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A company has developed an "easystart" mower that cranks the engine with the push of a button. The company claims that the probability the mower will start on any push of the button is 0.9. Assume for now that this claim is true. On the next 30 uses of the mower, let T = the number of times it starts on the first push of the button. What is the standard deviation of T? Interpret this value. O OT = 1.643. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 1.643, on average. O OT = 2.7. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 2.7, on average. O OT = 2.7. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 2.7. O GT = 2.7. If many rounds of 30 attempts were completed, the mower would start on the first push of the button 2.7% of the time. O OT = 1.643. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 1.643.

A company has developed an "easystart" mower that cranks the engine with the push of a button. The company claims that the probability the mower will start on any push of the button is 0.9. Assume for now that this claim is true. On the next 30 uses of the mower, let T = the number of times it starts on the first push of the button. What is the standard deviation of T? Interpret this value. O OT = 1.643. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 1.643, on average. O OT = 2.7. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 2.7, on average. O OT = 2.7. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 2.7. O GT = 2.7. If many rounds of 30 attempts were completed, the mower would start on the first push of the button 2.7% of the time. O OT = 1.643. If many many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 1.643.

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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A company has developed an “easystart” mower that cranks the engine with the push of a button. The company claims that the probability the mower will start on any push of the button is 0.9. Assume for now that this claim is true. On the next 30 uses of the mower, let \( T \) = the number of times it starts on the first push of the button. What is the standard deviation of \( T \)? Interpret this value. - \( \sigma_T = 1.643 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 1.643, on average. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 2.7, on average. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 2.7. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the mower would start on the first push of the button 2.7% of the time. - \( \sigma_T = 1.643 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 1.643.
Transcribed Image Text:A company has developed an “easystart” mower that cranks the engine with the push of a button. The company claims that the probability the mower will start on any push of the button is 0.9. Assume for now that this claim is true. On the next 30 uses of the mower, let \( T \) = the number of times it starts on the first push of the button. What is the standard deviation of \( T \)? Interpret this value. - \( \sigma_T = 1.643 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 1.643, on average. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would be 2.7, on average. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 2.7. - \( \sigma_T = 2.7 \). If many rounds of 30 attempts were completed, the mower would start on the first push of the button 2.7% of the time. - \( \sigma_T = 1.643 \). If many rounds of 30 attempts were completed, the number of times the mower would start on the first push of the button would typically vary from the mean (27) by about 1.643.
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