Stat 200 Week 4 Homework Problems: The Commuter Trains ✓ Solved

Stat 200 Week 4 Homework Problems The Commuter Trains On The

Cleaned assignment instructions (condensed): Solve the week 4 statistics problems related to normal and uniform distributions, sampling distributions, confidence/warranty concepts, and basic probability with real-data contexts (commuter trains, blood pressure, dishwasher lifetimes, rainfall, cholesterol tests, etc.). Provide calculations, interpretations, and brief explanations for each sub-question, covering variable definitions, distribution assumptions, probabilities, percentiles, and sampling distributions. Include a discussion of when normal approximations are appropriate, how to transform to standard normal, and how sample size affects the distribution of the sample mean. The problems involve uniform and normal models, z-scores, percentiles, and basic hypothesis- or decision-style reasoning about unusual values and warranty decisions.

Below are the cleaned assignment instructions, organized by problem sections as they appeared in the original prompt. Each item is preserved in its essential form to guide the full solution.

6.1 The Commuter Trains On The Red Line for the Regional Transit Authority (RTA) in Cleveland, OH have a waiting time during peak rush hour periods of eight minutes. a) State the random variable. b) Find the height of this uniform distribution. c) Find the probability of waiting between four and five minutes. d) Find the probability of waiting between three and eight minutes. e) Find the probability of waiting five minutes exactly.

6.3.2 Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean 0 and standard deviation 1. a) The area to the left of z is 15%. b) The area to the right of z is 65%. c) The area to the left of z is 10%. d) The area to the right of z is 5%. e) The area between 0 and z is 95%. (Hint: draw a picture and figure out the area to the left of the z.) f) The area between 0 and z is 99%.

6.3.4 According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is normally distributed. a) State the random variable. b) Find the probability that a person in China has blood pressure of 135 mmHg or more. c) Find the probability that a person in China has blood pressure of 141 mmHg or less. d) Find the probability that a person in China has blood pressure between 120 and 125 mmHg. e) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not? f) What blood pressure do 90% of all people in China have less than?

6.3.8 A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). Assume the life of a dishwasher is normally distributed. a) State the random variable. b) Find the probability that a dishwasher will last more than 15 years. c) Find the probability that a dishwasher will last less than 6 years. d) Find the probability that a dishwasher will last between 8 and 10 years. e) If you found a dishwasher that lasted less than 6 years, would you think that you have a problem with the manufacturing process? Why or why not? f) A manufacturer of dishwashers only wants to replace free of charge 5% of all dishwashers. How long should the manufacturer make the warranty period?

6.3.10 The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69 mm ("Annual maximums of," 2013). Assume rainfall is normally distributed. a) State the random variable. b) Find the probability that the yearly rainfall is less than 100 mm. c) Find the probability that the yearly rainfall is more than 240 mm. d) Find the probability that the yearly rainfall is between 140 and 250 mm. e) If a year has a rainfall less than 100 mm, does that mean it is an unusually dry year? Why or why not? f) What rainfall amount are 90% of all yearly rainfalls more than?

6.4.4 Annual rainfalls for Sydney, Australia are given in table #6.4.6. ("Annual maximums of," 2013). Can you assume rainfall is normally distributed? Table #6.4.6: Annual Rainfall in Sydney, Australia. A random variable is normally distributed with a mean of 245 and a standard deviation of 21. a) If you take a sample of size 10, can you say what the shape of the distribution for the sample mean is? Why? b) For a sample of size 10, state the mean of the sample mean and the standard deviation of the sample mean. c) For a sample of size 10, find the probability that the sample mean is more than 241. d) If you take a sample of size 35, can you say what the shape of the distribution of the sample mean is? Why? e) For a sample of size 35, state the mean of the sample mean and the standard deviation of the sample mean. f) For a sample of size 35, find the probability that the sample mean is more than 241. g) Compare your answers in part c and f. Why is one smaller than the other?

6.5.4 According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Blood pressure is normally distributed. a) State the random variable. b) Suppose a sample of size 15 is taken. State the shape of the distribution of the sample mean. c) Suppose a sample of size 15 is taken. State the mean of the sample mean. d) Suppose a sample of size 15 is taken. State the standard deviation of the sample mean. e) Suppose a sample of size 15 is taken. Find the probability that the sample mean blood pressure is more than 135 mmHg. f) Would it be unusual to find a sample mean of 15 people in China of more than 135 mmHg? Why or why not? g) If you did find a sample mean for 15 people in China to be more than 135 mmHg, what might you conclude?

