STAT 400

Fri. March 6th, 2020


Standard Deviation Ranges on Normal Distributions

Suppose XX follows N(μ,σ2)N(\mu, \sigma^2). What is the probability that XX is within 1 standard deviation of its mean?

We want to find P(μσXμ+σ)P(\mu-\sigma\le X\le \mu+\sigma). We can do this by standardizing our XX to produce a new RV ZZ that follows the standard normal distribution N(0,1)N(0,1).

P(μσXμ+σ)=P(μσμσXμσμ+σμσ)=P(1Z1)=Φ(1)Φ(1)0.6826.\begin{aligned} &P(\mu-\sigma\le X\le \mu+\sigma) \\&=P(\frac{\mu-\sigma-\mu}{\sigma}\le \frac{X-\mu}{\sigma}\le \frac{\mu+\sigma-\mu}{\sigma})\\&=P(-1\le Z\le 1)\\&=\Phi(1)-\Phi(-1)\\&\approx0.6826. \end{aligned} Since ZZ follows N(0,1)N(0,1), we can use a standard normal distribution table to calculate Φ(1)\Phi(1) and Φ(1)\Phi(-1) and get our final value of 0.68260.6826. Interestingly, this result holds true for all values of μ\mu and σ2\sigma^2.


Percentiles of N(μ,σ2)N(\mu,\sigma^2)

Proposition. Let η(p)\eta(p) be the (100p)(100p)th percentile for N(0,1)N(0,1) (i.e. P(Zη(p))=pP(Z\le \eta(p))=p). Suppose XX follows N(μ,σ2)N(\mu,\sigma^2). Then P(Xμ+ση(p))=p.P(X\le \mu+\sigma\eta(p))=p. This is yet another way to relate the standard normal distribution to nonstandard normal distributions.