Assume 2 is a random variable with a standard normal distribution and a is a positive number. If P(Z > a) = 0.1, then P(-a < Z
Assume 2 is a random variable with a standard normal distribution and a is a positive number. If P(Z > a) = 0.1, then P(-a < Z
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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![**Question Explanation and Concept:**
Assume \( Z \) is a random variable with a standard normal distribution and \( a \) is a positive number. If \(\mathbb{P}(Z > a) = 0.1\), then \(\mathbb{P}(-a < Z < a) = 0.8.\)
**Question Statement:**
True or False
- ( ) True
- ( ) False
**Explanation:**
Here, \( Z \) represents a standard normal random variable, which implies it follows a normal distribution with a mean of 0 and a standard deviation of 1.
Let’s analyze the given conditions:
1. \(\mathbb{P}(Z > a) = 0.1\)
This condition means the probability that \( Z \) is greater than \( a \) is 0.1. Because the standard normal distribution is symmetric about the mean (0), the probability \( \mathbb{P}(Z < -a) \) will also be 0.1.
2. The probability for the range \( -a < Z < a \) can be interpreted as the area under the normal curve within these limits. To find this, observe that:
\[
\mathbb{P}(Z > a) + \mathbb{P}(Z < -a) + \mathbb{P}(-a < Z < a) = 1
\]
Given \(\mathbb{P}(Z > a) = 0.1\) and \( \mathbb{P}(Z < -a) = 0.1 \), the combined probabilities for the tails are:
\[
\mathbb{P}(Z > a) + \mathbb{P}(Z < -a) = 0.1 + 0.1 = 0.2
\]
So, the probability within \( -a \) and \( a \):
\[
\mathbb{P}(-a < Z < a) = 1 - 0.2 = 0.8
\]
Given this calculation, the statement:
"If \(\mathbb{P}(Z > a) = 0.1\), then \(\mathbb{P}(-a < Z < a) = 0.8\)"
is **True**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc5d414e-1582-4cc7-86ec-08dd2c0a0909%2F4e4e79df-8996-4c25-8817-1072b46ff686%2Fhzi3h1pq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question Explanation and Concept:**
Assume \( Z \) is a random variable with a standard normal distribution and \( a \) is a positive number. If \(\mathbb{P}(Z > a) = 0.1\), then \(\mathbb{P}(-a < Z < a) = 0.8.\)
**Question Statement:**
True or False
- ( ) True
- ( ) False
**Explanation:**
Here, \( Z \) represents a standard normal random variable, which implies it follows a normal distribution with a mean of 0 and a standard deviation of 1.
Let’s analyze the given conditions:
1. \(\mathbb{P}(Z > a) = 0.1\)
This condition means the probability that \( Z \) is greater than \( a \) is 0.1. Because the standard normal distribution is symmetric about the mean (0), the probability \( \mathbb{P}(Z < -a) \) will also be 0.1.
2. The probability for the range \( -a < Z < a \) can be interpreted as the area under the normal curve within these limits. To find this, observe that:
\[
\mathbb{P}(Z > a) + \mathbb{P}(Z < -a) + \mathbb{P}(-a < Z < a) = 1
\]
Given \(\mathbb{P}(Z > a) = 0.1\) and \( \mathbb{P}(Z < -a) = 0.1 \), the combined probabilities for the tails are:
\[
\mathbb{P}(Z > a) + \mathbb{P}(Z < -a) = 0.1 + 0.1 = 0.2
\]
So, the probability within \( -a \) and \( a \):
\[
\mathbb{P}(-a < Z < a) = 1 - 0.2 = 0.8
\]
Given this calculation, the statement:
"If \(\mathbb{P}(Z > a) = 0.1\), then \(\mathbb{P}(-a < Z < a) = 0.8\)"
is **True**.
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