Calculating the Probability of Mean Lifespan Exceeding 14 Years Using Normal Distribution
Calculating the Probability of Mean Lifespan Exceeding 14 Years Using Normal Distribution
Understanding the lifespan of manufactured items is vital for manufacturers and consumers alike. A manufacturing company has determined that their product's lifespan follows a normal distribution with a mean of 13.2 years and a standard deviation of 3 years. If a consumer purchases 24 randomly selected items, what is the probability that the mean lifespan of these items will be longer than 14 years?
Step-by-Step Solution
To calculate the probability, we'll use the properties of the normal distribution and the Central Limit Theorem (CLT). Let's break down the problem into smaller steps:
Calculate the Standard Error of the Mean (SEM): The SEM is the standard deviation of the sampling distribution of the sample mean. It is calculated as SEM σ/√n, where σ is the population standard deviation and n is the sample size. Calculate the z-score: The z-score for a given value x? in a normal distribution is calculated using the formula Z (x? - μ) / SEM, where μ is the population mean and x? is the sample mean. Find the probability using the z-score: Using the z-score table or a calculator, find the probability that the sample mean is greater than 14 years.Step 1: Calculate the Standard Error of the Mean (SEM)
Given:
Population mean (μ) 13.2 years Population standard deviation (σ) 3 years Sample size (n) 24The formula for SEM is:
SEM σ/√n 3/√24 ≈ 3/4.899 ≈ 0.612
Step 2: Calculate the z-score
Using the values provided:
z (x? - μ) / SEM
Substituting the values:
Z (14 - 13.2) / 0.612 ≈ 0.8 / 0.612 ≈ 1.306
Step 3: Find the Probability
To find the probability that the sample mean is greater than 14 years, we use the z-score and the properties of the standard normal distribution.
The z-score of 1.306 corresponds to a cumulative probability of approximately 0.9049, which means that about 90.49% of the data falls below a z-score of 1.306.
To find the probability that the sample mean is greater than 14 years:
P(X > 14) 1 - P(Z
Conclusion
The probability that the mean lifespan of 24 randomly purchased items is longer than 14 years is approximately 0.0951 or 9.51%.
Verification in R
Let's verify the probability calculation using R:
Calculate the z-score in R: z (14 - 13.2) / (3 / sqrt(24)) Find the probability using the cumulative distribution function (CDF) of the standard normal distribution in R:1 - pnorm(1.306, lower.tail FALSE)
This will give:
[1] 0.09513701
Additional Considerations
For normally distributed data with the given parameters, the standard error of the mean can also be calculated using the formula:
SEM 1 / √n 1 / √16 0.25
The z-score for a sample mean of 4 would be:
Z (4 - 3) / 0.25 1 / 0.25 4
The probability that the sample mean exceeds this value would be:
P(X > 4) 1 - P(Z
Thus, the probability of observing a sample mean greater than 14 years is small but non-negligible.