- #1
azaharak
- 152
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In Frank Attix's book on Radialogical Physics
Equation 1.2a reads
σ = √(E) ≈ √(μ) σ= standard deviation of a single random measurement
Where E is the expecation value of a stochastic process which approaches μ (μ is the average of measured values) as the number of measured values becomes vary large (∞)
I agree with the how the mean and expectation value approach each other,
I do not see how the standard deviation is the square root of the expectation value.
Isn't the stdv σ = √[E(x^2)-E(x)^2] = √[E(x-μ)^2]
He continues with the following example:
A detector makes 10 measurements, for each measurement the average number of rays detected (counts) per measurement is 10^5.
He writes that the standard deviation of the mean is √[E(x)/n]≈√[μ/n] = √[(10^5)/10]
where n is the number of measurements.
I agree that the standard deviation of the mean is related to the standard deviation by σ'=σ/√(n), but once again I'm not sure how he gets the standard deviation itself.
Help!
Equation 1.2a reads
σ = √(E) ≈ √(μ) σ= standard deviation of a single random measurement
Where E is the expecation value of a stochastic process which approaches μ (μ is the average of measured values) as the number of measured values becomes vary large (∞)
I agree with the how the mean and expectation value approach each other,
I do not see how the standard deviation is the square root of the expectation value.
Isn't the stdv σ = √[E(x^2)-E(x)^2] = √[E(x-μ)^2]
He continues with the following example:
A detector makes 10 measurements, for each measurement the average number of rays detected (counts) per measurement is 10^5.
He writes that the standard deviation of the mean is √[E(x)/n]≈√[μ/n] = √[(10^5)/10]
where n is the number of measurements.
I agree that the standard deviation of the mean is related to the standard deviation by σ'=σ/√(n), but once again I'm not sure how he gets the standard deviation itself.
Help!