Finding X Value In Normal Distribution Given Probability

In the realm of statistics, understanding normal distributions is paramount. Normal distributions, often called Gaussian distributions, are ubiquitous in modeling real-world phenomena, from heights and weights to test scores and financial data. A key aspect of working with normal distributions involves calculating probabilities and finding values associated with specific probabilities. In this article, we will delve into a problem that requires us to find the approximate value x for a normally distributed variable X, given its mean, standard deviation, and a specific probability. We will use the properties of the standard normal distribution and z-scores to solve this problem. This exploration will not only enhance our understanding of normal distributions but also provide a practical approach to solving similar statistical problems.

Problem Statement

Let's consider a random variable X that is normally distributed with a mean (μ) of 250 and a standard deviation (σ) of 80. Our objective is to find the approximate value x such that the probability of X being less than or equal to x is 0.9394. Mathematically, we are looking for x such that P(X ≤ x) = 0.9394. This type of problem is fundamental in statistics and has applications in various fields, including quality control, finance, and risk management. Understanding how to solve it is crucial for anyone working with statistical data. We will explore the steps involved in solving this problem, including standardizing the normal distribution and using z-tables or statistical software to find the corresponding value of x.

Understanding the Normal Distribution

The normal distribution is a continuous probability distribution that is symmetrical around its mean. Its bell-shaped curve is defined by two parameters: the mean (μ) and the standard deviation (σ). The mean represents the central tendency of the distribution, while the standard deviation measures the spread or dispersion of the data. A larger standard deviation indicates a wider spread, while a smaller standard deviation indicates a tighter clustering around the mean. The total area under the normal curve is equal to 1, representing the total probability of all possible outcomes. Because of its symmetry, 50% of the data falls below the mean, and 50% falls above the mean. This property is essential for understanding probabilities associated with different ranges of values. The normal distribution is a cornerstone of statistical theory and is used extensively in hypothesis testing, confidence interval estimation, and predictive modeling. Its mathematical properties and widespread applicability make it a vital tool for statisticians and data analysts.

Standardizing the Normal Distribution

To solve the problem, we need to standardize the normal distribution. Standardizing involves transforming the variable X into a new variable Z, which follows a standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1. This transformation is crucial because it allows us to use standard normal tables (also known as z-tables) or statistical software to find probabilities and corresponding values. The formula for converting a value x from a normal distribution with mean μ and standard deviation σ to a z-score is: z = (x - μ) / σ. The z-score represents the number of standard deviations that x is away from the mean. A positive z-score indicates that x is above the mean, while a negative z-score indicates that x is below the mean. By standardizing, we can compare values from different normal distributions and calculate probabilities using the standard normal distribution, which is well-documented and widely used in statistical analysis.

Using Z-Scores and Z-Tables

Once we have standardized the normal distribution, we can use z-scores and z-tables to find the value of x corresponding to a given probability. A z-score represents the number of standard deviations a particular value is from the mean in a standard normal distribution. Z-tables provide the cumulative probabilities for the standard normal distribution, i.e., the probability that a standard normal variable is less than or equal to a given z-score. In our case, we are given P(X ≤ x) = 0.9394, which corresponds to a cumulative probability of 0.9394 in the standard normal distribution. We need to find the z-score that corresponds to this probability. By looking up 0.9394 in a z-table or using statistical software, we can find the corresponding z-score. This z-score will then allow us to reverse the standardization process and find the value of x in the original normal distribution. Understanding how to use z-scores and z-tables is a fundamental skill in statistics, enabling us to calculate probabilities and make inferences about normally distributed data.

Solution

1. Find the Z-score

First, we need to find the z-score that corresponds to a cumulative probability of 0.9394. We can use a z-table or a statistical calculator for this. Looking up 0.9394 in a standard z-table, we find that the z-score is approximately 1.55. This means that the value we are looking for is 1.55 standard deviations above the mean in the standard normal distribution. Alternatively, statistical software or online calculators can be used to find this z-score, providing a more precise value if needed. The z-score is a critical link between the standard normal distribution and our original distribution, allowing us to translate probabilities from one to the other. This step is a fundamental application of the z-table, a tool that is essential for anyone working with normal distributions and statistical probabilities. The ability to accurately find z-scores for given probabilities is crucial for solving a wide range of statistical problems.

2. Use the Z-score Formula

Now that we have the z-score, we can use the standardization formula to find the value of x. The formula is: z = (x - μ) / σ. We know z = 1.55, μ = 250, and σ = 80. We need to solve for x. Plugging in the values, we get: 1.55 = (x - 250) / 80. To solve for x, we multiply both sides by 80: 1.55 * 80 = x - 250, which gives us 124 = x - 250. Then, we add 250 to both sides: x = 124 + 250, which gives us x = 374. This calculation allows us to convert the z-score back into the original scale of the problem, providing us with the value of x that corresponds to the given probability. The algebraic manipulation of the z-score formula is a key step in connecting the standard normal distribution back to the original distribution, allowing us to make meaningful interpretations in the context of the problem.

3. Approximate Value of x

Therefore, the approximate value of x such that P(X ≤ x) = 0.9394 is 374. This means that there is a 93.94% probability that the random variable X will be less than or equal to 374. This result provides a concrete answer to the problem statement, giving us a specific value that corresponds to the desired probability. The interpretation of this result is crucial for understanding the implications of the normal distribution in the given context. It allows us to make informed decisions and predictions based on the statistical properties of the data. This final step of calculating x completes the problem-solving process, demonstrating the practical application of normal distribution concepts and techniques.

Conclusion

In conclusion, we have successfully found the approximate value x for a normally distributed variable X with a mean of 250 and a standard deviation of 80, such that P(X ≤ x) = 0.9394. By standardizing the normal distribution, using z-scores, and applying the z-score formula, we determined that x is approximately 374. This process illustrates the importance of understanding normal distributions and their properties in statistical analysis. The ability to solve such problems is crucial in various fields where data analysis and interpretation are essential. The steps we followed provide a clear and systematic approach to solving similar problems involving normal distributions, enabling us to make informed decisions based on statistical data. Mastering these techniques is fundamental for anyone working with statistical models and data analysis, providing a solid foundation for more advanced statistical concepts and applications.

Options Analysis

Now, let's analyze the given options:

A. 1.55 - This is the z-score, not the value of x. B. 126 - This value is significantly below the mean and would correspond to a much lower probability. C. 374 - This is the correct approximate value of x that we calculated. D. -1.55 - This is the negative of the z-score and does not represent the value of x.

Final Answer

The correct answer is C. 374.