Paper For Above instruction
The problem at hand involves understanding and calculating probabilities associated with normally distributed variables, specifically related to the lifespan of a certain refrigerator type. The normal distribution, characterized by its mean (average) and standard deviation, is extensively used in statistics to model continuous data that clusters around a central value. This paper systematically addresses the various probability questions posed, providing both calculations and contextual interpretations to enhance comprehension of the distribution's real-world implications.
Understanding the Normal Distribution and Its Parameters
The normal distribution, often depicted as the bell curve, is defined by two parameters: the mean (µ) and the standard deviation (σ). The mean indicates the central tendency or the expected value of the data, while the standard deviation measures the spread or variability around the mean. In this context, the lifespan of refrigerators follows a normal distribution with a specified mean and standard deviation, enabling the use of standard statistical tools to compute probabilities related to their longevity.
Part A: Calculating the Probability
P(2.9 ≤ x ≤ 5)
Given: mean µ = 3, standard deviation σ = 2.2. The goal is to find the probability that a refrigerator's lifespan lies between 2.9 and 5 years. To do this, first, convert these raw scores to z-scores using the formula:
z = (x - µ) / σ
Calculations:
For x = 2.9:
z = (2.9 - 3) / 2.2 ≈ -0.0455
For x = 5:
z = (5 - 3) / 2.2 ≈ 0.9091
Next, use standard normal distribution tables or computational tools to find the probabilities corresponding to these z-scores:
P(z ≤ 0.9091) ≈ 0.8186
P(z ≤ -0.0455) ≈ 0.4818
Therefore, the probability that x is between 2.9 and 5 years is:
P(2.9 ≤ x ≤ 5) = P(z ≤ 0.9091) - P(z ≤ -0.0455) ≈ 0.8186 - 0.4818 = 0.3368
This indicates that approximately 33.68% of refrigerators are expected to last between 2.9 and 5 years.
Part B: Probability of Replacement Under the Guarantee
The refrigerator's lifespan is modeled with µ = 4.8 years and σ = 1.3 years. The guarantee period is two years. To find the probability that a refrigerator needs replacement under this guarantee, we calculate the probability that its lifespan is less than 2 years:
z = (2 - 4.8) / 1.3 ≈ -2.1538
Using standard normal distribution tables or computational tools:
P(z ≤ -2.1538) ≈ 0.0156
Interpreted in the context of the problem, there's approximately a 1.56% chance that a randomly chosen refrigerator will fail within two years of use, thus requiring replacement under the guarantee.
Part C: Probability of Lasting Between 5 and 6 Years
Using the same parameters for the lifespan distribution (µ = 4.8, σ = 1.3), we aim to find the probability that a refrigerator lasts between 5 and 6 years. Calculate the z-scores:
For x = 5:
z = (5 - 4.8) / 1.3 ≈ 0.1538
For x = 6:
z = (6 - 4.8) / 1.3 ≈ 0.9231
Using standard normal distribution tables or software:
P(z ≤ 0.9231) ≈ 0.8238
P(z ≤ 0.1538) ≈ 0.5604
The probability that a refrigerator lasts between 5 and 6 years is:
P(5 ≤ x ≤ 6) = 0.8238 - 0.5604 = 0.2634
Thus, approximately 26.34% of refrigerators are expected to last between 5 and 6 years.
Interpretation of Probabilities in Context
The probabilities reflect the likelihood of certain lifespan outcomes for the refrigerators based on the normal distribution model. For example, the 1.56% chance of failure within two years signifies that most refrigerators are expected to function well beyond the guarantee period, indicating a reliable product. Conversely, the 33.68% probability for the lifespan between 2.9 and 5 years shows a moderate proportion of units that may fail earlier than average, providing insights for manufacturers regarding quality control and warranty planning. The probability of lasting between 5 and 6 years suggests a significant proportion of units demonstrate durability exceeding the mean lifespan, which can be used in marketing claims or warranty extensions to attract customers seeking longer-lasting appliances.
Conclusion
This analysis demonstrates how normal distribution calculations can be applied to real-world problems involving product lifespan. By translating raw scores into z-scores and utilizing standard normal distribution tables or software, we gain valuable probabilistic insights that assist manufacturers, consumers, and warranty analysts in understanding product reliability and expected performance. The capacity to interpret these probabilities in context underscores the importance of statistical literacy in making informed decisions in various applied settings.
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