Solving for Mean and Variance in a Normal Distribution Problem

This article provides a detailed step-by-step guide on solving for mean and variance in a normal distribution problem, often encountered in statistical analysis and probability theory. We will explore the use of given probabilities to determine the unknown mean and standard deviation of a normally distributed random variable. This approach is particularly valuable for understanding the application of the standard normal distribution and its cumulative distribution function (CDF).

The Given Problem

The problem we are tackling involves a normally distributed random variable (X), where the probability density function (PDF) is characterized by mean (mu) and standard deviation (sigma). Specifically, we know that (P(X and (P(X . We need to find the mean ((mu)) and variance ((sigma^2)) of this distribution.

Step-by-Step Solution

The first step is to transform the (X) variable to the standard normal variable (Z) using the formula:

[Z frac{X - mu}{sigma}]

Given that (Z sim N(0, 1)), we can use the cumulative distribution function (CDF) of the standard normal distribution, denoted by (Phi), to express our problem in terms of (Z).

Step 1: Using the CDF to Relate Probabilities to Standard Normal Distributions

From the given probabilities:

(P(X

We can write:

[Pleft( frac{X - mu}{sigma}

Similarly:

(P(X

This can be written as:

[Pleft( frac{X - mu}{sigma}

Step 2: Setting Up the Equations to Solve for Mean and Standard Deviation

We now have two equations:

(frac{35 - mu}{sigma} -1.4578)

(frac{63 - mu}{sigma} 1.2265)

These can be rearranged to solve for (sigma) and (mu):

[35 - mu -1.4578sigma]

[63 - mu 1.2265sigma]

Subtract the first equation from the second to eliminate (mu):

[28 2.6843sigma]

Thus:

[sigma frac{28}{2.6843} approx 10.45]

Substitute (sigma) back into one of the original equations to solve for (mu):

(35 - mu -1.4578 times 10.45)

[35 - mu -15.22]

[mu 50.22]

Step 3: Calculating the Variance

With (mu) and (sigma) known, we can now calculate the variance:

(sigma^2 (10.45)^2 approx 109.2025)

Thus, the mean ((mu)) is approximately 50.22 and the variance ((sigma^2)) is approximately 109.2025.

Verification and Conclusion

We can verify our solution by checking the given probabilities:

[P(X

The solution satisfies the problem's conditions. In practice, you can use R or other statistical software to directly compute these values, confirming our manual calculations.

Key Concepts and Practical Applications

This method is widely applicable in various fields such as finance, engineering, and social sciences, where the normal distribution model is used to describe and analyze data. Understanding how to solve for parameters given specific probabilities is crucial in statistical analysis and inference.