(2) The Bright Electric Company produces large quantities of light bulbs per day. For the 75 watt light bulb that it is currently producing, it is going to be advertised to have mean life of 500 hours and variance of 256 hours. In order to verify this claim, a consumer protection group plans to take a sample of size n = 100 and observe the sample average X. If the manufacturer's claim is taken to be accurate what is the probability that the sample average will differ from 500 by more than 5 hours? When the actual sample was taken and the resulting sample average was observed, it was found to be 473 hours. Is the manufacturer's claim trustworthy? Solve this problem by taking the following four steps one by one. Step (i) Even though the population distribution is not going to be normal, the sample size is large enough so that the sampling distribution of X 100 is approximately normal. This is due to TTE CBS By luck CLT GCVF N/A (i- Select One) Step (ii) The problem wants us to compute P(|X 100 – 500| 2 5). which is the same as X 100 - 500 -5 5 1 — Р 16/10 16/10 16/10 Why: Lucky guess Standardization TTE GCVF CLT N/A (ii- Select One) Step (iii) The probability expression of the last part is approximately equal to 1 - P(-3.125
(2) The Bright Electric Company produces large quantities of light bulbs per day. For the 75 watt light bulb that it is currently producing, it is going to be advertised to have mean life of 500 hours and variance of 256 hours. In order to verify this claim, a consumer protection group plans to take a sample of size n = 100 and observe the sample average X. If the manufacturer's claim is taken to be accurate what is the probability that the sample average will differ from 500 by more than 5 hours? When the actual sample was taken and the resulting sample average was observed, it was found to be 473 hours. Is the manufacturer's claim trustworthy? Solve this problem by taking the following four steps one by one. Step (i) Even though the population distribution is not going to be normal, the sample size is large enough so that the sampling distribution of X 100 is approximately normal. This is due to TTE CBS By luck CLT GCVF N/A (i- Select One) Step (ii) The problem wants us to compute P(|X 100 – 500| 2 5). which is the same as X 100 - 500 -5 5 1 — Р 16/10 16/10 16/10 Why: Lucky guess Standardization TTE GCVF CLT N/A (ii- Select One) Step (iii) The probability expression of the last part is approximately equal to 1 - P(-3.125
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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In case you're wondering what the abbreviations stand for:
CLT: central limit theorem
TTE: Time to the
GCVF: Generalized change of variable formula
CBS: Cauchy bunyakowski schwarz.
If there is an abbreviation you don't understand, please ignore it because it might not
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