- #1

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## Main Question or Discussion Point

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!