The Wall Street Journal Reported That Long Term Treasury Bon

The Wall Street Journal Reported That Long Term Treasury Bonds Had A M The Wall Street Journal reported that long term Treasury bonds had a mean return of 24.03% in 2008.

Assume that the returns for the long term Treasury bonds were distributed as a normal random variable, with a mean of 24.03 and a standard deviation of 10. If you select an individual Treasury bond from this population, what is the probability that it would have a return a. less than 0 (a loss)? b. between 10 and 20 ? c. greater than 10 ? If you select a random sample of 4 Treasury bonds from this population, what is the probability that the sample would have a mean return d. less than 0 (a loss)? e. between 10 and 20 ? f. greater than 10 ? g. Compare your results in parts (d) through (f) to those in parts (a) through (c). From long term experience it is known that the time required to answer a set of 10 computer-managed questions in the Data Analysis course follows a normal distribution with mean = 15 minutes and standard deviation = 2 minutes. If a randomly chosen off-campus student answers a test of 10 computer-managed questions, answer the following questions a. what is the probability that she would complete the test in less than 14 minutes? b. what is the probability that she would complete the test between 15 and 19 minutes? c. determine her completion time so that only 10% of the students doing the test will take longer than her? d. for a set of 5 randomly selected tests (each with 10 questions), what is the probability that her mean completion time will be 14 minutes or more?

Paper For Above instruction Analyzing the probability distributions related to financial returns and test completion times provides crucial insights into risk assessment and performance evaluation. This paper discusses the statistical probabilities concerning long-term Treasury bond returns and test durations, utilizing the properties of the normal distribution to quantify uncertainties and expectations in these contexts.

Probability Analysis of Treasury Bond Returns Given that the returns of long-term Treasury bonds are modeled as a normal distribution with a mean (µ) of 24.03% and a standard deviation (σ) of 10, we can compute the probabilities for various return levels using standard normal distribution techniques. The Z-score, which standardizes the value of interest, is calculated as: Z = (X - µ) / σ. a. The probability that a randomly selected bond yields a return less than 0 (a loss) is found by calculating the Z-score for X=0: Z = (0 - 24.03)/10 = -2.403. Consulting the standard normal table, the cumulative probability corresponding to Z = -2.403 is approximately 0.0082, or 0.82%. This indicates a very low

chance that a bond would produce a negative return in this scenario. b. For returns between 10 and 20, the respective Z-scores are Z■ = (10 - 24.03)/10 = -1.403 and Z■ = (20 - 24.03)/10 = -0.403. The respective cumulative probabilities are approximately 0.0803 and 0.3446. The probability that the return falls within this interval is the difference: 0.3446 - 0.0803 = 0.2643 or 26.43%. c. For returns greater than 10, Z = (10 - 24.03)/10 = -1.403, with a cumulative probability of about 0.0803.

The probability of returns exceeding 10 is thus 1 - 0.0803 = 0.9197 or 91.97%.

Probability for Sample Mean Return from Four Bonds When selecting a sample of four bonds, the sampling distribution of the mean has the same population mean but a reduced standard deviation, called the standard error: SE = σ / √n = 10 / √4 = 5. The probability that the sample mean is less than a particular value uses the Z-score with the standard error. d. The probability that the sample of four bonds has a mean return less than 0: Z = (0 - 24.03) / 5 = -4.806.

The cumulative probability for Z = -4.806 is nearly 0, indicating an almost zero likelihood of the sample mean being less than zero. This infers that it’s virtually impossible for the average return of a four-bond sample to be negative under these assumptions. e. The probability that the mean return is between 10 and 20 involves calculating Z for these points: Z■ = (10 - 24.03) / 5 = -2.806 and Z■ = (20 - 24.03) / 5 = -0.806. Their respective cumulative probabilities are approximately 0.0026 and 0.2090; so, the probability that the mean lies between these values is 0.2090 0.0026 = 0.2064 or 20.64%. f. The probability that the mean return exceeds 10 is 1 - 0.0026 = 0.9974 or 99.74%. This high probability reflects that the average return of a four-bond sample is very likely to be above 10% given the distribution parameters. g. Comparing the probabilities of parts (d) to (f) with those in (a) to (c) reveals how sample size influences the variability and likelihood of different outcomes. The probabilities of extreme outcomes diminish as the sample size increases, demonstrating the Law of Large Numbers—larger samples produce more stable and predictable average returns in this context.

Analysis of Test Completion Times The second part examines test durations modeled as a normal distribution with a mean of 15 minutes and a

standard deviation of 2 minutes. This analysis applies similar probabilistic reasoning to estimate how students perform under typical testing conditions. a. The probability that a student completes the test in less than 14 minutes is calculated with Z = (14 15)/2 = -0.5. The cumulative probability for Z = -0.5 is approximately 0.3085, indicating about 30.85% of students finish in less than 14 minutes. b. For completion times between 15 and 19 minutes, Z-scores are 0 (for 15 minutes) and 2 (for 19 minutes). The cumulative probabilities are 0.5 and 0.9772 respectively; thus, the probability of completion in this interval is 0.9772 - 0.5 = 0.4772 or 47.72%. c. To find the completion time exceeding which only 10% of students surpass, we find the 90th percentile: Z = 1.2816. Computing T = µ + Zσ yields T = 15 + 1.2816*2 ≈ 17.56 minutes. Hence, approximately 17.56 minutes is the cutoff time such that only 10% of students take longer. d. For five tests each with 10 questions, the mean completion time's distribution has a mean of 15 minutes and a standard error of σ / √n = 2 / √5 ≈ 0.8944. The probability that the sample mean exceeds 14 minutes is computed as: Z = (14 - 15) / 0.8944 ≈ -1.118. The cumulative probability for Z = -1.118 is about 0.1314, so the probability of a mean of 14 minutes or more is 1 - 0.1314 = 0.8686 or 86.86%. This indicates a high likelihood that the average of multiple tests exceeds 14 minutes, reflecting persistent typical performance time.

Conclusion This analysis illustrates how statistical methods, particularly normal distribution calculations, can effectively quantify and interpret uncertainties in financial returns and academic performance metrics.

Recognizing probabilities associated with various outcomes enables decision-makers to gauge risks and expectations more accurately, facilitating better planning and management in finance and education sectors.

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