Solving Probability For Normal Distribution X ≥ X = 0.1736

July 11, 2025 by Scholario Team 59 views

Understanding Normal Distribution and Probability Calculation for X ≥ x

In the realm of statistics, the normal distribution, also known as the Gaussian distribution, holds a paramount position. Its bell-shaped curve elegantly describes the distribution of numerous natural phenomena and serves as a cornerstone for various statistical analyses. Understanding normal distributions is crucial for solving probability-related problems, especially in fields like mathematics, finance, and engineering. This article delves into a specific problem involving a normally distributed variable and demonstrates how to find the value x for a given probability. We will explore the concepts of mean, standard deviation, and Z-scores, and use these tools to solve the problem effectively. Our main focus is to solve the following problem: Let $X$ be normally distributed with mean $\mu=25$ and standard deviation $\sigma=5$. Find the approximate value $x$ such that $P(X \geq x)=0.1736$. This problem highlights a practical application of normal distribution principles and offers valuable insights into probability calculations.

Key Concepts: Mean, Standard Deviation, and Z-Scores

Before diving into the solution, it's essential to grasp the fundamental concepts that govern normal distributions. The mean ( $\mu$ ) represents the average value of the distribution, while the standard deviation ( $\sigma$ ) measures the spread or dispersion of the data around the mean. A larger standard deviation indicates a wider spread, whereas a smaller standard deviation implies that the data points are clustered closer to the mean. In our problem, the mean ( $\mu$ ) is given as 25, and the standard deviation ( $\sigma$ ) is 5.

To solve problems involving normal distributions, we often use Z-scores. A Z-score represents the number of standard deviations a particular data point is away from the mean. It allows us to standardize the normal distribution, making it easier to calculate probabilities using a standard normal distribution table (also known as a Z-table). The formula for calculating the Z-score is:

Z = \frac{x - \mu}{\sigma}

Where:

$x$ is the value of the data point.
$\mu$ is the mean of the distribution.
$\sigma$ is the standard deviation of the distribution.

Z-scores are indispensable tools for determining probabilities associated with specific values in a normal distribution. By converting a raw score into a Z-score, we can easily find the corresponding probability using the standard normal distribution table. This table provides the cumulative probability for Z-scores, which is the probability of a value falling below a given Z-score. Understanding how to calculate and interpret Z-scores is pivotal for solving problems related to normal distributions.

Problem Breakdown and Solution Strategy

Now, let's revisit the problem statement: We are given a normally distributed variable $X$ with a mean ( $\mu$ ) of 25 and a standard deviation ( $\sigma$ ) of 5. Our objective is to find the approximate value $x$ such that $P(X \geq x) = 0.1736$. This means we are looking for the value x for which the probability of $X$ being greater than or equal to x is 0.1736.

To solve this, we can follow these steps:

Convert the probability $P(X \geq x)$ to a probability that can be looked up in the Z-table, which typically provides values for $P(Z < z)$. Since the total probability under the normal curve is 1, we can use the relationship: $P(X \geq x) = 1 - P(X < x)$.
Find the corresponding Z-score for the probability $P(X < x)$. We will use the Z-table to find the Z-score that corresponds to the probability $1 - 0.1736 = 0.8264$. The Z-table gives us the probability of a value being less than a certain Z-score.
Use the Z-score formula to find the value of x. Once we have the Z-score, we can rearrange the Z-score formula to solve for x:
$x = Z \cdot \sigma + \mu$

By following these steps, we can systematically find the value of x that satisfies the given probability condition. This approach highlights the importance of understanding the relationship between probabilities, Z-scores, and the parameters of the normal distribution.

Step-by-Step Solution

Let's now walk through the solution step by step:

Convert the probability:

We are given $P(X \geq x) = 0.1736$. To use the Z-table, we need to find $P(X < x)$. Using the complement rule:
$P(X < x) = 1 - P(X \geq x) = 1 - 0.1736 = 0.8264$
Find the corresponding Z-score:

We need to find the Z-score that corresponds to a cumulative probability of 0.8264. Looking at the Z-table, we find that a probability of approximately 0.8264 corresponds to a Z-score of 0.94. Therefore, $Z \approx 0.94$.
Use the Z-score formula to find x:

Now we use the Z-score formula to solve for x:
$x = Z \cdot \sigma + \mu$
Substitute the values: $Z = 0.94$, $\sigma = 5$, and $\mu = 25$:
$x = 0.94 \cdot 5 + 25$ $x = 4.7 + 25$ $x = 29.7$

Thus, the approximate value of x such that $P(X \geq x) = 0.1736$ is 29.7.

