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MATLAB: An Introduction with Applications
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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**Transcription of Text for Educational Purposes** --- 4. For X̄ = 55.00 and S = 7.00, between what two values do approximately 95% of the cases lie in a normal distribution? (41, 69) 5. For X̄ = 90.00 and S = 15.00, between what two values do 99.7% of the cases lie in a normal distribution? 6. What is the multiplier for the estimate 95% rule? --- **Explanation:** The questions pertain to statistical concepts involving normal distribution. Question 4 uses the empirical rule to find the range within which 95% of data lies, typically about two standard deviations from the mean. The given values are X̄ (mean) = 55.00 and S (standard deviation) = 7.00, resulting in a range from 41 to 69. Question 5 refers to the same rule, extended to cover 99.7% of data points, usually within three standard deviations from the mean. The calculations will involve X̄ = 90.00 and S = 15.00. The sixth question inquires about the multiplier used to apply the 95% rule, generally involving a Z-score of approximately 1.96 for a normal distribution.
Transcribed Image Text:**Transcription of Text for Educational Purposes** --- 4. For X̄ = 55.00 and S = 7.00, between what two values do approximately 95% of the cases lie in a normal distribution? (41, 69) 5. For X̄ = 90.00 and S = 15.00, between what two values do 99.7% of the cases lie in a normal distribution? 6. What is the multiplier for the estimate 95% rule? --- **Explanation:** The questions pertain to statistical concepts involving normal distribution. Question 4 uses the empirical rule to find the range within which 95% of data lies, typically about two standard deviations from the mean. The given values are X̄ (mean) = 55.00 and S (standard deviation) = 7.00, resulting in a range from 41 to 69. Question 5 refers to the same rule, extended to cover 99.7% of data points, usually within three standard deviations from the mean. The calculations will involve X̄ = 90.00 and S = 15.00. The sixth question inquires about the multiplier used to apply the 95% rule, generally involving a Z-score of approximately 1.96 for a normal distribution.
Expert Solution
Step 1

Let X be a random variable which follows normal distribution.

Here, 

mean(X¯) = 90

standard deviation(S) = 15

So, X~N(90,152)

Z = X-9015~N(0,1)

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