Calculating the Probability of Light Bulb Lifespan Exceeding Advertised Hours
In the world of lighting products, consumers often rely on manufacturers' claims of bulb lifespan to ensure they meet their needs. For instance, a particular lightbulb is advertised to last for 4400 hours. However, the true lifespan can vary from unit to unit. This article will walk you through calculating the probability that a bulb lasts longer than the advertised figure, given that the lifespans are normally distributed with a specific mean and standard deviation.Understanding the Problem
To solve this, we need to follow these steps: 1. Recognize that the lightbulb lifespan is normally distributed with a mean of 4600 hours and a standard deviation of 250 hours. 2. Determine the probability that a bulb lasts longer than 4400 hours.Applying the Normal Distribution
The first solution provided in the reference text simplifies the problem as follows: The advertised lifespan is 4400 hours. The mean lifespan of the lightbulbs is 4600 hours. The standard deviation is 250 hours. The probability that a lightbulb lasts longer than the advertised figure can be calculated using the cumulative distribution function of the normal distribution. This is given by:P(X > 4400) 1 - P(X ≤ 4400)
First, we need to standardize the value by converting 4400 hours into a Z-score using the formula:Z (frac{X - mu}{sigma}) Where X is the value we are interested in (4400 hours), (mu) is the mean (4600 hours), and (sigma) is the standard deviation (250 hours). Substituting the values, we get:Z (frac{4400 - 4600}{250} frac{-200}{250} -0.8)
Using a standard normal distribution table, we find that the probability of a Z-score being less than -0.8 is approximately 0.21186. Therefore, the probability that a bulb lasts longer than 4400 hours is:P(X > 4400) 1 - 0.21186 0.78814
Another Approach
The second solution re-evaluates the problem by considering a different set of values, which are not directly stated in the reference. Here, we assume that the mean lifespan is 4500 hours and the standard deviation is 150 hours. The solution follows a similar path but with different numerical inputs: The advertised lifespan is still 4400 hours. The mean lifespan of the lightbulbs is adjusted to 4500 hours. The standard deviation is now 150 hours. Using the same steps, we calculate the Z-score as:Z (frac{4400 - 4500}{150} frac{-100}{150} -0.67)
From the standard normal distribution table, the probability of a Z-score being less than -0.67 is approximately 0.25143. Therefore, the probability that a bulb lasts longer than 4400 hours is:P(X > 4400) 1 - 0.25143 0.74857