Detailed Explanation of Each Step

Step 1: Converting the Probability

The first step in solving this problem involves converting the given probability $P(X \geq x)$ into a form that can be easily used with the standard normal distribution table (Z-table). The Z-table provides cumulative probabilities, which are the probabilities of a value falling below a certain Z-score. Therefore, we need to express the probability $P(X \geq x)$ in terms of $P(X < x)$.

The fundamental concept we use here is the complement rule of probability. The complement rule states that the probability of an event not occurring is equal to 1 minus the probability of the event occurring. In mathematical terms:

P(A') = 1 - P(A)

Where $P(A')$ is the probability of the complement of event A, and $P(A)$ is the probability of event A.

In our case, the event A is $X \geq x$, and its complement A' is $X < x$. Therefore, we can write:

P(X \geq x) = 1 - P(X < x)

We are given $P(X \geq x) = 0.1736$. Substituting this into the equation:

0.1736 = 1 - P(X < x)

Now, we solve for $P(X < x)$:

P(X < x) = 1 - 0.1736 = 0.8264

This conversion is crucial because it allows us to use the Z-table effectively. The Z-table gives us the probability of a value being less than a certain Z-score, which is exactly what we have calculated here. This step demonstrates a key principle in probability: understanding how to manipulate probabilities using rules like the complement rule to fit the tools and resources available.

Step 2: Finding the Corresponding Z-score

Once we have the probability $P(X < x) = 0.8264$, the next step is to find the Z-score that corresponds to this probability. The Z-score represents the number of standard deviations a particular value is away from the mean. To find the Z-score, we use the standard normal distribution table (Z-table), which provides the cumulative probabilities for various Z-scores.

The Z-table is structured in such a way that the rows represent the integer part and the first decimal place of the Z-score, while the columns represent the second decimal place. To find the Z-score for a given probability, we look for the probability value in the table and then read off the corresponding Z-score from the row and column headers.

In our case, we are looking for the Z-score that corresponds to a probability of 0.8264. By examining the Z-table, we find that the probability 0.8264 falls approximately at the intersection of the row corresponding to 0.9 and the column corresponding to 0.04. This means the Z-score is approximately 0.94.

Z \approx 0.94

This Z-score tells us that the value x is 0.94 standard deviations above the mean. The process of finding the Z-score from a probability is a fundamental skill in statistics, allowing us to link probabilities to specific values in a normal distribution. Understanding how to use the Z-table is essential for solving a wide range of problems related to normal distributions.

Step 3: Using the Z-score Formula to Find x

After finding the Z-score, the final step is to calculate the value of x using the Z-score formula. The Z-score formula relates the Z-score to the original value x, the mean ( $\mu$ ), and the standard deviation ( $\sigma$ ) of the distribution. The formula is given by:

Z = \frac{x - \mu}{\sigma}

We need to rearrange this formula to solve for x. Multiplying both sides by $\sigma$ gives:

Z \cdot \sigma = x - \mu

Then, adding $\mu$ to both sides gives:

x = Z \cdot \sigma + \mu

This is the formula we will use to find the value of x. We have the following values:

Z-score ($Z$) = 0.94
Standard deviation ($\sigma$) = 5
Mean ($\mu$) = 25

Substituting these values into the formula:

x = 0.94 \cdot 5 + 25

First, we multiply 0.94 by 5:

0.94 \cdot 5 = 4.7

Then, we add 4.7 to 25:

x = 4.7 + 25 = 29.7

Therefore, the approximate value of x such that $P(X \geq x) = 0.1736$ is 29.7. This calculation demonstrates the practical application of the Z-score formula in converting a standardized value (Z-score) back into the original scale of the data. By understanding this formula, we can easily find specific values corresponding to given probabilities in a normal distribution.

Final Answer and Conclusion

In conclusion, by systematically applying the concepts of mean, standard deviation, Z-scores, and the Z-table, we have successfully found the approximate value of x such that $P(X \geq x) = 0.1736$. The step-by-step solution involved converting the probability, finding the corresponding Z-score, and using the Z-score formula to calculate x. The final answer is:

x = 29.7

This problem exemplifies the importance of understanding normal distributions and their properties in solving practical statistical problems. The normal distribution is a fundamental concept in statistics, and the ability to work with probabilities and Z-scores is essential for various applications in science, engineering, and finance. The methods demonstrated here can be applied to a wide range of problems involving normally distributed variables, providing a solid foundation for further statistical analysis.

The correct answer is (d). 29.70

By breaking down the problem into smaller, manageable steps and providing detailed explanations, we have made the solution process clear and accessible. This approach not only helps in understanding the specific problem but also reinforces the underlying statistical concepts, making it easier to tackle similar problems in the future.