Uniform distribution and standard deviation

[r0bHadz]

Homework Statement

4.19:
According to one of the Western Electric rules for quality control, a produced item is considered conforming if its measurement falls within three standard deviations from the target value. Suppose that the process is in control so that the expected value of each measurement equals the target value. What percent of items will be considered conforming, if the distribution of measurements is: Uniform(a,b)?

4.20:
Refer to exercise 4.19, what percent of items falls beyond 1.5 standard deviations from the mean, if the distribution of measurements is Uniform(a,b)?

Relevant Equations

density f(x) = 1/(b-a), a<x<b
μ = (a+b)/2
σ = (b-a)/sqrt(12)

+(3/2) standard deviations from the mean = $\frac {a+b}{12} + \frac{\sqrt3}{4} (b-a)$
-(3/2) standard deviations from the mean = $\frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)$

$\frac {1}{b-a} \int_a^{\frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)} dx$ = m_1= $\frac {(-11+3\sqrt3)a + (1-3\sqrt3)b}{12(b-a)}$

$\frac {1}{b-a} \int_{\frac {a+b}{12} + \frac{\sqrt3}{4} (b-a)}^{b} dx$ = m_2 = $\frac {(11-3\sqrt3)b + (-1+3\sqrt3)a}{12(b-a)}$

adding m_1+m_2 I get:

$\frac{(-12+6\sqrt3)a + (12-6\sqrt3)b}{12(b-a)}$

I got up to this point where I quit the problem.

Now if I let b = 1 and a = 0 I get the answer to my problem which is .13397

I understand that this is the standard uniform distribution. But I don't understand what part of the problem statement let's me assign these values to a and b. Nothing in the problem statement tells me that this is not a general distribution, so why does it take values a = 0 and b = 1 to make it standard?

 

[Orodruin]

You can always rescale to a standard distribution using a linear transformation.

 

[LCKurtz]

+(3/2) standard deviations from the mean = $\frac {a+b}{12} + \frac{\sqrt3}{4} (b-a)$
-(3/2) standard deviations from the mean = $\frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)$

You shouldn't have $12$ in those two means above, or below.

$\frac {1}{b-a} \int_a^{\frac {a+b}{12} - \frac{\sqrt3}{4} (b-a)} dx$ = m_1= $\frac {(-11+3\sqrt3)a + (1-3\sqrt3)b}{12(b-a)}$

$\frac {1}{b-a} \int_{\frac {a+b}{12} + \frac{\sqrt3}{4} (b-a)}^{b} dx$ = m_2 = $\frac {(11-3\sqrt3)b + (-1+3\sqrt3)a}{12(b-a)}$

adding m_1+m_2 I get:

$\frac{(-12+6\sqrt3)a + (12-6\sqrt3)b}{12(b-a)}$

Factor a minus out of the first term and simplify it.

I got up to this point where I quit the problem.

Now if I let b = 1 and a = 0 I get the answer to my problem which is .13397

Fix the algebra and you will get the same answer.

 

[r0bHadz]

Fix the algebra and you will get the same answer.

ended up fixing it, and got the same answer >.> damn I hate algebra