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Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities.$$P(x \geq 2) ; \mu=3 ; \sigma=0.25$$

Short Answer

Expert verified
The probability \( P(x \geq 2) \) is approximately 1.

Step by step solution

01

Understand the Problem

We need to find the probability that a normally distributed random variable, \( x \), is greater than or equal to 2. The distribution has a mean \( \mu \) of 3 and a standard deviation \( \sigma \) of 0.25.
02

Convert to Standard Normal Variable

To find this probability, we first convert \( x \) to a standard normal variable \( z \) using the formula \( z = \frac{x - \mu}{\sigma} \). For \( x = 2 \), \( z = \frac{2 - 3}{0.25} = -4 \).
03

Find the Probability from the Z-Table

Use the standard normal distribution table (Z-table) to find the probability of \( z \leq -4 \). The Z-table gives the probability for \( P(Z \leq z) \). However, a \( z \)-score of -4 is off the table, typically considered as 0 for practical purposes.
04

Calculate the Desired Probability

Since we seek \( P(x \geq 2) \), which is equivalent to \( P(z \geq -4) \), we use the complement rule. Knowing \( P(z < -4) \) is approximately 0, \( P(z \geq -4) = 1 - P(z < -4) = 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Standard Normal Variable
When working with normal distributions, the standard normal variable, often denoted as \( z \), plays a crucial role. The standard normal variable is a way to transform a normal distribution with any given mean \( \mu \) and standard deviation \( \sigma \) into a standardized form with a mean of 0 and a standard deviation of 1.
This process involves calculating the \( z \)-score, which is a measure that describes how many standard deviations an element \( x \) is from the mean. The transformation formula is:
  • \( z = \frac{x - \mu}{\sigma} \)
This standardized form allows easier calculation of probabilities. It effectively maps our specific distribution to a reference distribution, making it easier to use universal tools like the Z-table.
The Concept of Z-Score
The \( z \)-score is an essential statistic in probability because it allows comparison between different normal distributions. By converting an individual data point into this standardized form, we can compare different datasets or variables. The \( z \)-score indicates exactly how many standard deviations away from the mean a particular score is.
When you have a positive \( z \)-score, it means the value \( x \) is above the mean. Conversely, a negative \( z \)-score shows the value is below the mean.
  • Example: A \( z \)-score of -4 suggests that the observed value is 4 standard deviations below the mean.
  • Conversely, if the \( z \)-score were +4, it would mean the value is 4 standard deviations above the mean.
Therefore, calculating the \( z \)-score allows us to address probability questions effectively, by referring these scores to a standardized table or distribution.
Probability Calculation in Standard Normal Distribution
Once the normal random variable is converted to a standard normal variable (\( z \)), calculating probabilities becomes significantly easier. The goal of probability calculation in this context is to find the likelihood of an event occurring under the normal distribution.
This is often executed with the help of the standard normal distribution table, commonly known as the Z-table, which provides the probability that a standard normal variable \( Z \) will be less than or equal to a certain value.
When you have the \( z \)-score:
  • Locate it in the Z-table to find \( P(z \leq z) \).
  • If \( z \) is not found in the table (extreme \( z \)-scores), the probability is often approximated. For example, for \( z \leq -4 \), the probability is effectively 0.
For probabilities like \( P(x \geq a) \), where \( x \) is a normally distributed variable, use the complement rule. If \( P(z < a) \) is found, then \( P(z \geq a) = 1 - P(z < a) \), which helps determine the desired probability efficiently.

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Most popular questions from this chapter

Find the \(z\) value described and sketch the area described.Find \(z\) such that \(55 \%\) of the standard normal curve lies to the left of \(z\).

Sketch the areas under the standard normal curve over the indicated intervals and find the specified areas. Between \(z=0\) and \(z=3.18\)

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(z \geq-1.20)$$

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