6.5.6 The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the standard deviation is 1.0 mmol/l (Lawes, Hoorn, Law & Rodgers, 2004). Assume that cholesterol levels are normally distributed. a) State the random variable. b) Find the probability that a woman age 45-59 in Ghana has a cholesterol level above 6.2 mmol/l (considered a high level). c) Suppose doctors decide to test the woman’s cholesterol level again and average the two values. Find the probability that this woman’s mean cholesterol level for the two tests is above 6.2 mmol/l. d) Suppose doctors being very conservative decide to test the woman’s cholesterol level a third time and average the three values. Find the probability that this woman’s mean cholesterol level for the three tests is above 6.2 mmol/l. e) If the sample mean cholesterol level for this woman after three tests is above 6.2 mmol/l, what could you conclude?

6.5.8 A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). The life of a dishwasher is normally distributed. Suppose you are a manufacturer and you take a sample of 10 dishwashers that you made. a) State the random variable. b) Find the mean of the sample mean. c) Find the standard deviation of the sample mean. d) What is the shape of the sampling distribution of the sample mean? Why? e) Find the probability that the sample mean of the dishwashers is less than 6 years. f) If you found the sample mean life of the 10 dishwashers to be less than 6 years, would you think that you have a problem with the manufacturing process? Why or why not?

Note: Some symbols at the end of the original prompt appeared garbled (for example, “µ = 0 m=0 σ =1 s=1”). Those were omitted in the cleaned version, as they do not affect the core assignment tasks.

Paper For Above Instructions

6.1 The Commuter Trains On The Red Line (Uniform Distribution)

6.3.2 Z-scores for Given Areas

Understanding z-scores requires converting areas under the standard normal curve to cutoff values. a) If the area to the left of z is 0.15, z ≈ −1.04. b) If the area to the right of z is 0.65, the left-tail area is 0.35, so z ≈ −0.39. c) If the area to the left of z is 0.10, z ≈ −1.28. d) If the area to the right of z is 0.05, left area is 0.95, so z ≈ 1.64. e) If the area between 0 and z is 0.95, z ≈ 1.64 (assuming symmetry and standard normal properties). f) If the area between 0 and z is 0.99, z ≈ 2.33. In practice, exact values are taken from z-tables or standard normal software; the key is translating areas into unique z-scores and using Phi(z) to determine probabilities.

6.3.4 Blood Pressure for China (Normal Distribution)

Let X be blood pressure for a Chinese adult, X ~ N(μ, σ²) with μ = 128 and σ = 23. a) Random variable: X = systolic blood pressure (mmHg). b) P(X ≥ 135) = 1 − Φ((135 − 128)/23) ≈ 1 − Φ(0.304) ≈ 1 − 0.618 ≈ 0.382. c) P(X ≤ 141) = Φ((141 − 128)/23) ≈ Φ(0.565) ≈ 0.714. d) P(120 ≤ X ≤ 125) = Φ((125 − 128)/23) − Φ((120 − 128)/23) ≈ Φ(−0.130) − Φ(−0.348) ≈ 0.448 − 0.363 ≈ 0.085. e) 135 mmHg is not unusual for this population (z ≈ 0.30). f) The 90th percentile: x0.90 = μ + z0.90 σ ≈ 128 + 1.2816(23) ≈ 157.5 mmHg. This demonstrates how normal models provide simple probabilistic answers for physiological measurements when the assumptions are reasonable.

