Finding X For Normal Distribution P(X ≤ X) = 0.9394

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#Introduction

In the realm of statistics, the normal distribution, often called the Gaussian distribution, stands as a cornerstone for modeling continuous random variables. Its bell-shaped curve elegantly describes the distribution of numerous natural phenomena, from heights and weights to test scores and financial data. Understanding the properties of normal distributions is crucial for making informed decisions and predictions across various fields. The normal distribution is characterized by two parameters: the mean (µ), which represents the central tendency of the distribution, and the standard deviation (σ), which quantifies the spread or variability of the data. With these parameters, we can determine the probability of a random variable falling within a specific range or, conversely, find the value of the variable corresponding to a given probability. The ability to solve these types of problems is vital in statistical inference, hypothesis testing, and risk assessment. In this article, we will delve into the practical application of normal distribution properties to solve a specific problem. We are given a normally distributed random variable X with a mean (µ) of 250 and a standard deviation (σ) of 80. Our task is to find the approximate value x such that the probability P(X ≤ x) = 0.9394. This means we need to find the value x below which 93.94% of the data falls. This problem exemplifies the common task of finding a percentile or quantile in a normal distribution, which has numerous practical applications, such as determining cutoff scores in tests or identifying risk thresholds in financial markets. We will explore the step-by-step process of solving this problem, highlighting the use of the standard normal distribution (Z-distribution) and Z-tables (or statistical software) to find the corresponding value of x. By understanding this process, you can apply the same principles to solve a wide array of problems involving normal distributions. The standard normal distribution (Z-distribution) is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It serves as a reference distribution for all normal distributions, allowing us to standardize any normal distribution by transforming its values into Z-scores. This standardization process simplifies probability calculations and allows us to use Z-tables, which provide probabilities associated with different Z-scores. Understanding and utilizing the Z-distribution is paramount for solving problems involving normal distributions, as it provides a standardized framework for probability calculations. The concept of Z-scores is also essential, as they represent the number of standard deviations a particular value is away from the mean. This allows us to compare values from different normal distributions and assess their relative positions. In the following sections, we will leverage these concepts to solve the given problem and provide a comprehensive explanation of the solution process.

Problem Statement

Let $X$ be normally distributed with mean $\\mu=250$ and standard deviation $\\\sigma=80$. Find the approximate value $x$ such that $P(X \leq x)=0.9394$. The options provided are:

A. 1.55

B. 126

C. 374

D. -1.55

This problem requires us to find the value x in a normal distribution such that the cumulative probability up to x is 0.9394. In simpler terms, we need to find the value x below which 93.94% of the data falls. This is a common problem in statistics, often encountered in scenarios such as determining cutoff scores, assessing risk levels, or calculating confidence intervals. To solve this problem, we will utilize the properties of the normal distribution and the standard normal distribution (Z-distribution). The key idea is to transform the given problem into a standard normal distribution problem, which we can then solve using Z-tables or statistical software. The process involves calculating the Z-score corresponding to the given probability (0.9394) and then converting this Z-score back to the original scale of the variable X. This transformation allows us to leverage the readily available Z-tables, which provide the cumulative probabilities for the standard normal distribution. Z-tables are an essential tool in statistics, providing a quick and easy way to find probabilities associated with different Z-scores. They typically display the cumulative probability to the left of a given Z-score, which is precisely what we need for this problem. Understanding how to use Z-tables is crucial for solving problems involving normal distributions. In the following sections, we will outline the steps involved in solving this problem, including the transformation to the Z-distribution, finding the Z-score, and converting it back to the original value x. By following this step-by-step approach, you can apply the same principles to solve a wide range of similar problems. The ability to find values corresponding to specific probabilities in a normal distribution is a fundamental skill in statistics, with applications in various fields such as finance, engineering, and healthcare. The use of Z-scores and Z-tables simplifies this process, making it accessible even without advanced statistical software. The Z-score, often called the standard score, is a measure of how many standard deviations an element is from the mean. A positive Z-score indicates that the element is above the mean, while a negative Z-score indicates that it is below the mean. The magnitude of the Z-score indicates the distance from the mean in terms of standard deviations. A Z-score of 0 indicates that the element is exactly at the mean. The Z-score is a dimensionless quantity, making it useful for comparing values from different normal distributions. By converting values to Z-scores, we can assess their relative positions within their respective distributions. In the context of our problem, we will use the Z-score to find the value x corresponding to the given probability. The formula for calculating the Z-score is Z = (X - µ) / σ, where X is the value of the variable, µ is the mean, and σ is the standard deviation. Conversely, we can rearrange this formula to find the value of X given the Z-score: X = µ + Zσ. This formula will be crucial in converting the Z-score we find from the Z-table back to the original value x. In the following sections, we will demonstrate how to use these formulas and concepts to solve the given problem.

