Standard Deviation Ranges on Normal Distributions
Suppose X follows N(μ,σ2). What is the probability that X is within 1 standard deviation of its mean?
We want to find P(μ−σ≤X≤μ+σ). We can do this by standardizing our X to produce a new RV Z that follows the standard normal distribution N(0,1).
P(μ−σ≤X≤μ+σ)=P(σμ−σ−μ≤σX−μ≤σμ+σ−μ)=P(−1≤Z≤1)=Φ(1)−Φ(−1)≈0.6826. Since Z follows N(0,1), we can use a standard normal distribution table to calculate Φ(1) and Φ(−1) and get our final value of 0.6826. Interestingly, this result holds true for all values of μ and σ2.
Proposition. Let η(p) be the (100p)th percentile for N(0,1) (i.e. P(Z≤η(p))=p). Suppose X follows N(μ,σ2). Then P(X≤μ+ση(p))=p. This is yet another way to relate the standard normal distribution to nonstandard normal distributions.