Find The Following Probabilities P Z = 5, B Z = 5, C P

Find The Following Probabilitiesap 0 Z 5bp Z 5c P

1) Find the following probabilities: a) P (0 ≤ Z ≤ .5) b) P (Z ≤ .5) c) P (Z ≤ -.5) d) P (Z ≥ .5) e) P (- .5 ≤ Z ≤ .5)

2) Dearborn Tech produces a circuit board that has an average life span of 4,500 hours. The life span is normally distributed with a standard deviation of 500 hours. The firm is considering a 3,800 hours’ warranty on the circuit board. If this warranty policy is adopted, what proportion of circuit boards should the firm expect to replace under warranty?

Paper For Above instruction

In this paper, we analyze two statistical probability problems: first, calculating specific probabilities related to the standard normal distribution (Z-distribution), and second, applying the properties of the normal distribution to determine the proportion of circuit boards that will require replacement under a given warranty period. Both problems involve foundational understanding and application of the normal distribution, which is central to many statistical analyses in quality control and reliability engineering.

Part 1 addresses probability calculations involving the standard normal distribution. The standard normal distribution, characterized by a mean of 0 and a standard deviation of 1, is a widely used model for continuous data that is symmetric and bell-shaped. The specific probabilities requested include cumulative probabilities like P(Z ≤ .5) and P(Z ≥ .5), as well as probabilities within a range such as P(-.5 ≤ Z ≤ .5). To compute these probabilities, we typically consult standard normal distribution tables or utilize statistical software or calculators that provide cumulative distribution function (CDF) values for Z-scores.

For instance, the probability P(0 ≤ Z ≤ .5) is the difference between P(Z ≤ .5) and P(Z ≤ 0). The symmetry of the normal distribution simplifies some calculations, especially P(Z ≤ -.5), which equals P(Z ≥ .5) due to the distribution's symmetry about zero. The range probability P(-.5 ≤ Z ≤ .5) can be calculated as P(Z ≤ .5) - P(Z ≤ -.5).

Part 2 applies the normal distribution to a practical reliability problem. The machine produces circuit boards with an average lifespan (mean) of 4,500 hours and a standard deviation of 500 hours. The question is what proportion of these circuit boards will need replacements if the warranty covers 3,800 hours. This involves calculating the Z-score for 3,800 hours: Z = (X - μ) / σ = (3800 - 4500)/500 = -700/500 = -1.4. Using standard normal tables or software, the cumulative probability P(Z ≤ -1.4) gives the proportion of circuit boards expected to fail before 3,800 hours.

This probability indicates the expected failure rate under the warranty policy, which helps the firm in financial planning and risk assessment. Typically, a P(Z ≤ -1.4) ≈ 0.0808, meaning approximately 8.08% of circuit boards will require replacement under the 3,800-hour warranty period.

In conclusion, these problems demonstrate how statistical tools like the standard normal distribution are essential in both theoretical probability calculations and real-world applications such as warranty analysis, quality control, and reliability engineering. Accurate probability assessments assist companies in making informed decisions about product warranties, quality standards, and process improvements, ultimately leading to better resource allocation and customer satisfaction.

References

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