Solution

To find the value $x$ such that $P(X \leq x) = 0.9394$, we need to use the properties of the normal distribution and the Z-table. The process involves several key steps:

  1. Standardize the Normal Distribution: We need to convert the given normal distribution to the standard normal distribution (Z-distribution). The Z-distribution has a mean of 0 and a standard deviation of 1. The formula to convert a value from a normal distribution to a Z-score is:

    Z=fracX−musigmaZ = \\frac{X - \\mu}{\\\\sigma}

    Where:

    • X$ is the value we want to convert.

    • \\\\mu$ is the mean of the normal distribution (250 in this case).

    • \\\\\\sigma$ is the standard deviation of the normal distribution (80 in this case).

  2. Find the Z-score: We are given the probability $P(X \leq x) = 0.9394$. We need to find the Z-score that corresponds to this cumulative probability in the Z-table. The Z-table gives the area under the standard normal curve to the left of a given Z-score. So, we look for the value 0.9394 in the Z-table. The closest value we find in the Z-table is 0.9394, which corresponds to a Z-score of approximately 1.55.

    Using a Z-table, we find that the Z-score corresponding to a cumulative probability of 0.9394 is approximately 1.55.

    The Z-table is a crucial tool for working with normal distributions. It provides the cumulative probability for a given Z-score, which is the probability that a standard normal random variable is less than or equal to that Z-score. Z-tables are typically structured with Z-scores listed in the margins, and the corresponding probabilities in the body of the table. To find the probability associated with a specific Z-score, you locate the Z-score in the table and read off the corresponding probability. Conversely, to find the Z-score associated with a specific probability, you search for the probability in the body of the table and read off the corresponding Z-score. There are various types of Z-tables, but they all provide the same fundamental information. Some tables may include negative Z-scores, while others may only include positive Z-scores and rely on the symmetry of the standard normal distribution to find probabilities for negative Z-scores. Understanding how to use Z-tables is essential for solving problems involving normal distributions, as it allows us to easily find probabilities and Z-scores. The cumulative probability is the probability that a random variable takes on a value less than or equal to a specified value. In the context of the normal distribution, the cumulative probability at a particular value x is the area under the normal curve to the left of x. This area represents the proportion of the distribution that falls below x. Cumulative probabilities are crucial for understanding the distribution of data and for making inferences about populations based on samples. They are also used extensively in hypothesis testing, where we compare the cumulative probability of an observed result to a predetermined significance level. In our problem, the cumulative probability P(X ≤ x) = 0.9394 represents the proportion of the distribution that falls below the value x. By finding the Z-score corresponding to this cumulative probability, we can then determine the value of x in the original normal distribution.

  3. Convert the Z-score back to X: Now that we have the Z-score, we can use the Z-score formula rearranged to solve for X:

    X=mu+ZtimesσX = \\mu + Z \\times \\\sigma

    Substitute the values:

    X=250+1.55times80X = 250 + 1.55 \\times 80

    X=250+124X = 250 + 124

    X=374X = 374

    Therefore, the approximate value of $x$ such that $P(X \leq x) = 0.9394$ is 374.

    The process of converting the Z-score back to the original value X is a critical step in solving problems involving normal distributions. It allows us to translate the information we obtained from the Z-table (which is based on the standard normal distribution) back to the original scale of the variable we are working with. This conversion is essential for making meaningful interpretations and decisions based on the statistical analysis. The formula for converting the Z-score back to X is derived from the Z-score formula itself, and it highlights the relationship between the Z-score, the original value, the mean, and the standard deviation. By understanding this relationship, we can easily move between the standard normal distribution and the original distribution, allowing us to solve a wide range of problems. In our problem, we found the Z-score corresponding to the given probability and then used this Z-score to calculate the value of X. This demonstrates the power of using the standard normal distribution as a reference point for solving problems involving other normal distributions. The ability to convert between Z-scores and original values is a fundamental skill in statistics, with applications in various fields such as finance, engineering, and healthcare. In summary, this problem highlights the importance of understanding the properties of the normal distribution, the Z-distribution, and the Z-table. By following the steps outlined above, we can effectively solve problems involving probabilities and values in normal distributions. This knowledge is invaluable in various fields that rely on statistical analysis.

Final Answer

The approximate value of $x$ such that $P(X \leq x)=0.9394$ is 374.

Therefore, the correct answer is:

C. 374