Let x be a continuous random variable with a standard normal distribution. Using the accompanying standard normal distribution table, find P(1.22 ≤x≤ 1.77). Click the icon to view the standard normal distribution table. P(1.22 ≤x≤ 1.77)= (Round to four decimal places as needed.)
Let x be a continuous random variable with a standard normal distribution. Using the accompanying standard normal distribution table, find P(1.22 ≤x≤ 1.77). Click the icon to view the standard normal distribution table. P(1.22 ≤x≤ 1.77)= (Round to four decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Understanding Standard Normal Distribution
To enhance your grasp of standard normal distribution, let's walk through this scenario:
**Problem Statement:**
Let \( x \) be a continuous random variable with a standard normal distribution. Using the standard normal distribution table, we need to find the probability that \( x \) lies between 1.22 and 1.77, i.e., find \( P(1.22 \leq x \leq 1.77) \).
**Steps to Solve:**
1. **Access the Standard Normal Distribution Table:**
To find probabilities associated with the standard normal distribution, you need to refer to the Z-table (standard normal distribution table). This table provides the cumulative probability up to a given z-value.
2. **Look Up Z-Values:**
- Locate the probability corresponding to \( z = 1.22 \).
- Locate the probability corresponding to \( z = 1.77 \).
3. **Calculate the Probability:**
- Subtract the cumulative probability at \( z = 1.22 \) from the cumulative probability at \( z = 1.77 \).
**What You Will Find:**
The result will be a decimal value that represents the probability that the random variable \( x \) falls between 1.22 and 1.77 standard deviations above the mean.
**Example Calculation:**
If the cumulative probability (using the Z-table) for:
- \( z = 1.22 \) is approximately 0.8888
- \( z = 1.77 \) is approximately 0.9616
Then,
\[ P(1.22 \leq x \leq 1.77) = 0.9616 - 0.8888 = 0.0728 \]
This means the probability that \( x \) falls between 1.22 and 1.77 standard deviations from the mean is approximately 0.0728 (rounded to four decimal places).
By understanding and using the standard normal distribution table, you can find specific probabilities for any ranges of a standard normal variable. This is a fundamental skill in statistics, useful in various analyses such as hypothesis testing and confidence interval estimation.
**Exercise for Practice:**
Utilize the Z-table to find the probability for the following:
\[ P(-1.96 \leq x \leq 1.96) \]
This will give you the probability that](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8bde779e-c68e-4dd4-a0b8-70b65f5ae251%2F0fe1fcb4-63cd-4aea-8a32-5ac76005369f%2F28s1tic_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Standard Normal Distribution
To enhance your grasp of standard normal distribution, let's walk through this scenario:
**Problem Statement:**
Let \( x \) be a continuous random variable with a standard normal distribution. Using the standard normal distribution table, we need to find the probability that \( x \) lies between 1.22 and 1.77, i.e., find \( P(1.22 \leq x \leq 1.77) \).
**Steps to Solve:**
1. **Access the Standard Normal Distribution Table:**
To find probabilities associated with the standard normal distribution, you need to refer to the Z-table (standard normal distribution table). This table provides the cumulative probability up to a given z-value.
2. **Look Up Z-Values:**
- Locate the probability corresponding to \( z = 1.22 \).
- Locate the probability corresponding to \( z = 1.77 \).
3. **Calculate the Probability:**
- Subtract the cumulative probability at \( z = 1.22 \) from the cumulative probability at \( z = 1.77 \).
**What You Will Find:**
The result will be a decimal value that represents the probability that the random variable \( x \) falls between 1.22 and 1.77 standard deviations above the mean.
**Example Calculation:**
If the cumulative probability (using the Z-table) for:
- \( z = 1.22 \) is approximately 0.8888
- \( z = 1.77 \) is approximately 0.9616
Then,
\[ P(1.22 \leq x \leq 1.77) = 0.9616 - 0.8888 = 0.0728 \]
This means the probability that \( x \) falls between 1.22 and 1.77 standard deviations from the mean is approximately 0.0728 (rounded to four decimal places).
By understanding and using the standard normal distribution table, you can find specific probabilities for any ranges of a standard normal variable. This is a fundamental skill in statistics, useful in various analyses such as hypothesis testing and confidence interval estimation.
**Exercise for Practice:**
Utilize the Z-table to find the probability for the following:
\[ P(-1.96 \leq x \leq 1.96) \]
This will give you the probability that
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