/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(x\) be a continuous random... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(0.76 \leq x \leq 1.45) $$

Short Answer

Expert verified
The probability that \( x \) falls between 0.76 and 1.45 is 0.1501.

Step by step solution

01

Understanding the Problem Requirements

We need to find the probability that the random variable \(x\), which follows a standard normal distribution, falls between 0.76 and 1.45. We can use the standard normal distribution table (Table A) to find the probabilities associated with these z-scores.
02

Locate â‹€ in the Standard Normal Table

Locate the z-score of 0.76 in the standard normal distribution table to find the probability \(P(z \leq 0.76)\). According to the table, \(P(z \leq 0.76) \approx 0.7764\).
03

Locate â‹€ in the Standard Normal Table

Locate the z-score of 1.45 in the standard normal distribution table to find the probability \(P(z \leq 1.45)\). According to the table, \(P(z \leq 1.45) \approx 0.9265\).
04

Calculate the Desired Probability

To find \(P(0.76 \leq x \leq 1.45)\), we subtract the probability of \(P(z \leq 0.76)\) from \(P(z \leq 1.45)\). The calculation is as follows: \[P(0.76 \leq x \leq 1.45) = P(z \leq 1.45) - P(z \leq 0.76) = 0.9265 - 0.7764 = 0.1501\].

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Key Concepts

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

Continuous Random Variable
A continuous random variable is a variable that can take an infinite number of possible values. It is characterized by the fact that its outcomes are not countable, and it can assume any value in a given range. For example, a continuous random variable might represent the height of students in a class, temperature readings, or, as in our exercise, values that follow a standard normal distribution. Unlike discrete random variables that have distinct separate values, continuous random variables can take on intervals of real numbers. This means measurements will always be estimates when dealing with continuous data. For the standard normal distribution, these values are described by the probability density function. The area under the curve of this function over a certain interval gives us the probability that the random variable falls within that range.
Z-Scores
Z-scores play a vital role in the world of statistics, especially when working with standard normal distributions. A z-score gives you an idea of how far from the mean a particular data point is. In a standard normal distribution, the mean () is zero, and the standard deviation () is one.To find the z-score for a particular data point, you use the formula:\[z = \frac{(x - \mu)}{\sigma}\]In simpler terms, z-scores tell you how many standard deviations away an observation is from the mean:- A z-score of 0 means the score is exactly at the mean.- Positive z-scores indicate the score is above the mean.- Negative z-scores mean the score is below the mean. For example, in our exercise, we use z-scores to determine where 0.76 and 1.45 lie in relation to the mean of the standard normal distribution. By referring to the standard normal distribution table (Table A), we can find probabilities that aid in further probability calculations.
Probability Calculation
Probability calculation in the context of a continuous random variable involves determining the likelihood of the variable falling within a certain range. One common tool used for this is the standard normal distribution table, which provides the probabilities of a z-score being less than or equal to a given value. This table is based on the properties of the standard normal distribution.To calculate probabilities like the one in our exercise, follow these steps:- **Identify the range of interest.** In our case, it is between the z-scores 0.76 and 1.45.- **Use the standard normal distribution table** to find the cumulative probabilities of these z-scores. Cumulative probability for a z-score gives the area under the normal distribution curve to the left of the z-score.- **Subtract the smaller probability from the larger one.** This gives the probability of the variable falling between those z-scores: \[P(0.76 \leq x \leq 1.45) = P(z \leq 1.45) - P(z \leq 0.76)\]This simple subtraction helps us isolate the probability that the random variable lies between the two z-scores, providing us with the solution to our problem.

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