6.3.8 Dishwasher Lifetimes (Normal Distribution)

Let X be the lifetime (years) of a dishwasher, X ~ N(μ, σ²) with μ = 12 and σ = 1.25. a) Random variable: X = dishwasher lifetime in years. b) P(X > 15) = 1 − Φ((15 − 12)/1.25) = 1 − Φ(2.40) ≈ 1 − 0.9918 ≈ 0.0082. c) P(X

6.3.10 Sydney Rainfall (Normal Distribution)

Let X be yearly rainfall (mm) in Sydney, X ~ N(μ, σ²) with μ = 137 and σ = 69. a) Random variable: X = yearly rainfall in mm. b) P(X 240) = 1 − Φ((240 − 137)/69) ≈ 1 − Φ(1.482) ≈ 0.069. d) P(140 ≤ X ≤ 250) = Φ((250 − 137)/69) − Φ((140 − 137)/69) ≈ Φ(1.636) − Φ(0.043) ≈ 0.948 − 0.516 ≈ 0.432. e) A year with rainfall

6.4.4 Normality of Sydney Rainfall Data

Annual rainfall data for Sydney can exhibit skewness and heavy tails, especially due to extreme wet or dry years. A formal normality check—such as visual histograms, Q-Q plots, or normality tests (e.g., Shapiro–Wilk)—is essential before applying a normal-model approximation to rainfall daily or yearly totals. In practice, rainfall data often deviate from normality, so while normal approximations can be useful for quick estimates, they should be validated against data-driven diagnostics. This discussion emphasizes the importance of graphical diagnostics and goodness-of-fit tests in model selection for environmental measurements.

6.5.4 Sampling Distributions and Mean Life (MONICA Blood Pressure Example)

a) Random variable: X = cholesterol level (mmol/L) for a woman aged 45–59 in Ghana/Nigeria/Seychelles. b) For Ghana with μ = 5.1 and σ = 1.0, P(X > 6.2) = 1 − Φ((6.2 − 5.1)/1.0) = 1 − Φ(1.1) ≈ 0.136. c) Two measurements averaged: X̄2 ~ N(μ, σ²/2) with σ/√2 ≈ 0.707. P(X̄2 > 6.2) = 1 − Φ((6.2 − 5.1)/0.707) ≈ 1 − Φ(1.556) ≈ 0.060. d) Three measurements averaged: X̄3 ~ N(μ, σ²/3) with σ/√3 ≈ 0.577. P(X̄3 > 6.2) ≈ 1 − Φ((6.2 − 5.1)/0.577) ≈ 1 − Φ(1.907) ≈ 0.028. e) If the three-test mean exceeds 6.2, this is relatively unlikely under the baseline, but not impossible; it could indicate a higher true mean or measurement variability, warranting follow-up testing or clinical review. These calculations illustrate how increasing sample size sharpens inference by reducing the standard error, thereby changing the tail probabilities in a sampling distribution.

6.5.6 Cholesterol Levels Across Countries (Two- and Three-Test Averages)

a) Random variable: X = cholesterol level (mmol/L) for a woman aged 45–59 in the given country. b) Ghana example: P(X > 6.2) ≈ 0.136 (as above). c) Two-test mean: X̄2 ~ N(μ, σ²/2); P(X̄2 > 6.2) ≈ 0.060. d) Three-test mean: X̄3 ~ N(μ, σ²/3); P(X̄3 > 6.2) ≈ 0.029. e) If the three-test mean exceeds 6.2, it provides stronger evidence that the true mean is higher than 6.2 than a single measurement; however, conclusions should consider multiple comparisons, measurement reliability, and population differences. The exercise shows how repeated sampling reduces measurement error and refines inference about whether a patient’s mean cholesterol level crosses a clinically meaningful threshold.

6.5.8 Sampling Distribution of the Mean Life for Dishwashers

a) Random variable: X̄10, the mean life (years) of a sample of 10 dishwashers. b) The mean of the sampling distribution is μX̄ = μ = 12 years. c) The standard deviation is σX̄ = σ/√n = 1.25/√10 ≈ 0.395. d) Since the population is normal, the sampling distribution of X̄10 is Normal for any n. e) P(X̄10

Summary and Implications

Across these problems, the core ideas include correctly identifying the underlying distribution (uniform vs. normal), computing probabilities from interval lengths for uniform models, converting tail areas to z-scores for normal models, and understanding how sampling distributions change with sample size. Normality justifies many convenient inferences in sampling distributions of the mean, while real-world data demand diagnostic checks for normality and appropriateness of models. The examples also illustrate practical decision-making implications, such as warranty-setting and screening thresholds, where tail probabilities guide policy and quality control decisions. The connection between simple formulas and real-world interpretation remains central to statistical thinking in applied contexts.

References

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