**Transcription for Educational Website:**An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with $ \mu = 44 $ and $ \sigma = 5.5 $.**(a)** What is the probability that yield strength is at most 38? Greater than 66? (Round your answers to four decimal places.)- at most 38: **0.2336**- greater than 66: **[Input Required]****(b)** What yield strength value separates the strongest 75% from the others? (Round your answer to four decimal places.)- **0.0001 ksi**You may need to use the appropriate table in the [Appendix of Tables](#) to answer this question.**Explanation:**This section illustrates a problem regarding the probability distribution of yield strength for A36 grade steel. The mean ($ \mu $) is 44, and the standard deviation ($ \sigma $) is 5.5, indicating the data follows a normal distribution. It includes probability calculations for specific values and a critical value determination separating the top 75% from the rest.

Let "X" be the yield strength (ksi) for A36 grade steel. Use excel formula…

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with u = 44 and o = 5.5. A USE SALT (a) What is the probability that yield strength is at most 387 Greater than 66? (Round your answers to four decimal places.) at most 38 0.2336 greater than 66 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to four decimal places.) 0.0001 |× ksi

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with u = 44 and o = 5.5. A USE SALT (a) What is the probability that yield strength is at most 387 Greater than 66? (Round your answers to four decimal places.) at most 38 0.2336 greater than 66 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to four decimal places.) 0.0001 |× ksi

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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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

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**Transcription for Educational Website:**

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with $ \mu = 44 $ and $ \sigma = 5.5 $.

**(a)** What is the probability that yield strength is at most 38? Greater than 66? (Round your answers to four decimal places.)

- at most 38: **0.2336**
- greater than 66: **[Input Required]**

**(b)** What yield strength value separates the strongest 75% from the others? (Round your answer to four decimal places.)

- **0.0001 ksi**

You may need to use the appropriate table in the [Appendix of Tables](#) to answer this question.

**Explanation:**

This section illustrates a problem regarding the probability distribution of yield strength for A36 grade steel. The mean ($ \mu $) is 44, and the standard deviation ($ \sigma $) is 5.5, indicating the data follows a normal distribution. It includes probability calculations for specific values and a critical value determination separating the top 75% from the rest.

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