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The equation for the normal distribution is given by:

P(x) = 1/(σ√(2π)) * e^(-(x-µ)²/(2σ²))

If we want to find P(x) maximum when µ = 0 and σ =1, then we substitute x =0. Giving:

P(0) = 1/(1√(2π)) * e^(-(0-0)²/(2(1)²))

= 1/(√(2π))

≈ 0.3989

This is also the value of the standard deviation if you were to produce a normal distribution with a maximum of 1. In order to find this σ, you must solve σ in P(x) = 1/(σ√(2π)) * e^(-(x-µ)²/(2σ²)) = 1. As the largest value that e^(-(x-µ)²/(2σ²)) can take is 1, (when (-(x-µ)²/(2σ²)) = 0), you can solve σ in:

1/(σ√(2π)) = 1

σ = 1/(√(2π))

Q: What is this 0.3989?

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