Questiol C7 A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1. Find the probability that a randomly selected value is greater than -196. P(X > -196) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or scores rounded to 3 decimal places are accepted. Submit Question
Questiol C7 A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1. Find the probability that a randomly selected value is greater than -196. P(X > -196) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or scores rounded to 3 decimal places are accepted. Submit Question
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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![**Statistics and Probability: Working with Normal Distributions**
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**Question 25**
A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1.
Find the probability that a randomly selected value is greater than -196.
\[ P(X > -196) = \boxed{\phantom{number}} \]
Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
---
**Instructions for Answering:**
1. Calculate the z-score for the value of -196 using the formula:
\[ z = \frac{X - \mu}{\sigma} \]
where:
- \( X \) is the value (-196 in this case)
- \( \mu \) is the mean (8.9 in this case)
- \( \sigma \) is the standard deviation (66.1 in this case)
2. Use the z-score to find the probability from the standard normal distribution table or using a calculator with statistical functions.
3. Enter your answer as a decimal rounded to 4 decimal places.
---
**Submit Your Answer:**
Click the "Submit Question" button to record your answer once you have calculated the probability.
---
**Example Text Explanation:**
1. Calculate the z-score:
\[ z = \frac{-196 - 8.9}{66.1} \approx -3.1097 \]
2. Find the probability corresponding to the z-score \(-3.1097\):
From the standard normal distribution table, the probability (P) corresponding to \( z \approx -3.11 \) is approximately \( 0.0009 \).
3. Since we want the probability \( P(X > -196) \):
\[ P(X > -196) = 1 - P(X \leq -196) = 1 - 0.0009 = 0.9991 \]
4. Finally, enter \( 0.9991 \) as your answer in the provided field.
---
**Reminder:**
Ensure your answers are precise and rounded as required. This exercise helps in understanding the application of the normal distribution and z-scores in determining probabilities.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1f51412a-93b8-4013-9035-1ebf79c0a718%2Fa99748c0-fc36-4d70-8f59-c7459ffae999%2Fa2nyi9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Statistics and Probability: Working with Normal Distributions**
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**Question 25**
A distribution of values is normal with a mean of 8.9 and a standard deviation of 66.1.
Find the probability that a randomly selected value is greater than -196.
\[ P(X > -196) = \boxed{\phantom{number}} \]
Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
---
**Instructions for Answering:**
1. Calculate the z-score for the value of -196 using the formula:
\[ z = \frac{X - \mu}{\sigma} \]
where:
- \( X \) is the value (-196 in this case)
- \( \mu \) is the mean (8.9 in this case)
- \( \sigma \) is the standard deviation (66.1 in this case)
2. Use the z-score to find the probability from the standard normal distribution table or using a calculator with statistical functions.
3. Enter your answer as a decimal rounded to 4 decimal places.
---
**Submit Your Answer:**
Click the "Submit Question" button to record your answer once you have calculated the probability.
---
**Example Text Explanation:**
1. Calculate the z-score:
\[ z = \frac{-196 - 8.9}{66.1} \approx -3.1097 \]
2. Find the probability corresponding to the z-score \(-3.1097\):
From the standard normal distribution table, the probability (P) corresponding to \( z \approx -3.11 \) is approximately \( 0.0009 \).
3. Since we want the probability \( P(X > -196) \):
\[ P(X > -196) = 1 - P(X \leq -196) = 1 - 0.0009 = 0.9991 \]
4. Finally, enter \( 0.9991 \) as your answer in the provided field.
---
**Reminder:**
Ensure your answers are precise and rounded as required. This exercise helps in understanding the application of the normal distribution and z-scores in determining probabilities.
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