A lawn mower company produces 1,500 machines lawn mowers in 2020. In an effort to determine how much maintenance-free that consumers will do when they have purchased a lawn mower, the company decided to conduct a study for the last 1 year from the lawn mower. First, be randomly sampled with contact the owner of 200 lawn mowers. Company owner give 800 and asked to contact the company during repairs first needed for lawn mowers. Owners who are no longer using lawn mowers to mow their disqualified lawns or not sampled. After 1 year passed, it turns out that there are 187 owners have report for the first fix. However, the other 51 owners disqualified themselves. The average amount per year until the first fix is 5.3 times for samples deemed worthy of taking. It is believed that the standard deviation is 1.28 per Years. If the company wants to advertise the lawn mower and with confidence level of 95%, what is the average estimated value per year of repair-free lawn mowing for these lawn mowers? What are the benefits of estimating estimated value in this case? (analyze the answer you have obtained)

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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A lawn mower company produces 1,500 machines
lawn mowers in 2020. In an effort to determine how much
maintenance-free that consumers will do when they have purchased a lawn mower, the company decided to conduct a study for the last 1 year
from the lawn mower. First, be randomly sampled with contact the owner of 200 lawn mowers. Company owner give 800 and asked to contact the company during repairs first needed for lawn mowers. Owners who are no longer using lawn mowers to mow their disqualified lawns or not sampled. After 1 year passed, it turns out that there are 187 owners have report for the first fix. However, the other 51 owners disqualified themselves. The average amount per year until the first fix is 5.3 times for samples deemed worthy of taking. It is believed that the standard deviation is 1.28 per Years. If the company wants to advertise the lawn mower and with confidence level of 95%, what is the average estimated value per year of repair-free lawn mowing for these lawn mowers? What are the benefits of estimating estimated value in this case? (analyze the answer you have obtained)

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