0.3 = 110,224 * 0.3 = 33,067.2
  • (1320 - 1392)^2 0.2 = (72)^2 0.2 = 5,184 * 0.2 = 1,036.8
  • (1620 - 1392)^2 0.5 = (228)^2 0.5 = 51,984 * 0.5 = 25,992

Summing these gives the variance:

Variance = 33,067.2 + 1,036.8 + 25,992 = 60,096.8

The standard deviation is the square root of the variance:

Std Dev = √60,096.8 ≈ 245.11

Therefore, the expected number of admissions is 1,392 with a standard deviation of approximately 245.11, indicating the typical fluctuation around this mean.

Hypergeometric Distribution and Charitable Contributions

The second scenario examines the probability that exactly one of the four audited returns claims a charitable deduction of over $1,000, based on a sample where 6 out of 20 returns exceed this amount. Since sampling is without replacement from a finite population, the hypergeometric distribution applies.

Parameters include:

  • Population size, N = 20
  • Number of successes in population, K = 6
  • Number of draws, n = 4
  • Number of successes in sample, k = 1

The probability of exactly one success is calculated as:

P(X = 1) = [C(K, k) * C(N - K, n - k)] / C(N, n)

Where C(n, k) is the combination function ("n choose k"). Computing each component:

  • C(6, 1) = 6
  • C(14, 3) = 364
  • C(20, 4) = 4845

Thus, the probability is:

P(X=1) = (6 * 364) / 4845 ≈ 2184 / 4845 ≈ 0.4503

Next, the probability that at least one return has a high charitable contribution is:

P(X ≥ 1) = 1 - P(X=0)

Where P(X=0) (no successes) is:

P(X=0) = C(6, 0) C(14, 4) / C(20, 4) = 1 1001 / 4845 ≈ 0.2066

Therefore, the probability that at least one return claims over $1,000 is:

P(X ≥ 1) = 1 - 0.2066 = 0.7934

Testing the "January Theory" in Stock Market Trends

The "January effect" suggests that the market's behavior in January can predict its performance for the entire year. Historically, this has held true 24 times out of 34 years. Assuming no inherent trend (probability of success = 0.5), we analyze the likelihood of this outcome occurring by chance using the binomial distribution.

The probability of observing 24 or more successes in 34 trials under the null hypothesis (p=0.5) is calculated by summing the binomial probabilities for X=24 to X=34. Using a calculator or software, the cumulative probability P(X ≥ 24) is approximately 0.007633, indicating that such an occurrence is quite unlikely under randomness.

Uniform Distribution in Support Time for Technical Support

The time to resolve internet support issues, modeled by a uniform distribution, ranges from a minimum of 1.5 minutes to a maximum of 12 minutes. The parameters a and b are derived directly from this data:

  • a = 1.5 minutes
  • b = 12 minutes

The mean support time, μ, for a uniform distribution is:

μ = (a + b) / 2 = (1.5 + 12) / 2 = 13.5 / 2 = 6.75 minutes

The standard deviation, σ, is calculated as:

σ = (b - a) / √12 = (12 - 1.5) / 3.464 ≈ 10.5 / 3.464 ≈ 3.03 minutes

Approximately 66.67% of the support times are expected to take longer than 5 minutes, computed by the proportion of the interval exceeding this time:

Since the distribution is uniform, the probability that a time exceeds 5 minutes is:

P > 5 = (b - 5) / (b - a) = (12 - 5) / 10.5 ≈ 7 / 10.5 ≈ 0.6667 or 66.67%

Similarly, the times corresponding to the middle 50% of support durations (interquartile range) are determined by calculating the 25th and 75th percentiles:

Q1 = a + 0.25(b - a) = 1.5 + 0.2510.5 = 1.5 + 2.625 = 4.125 minutes

Q3 = a + 0.75(b - a) = 1.5 + 0.7510.5 = 1.5 + 7.875 = 9.375 minutes

Normal Distribution in Population and Measurement Analysis

A population with a mean of 21 and standard deviation of 6 is analyzed to find the z-score for 24:

  • Z = (X - μ) / σ = (24 - 21) / 6 = 0.5

The proportion of the population between 21 and 24 is obtained from the standard normal distribution table, corresponding to Z between 0 and 0.5, which is approximately 0.1915. Since the total area between 21 and 24 is twice this value, the proportion is about 0.3830. The proportion less than 17 is calculated by:

  • Z = (17 - 21) / 6 = -0.6667

Looking up this Z-score, the cumulative probability is approximately 0.2525, indicating about 25.25% of the population scores below 17.

Operating Costs and Material Lifespan Using Normal Distribution

The hourly operating cost follows a normal distribution with mean $5,957 and standard deviation $288. To find the minimum cost at which only the lowest 4% of airplanes operate, we locate the corresponding z-value for 0.04 in the standard normal distribution table, which is approximately -1.75. Therefore, the operating cost is:

Cost = μ + z σ = 5957 + (-1.75) 288 ≈ 5957 - 504 ≈ $5453

Similarly, the manufacturer of laser cartridges can determine the page count threshold to ensure 90% of shipments last at least a certain number of pages. Given a mean of 12,375 pages and a standard deviation of 645, the z-value corresponding to the 10th percentile (since the lower 10%) is approximately -1.28. The threshold is:

Page Count = 12375 + (-1.28) * 645 ≈ 12375 - 826 ≈ 11549 pages

Conclusion

This comprehensive analysis demonstrates the importance of understanding various probability distributions—discrete, uniform, hypergeometric, binomial, and normal—in addressing real-world problems. The expected values, variances, and probabilities derived from these distributions facilitate informed decision-making in education, finance, quality control, and manufacturing. Accurate application of these concepts, supported by credible statistical data and calculations, enhances the reliability of predictions and strategizing across diverse fields of study.

References

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