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Answer:
First, determine how many standard deviations above the mean one would have to be to be in the 75th percentile. This can be found by using a z table and finding the z associated with 0.75. The value of z is 0.674. Thus, one must be .674 standard deviations above the mean to be in the 75th percentile
Complete Question
A measurement template with a historical value of 25.500 mm is measured 27 times and a mean value of 25.301 mm is recorded. What is the percent bias when the tolerance is +/- 0.3?
Answer:
The percent bias is
Step-by-step explanation:
From the question we are told that
The historical value is ![S = 25.00 \ mm](https://tex.z-dn.net/?f=S%20%3D%2025.00%20%5C%20mm)
The number of times it is measured is n = 7
The mean value is ![\= x = \frac{\sum x_i }{n} = 25.301 \ mm](https://tex.z-dn.net/?f=%5C%3D%20x%20%3D%20%5Cfrac%7B%5Csum%20x_i%20%7D%7Bn%7D%20%3D%2025.301%20%5C%20mm)
The tolerance is ![t =\pm 0.3 = 0.3 - (-0.3) = 0.6](https://tex.z-dn.net/?f=t%20%3D%5Cpm%200.3%20%3D%200.3%20-%20%28-0.3%29%20%3D%200.6)
Generally the percent bias is mathematically represented as
![B = 100 * \frac{\= x - \tau }{ t}](https://tex.z-dn.net/?f=B%20%3D%20100%20%2A%20%5Cfrac%7B%5C%3D%20x%20-%20%5Ctau%20%7D%7B%20t%7D)
=> ![B = 100 * \frac{25.301 - 25.5000 }{0.6}](https://tex.z-dn.net/?f=B%20%3D%20100%20%2A%20%5Cfrac%7B25.301%20-%2025.5000%20%7D%7B0.6%7D)
=>
This can be written as (x+4)(x+4)
Multiply everything in the second parenthesis by the x.
x^2 + 4x
Multiply everything in the second parenthesis by 4.
4x + 16
Add these two equations together.
x^2 + 4x + 4x + 16
Combine like terms.
x^2 + 8x + 16
Hope this helps!