Problem 175 - Applied Statistics and Probability for Engineers

Question

The lifetime of a semiconductor laser has a lognormal distribution, and it is known that the mean and standard deviation of lifetime are 10,000 and \(20,000,\) respectively. (a) Calculate the parameters of the lognormal distribution. (b) Determine the probability that a lifetime exceeds 10,000 hours.

Step-by-Step Solution

Answer

(a) \( \mu \approx 7.6009 \, , \sigma \approx 1.1513 \). (b) Probability that lifetime > 10,000 hours is 0.5.

Step 1: Understand the Lognormal Distribution

A lognormal distribution is a probability distribution of a random variable whose logarithm is normally distributed. If \( X \) is a lognormally distributed random variable, then \( \ln(X) \) is normally distributed with mean \( \mu \) and standard deviation \( \sigma \). The mean \( m \) and standard deviation \( s \) of a lognormal distribution are related to \( \mu \) and \( \sigma \).

Step 2: Use the Relationship Between Mean, Standard Deviation, and Parameters

The mean \( m \) and standard deviation \( s \) for a lognormal distribution have the following relations to its parameters:\[m = e^{\mu + \sigma^2/2} \s^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}\]For a lognormal distribution with mean 10,000 and standard deviation 20,000, substitute these into the above equations to find \( \mu \) and \( \sigma \).

Step 3: Solve for \( \mu \) and \( \sigma \)

Use the formula for the mean: \[10000 = e^{\mu + \sigma^2/2}\]And the variance relation:\[20000^2 = s^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}\]By solving these equations simultaneously, we find: \[\mu = \ln(10000) - \frac{1}{2} \ln\left(1 + \left(\frac{20000}{10000}\right)^2 \right)\]\[\sigma = \sqrt{\ln \left(1 + \left(\frac{20000}{10000}\right)^2\right)}\]Substituting, we find \( \mu \approx 7.6009 \) and \( \sigma \approx 1.1513 \).

Step 4: Calculate Probability of Exceeding 10,000 Hours

To find the probability that the lifetime exceeds 10,000 hours, we use the cumulative distribution function (CDF) of the lognormal distribution. The probability that \( X \) exceeds 10,000 is given by:\[ P(X > 10000) = 1 - P(X \leq 10000) = 1 - \Phi\left(\frac{\ln(10000) - \mu}{\sigma}\right) \]where \( \Phi \) is the CDF of the standard normal distribution. Substituting \( \mu \) and \( \sigma \),\[ P(X > 10000) \approx 1 - \Phi(0) = 0.5 \] since \( \ln(10000) = \mu \).

Key Concepts

Mean and Standard Deviation

In a lognormal distribution, understanding the relationship between the mean and standard deviation is crucial for determining the underlying parameters of the distribution. A lognormal distribution describes a scenario where the natural logarithm of the variable of interest follows a normal distribution, or Gaussian distribution.
The mean (\(m\)) and standard deviation (\(s\)) in a lognormal distribution are derived from the log-transformed variables. They relate to the normal distribution parameters, \(\mu\) (mean of the logged variable) and \(\sigma\) (standard deviation of the logged variable), through these formulas:

• Mean: \(m = e^{\mu + \sigma^2/2}\)
• Variance: \(s^2 = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2}\)

In a practical sense, this means that although the variable itself isn’t normally distributed, its log values are. This is why the formulas use logarithms. To find \(\mu\) and \(\sigma\), we rearrange the formulas using the known mean and standard deviation, solving them simultaneously to understand the distribution's shape and spread.

Probability Calculation

Calculating probabilities within a lognormal distribution often involves understanding the behavior of the variable in question relative to its probability distribution. Here, we need to assess the likelihood of the semiconductor laser outliving a certain time—in this case, 10,000 hours.
To do this, we use the characteristic property of the lognormal distribution: the natural log of the variable being normally distributed. Thus, calculating a probability from this distribution involves converting our variable into a normally distributed variable, then using the properties of the normal distribution to find our probability.
For a value \(X\), we convert our problem to the normal domain using:

• \( \text{Find } P(X > a) \text{ by converting to } P(\ln X > \ln a) \)
• \( \text{Then use: } P(\ln X > \ln a) = 1 - \Phi\left( \frac{\ln a - \mu}{\sigma} \right) \)

This approach allows us to work with our known parameters \(\mu\) and \(\sigma\) to determine the probability of exceeding a specific point in time.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) of a lognormal distribution plays a pivotal role in finding out the probability of outcomes within this type of distribution. The CDF helps in understanding the probability that a random variable is less than or equal to a particular value.
For lognormal distributions, the CDF can be expressed in terms of the CDF of a standard normal distribution, \(\Phi\). By converting the problem into the realm of the standard normal distribution, we can utilize known tables or computational functions to find probabilities efficiently:

• Use the transformation: \(P(X \leq a) = \Phi\left(\frac{\ln a - \mu}{\sigma}\right)\)
• If we need \(P(X > a)\), use \(1 - P(X \leq a)\)

In practice, this involves computing \(\Phi\left(\frac{\ln a - \mu}{\sigma}\right)\) using standard statistical software or Z-tables. In our example, where \(\ln(10000)\) equals \(\mu\), this results in symmetry about zero, simplifying our calculation to finding that the probability of exceeding 10,000 hours is roughly 0.5, given the standard properties of the normal distribution.