Suppose x is a normally distributed random variable with mean μ = 0 and standard deviation o = 1. Find the probability that x < 0.64. 26.11% 76.11% 23.89% 73.89%
Suppose x is a normally distributed random variable with mean μ = 0 and standard deviation o = 1. Find the probability that x < 0.64. 26.11% 76.11% 23.89% 73.89%
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Probability Calculation in a Normal Distribution**
**Problem:**
Suppose \( x \) is a normally distributed random variable with mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \). Find the probability that \( x < 0.64 \).
**Options:**
- \( 26.11\% \)
- \( 76.11\% \)
- \( 23.89\% \)
- \( 73.89\% \)
**Explanation:**
To find the probability \( P(x < 0.64) \) when \( x \) follows a normal distribution with a mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \), one would typically use a standard normal distribution table (Z-table) or statistical software.
1. **Standardize the variable:** Convert the value \( 0.64 \) to a Z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Since \( \mu = 0 \) and \( \sigma = 1 \):
\[ Z = \frac{0.64 - 0}{1} = 0.64 \]
2. **Consult the Z-table:** Locate the Z-score (0.64) in the Z-table to find the corresponding cumulative probability.
Based on standard Z-tables, the cumulative probability for \( Z = 0.64 \) is approximately \( 0.7389 \).
Therefore, the probability \( P(x < 0.64) \) is:
\[ P(x < 0.64) = 0.7389 \]
or \( 73.89\% \).
Hence, the correct option is:
- \( 73.89\% \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9d14dc9-6899-40d3-8d6b-066b6648b231%2Fcb6e697c-ed74-4906-9e49-90ea7538a99c%2Fuv6ceil_processed.png&w=3840&q=75)
Transcribed Image Text:**Probability Calculation in a Normal Distribution**
**Problem:**
Suppose \( x \) is a normally distributed random variable with mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \). Find the probability that \( x < 0.64 \).
**Options:**
- \( 26.11\% \)
- \( 76.11\% \)
- \( 23.89\% \)
- \( 73.89\% \)
**Explanation:**
To find the probability \( P(x < 0.64) \) when \( x \) follows a normal distribution with a mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \), one would typically use a standard normal distribution table (Z-table) or statistical software.
1. **Standardize the variable:** Convert the value \( 0.64 \) to a Z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Since \( \mu = 0 \) and \( \sigma = 1 \):
\[ Z = \frac{0.64 - 0}{1} = 0.64 \]
2. **Consult the Z-table:** Locate the Z-score (0.64) in the Z-table to find the corresponding cumulative probability.
Based on standard Z-tables, the cumulative probability for \( Z = 0.64 \) is approximately \( 0.7389 \).
Therefore, the probability \( P(x < 0.64) \) is:
\[ P(x < 0.64) = 0.7389 \]
or \( 73.89\% \).
Hence, the correct option is:
- \( 73.89\% \)
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