Question

according to the empirical rule, in a standard normal distribution, approximately 68% of all values fall between -1 and 1. find the percentage of observations that fall between ( z = -1 ) and ( z = 1 ).
% (round to 2 decimal places.)

Explanation:

Step1: Use the standard normal distribution formula

The standard normal distribution is symmetric about \(z = 0\). The probability density function is \(f(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\).
The cumulative - distribution function \(P(Z\leq z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{-\frac{t^{2}}{2}}dt\).
We want to find \(P(- 1<Z<1)\), which is \(P(Z < 1)-P(Z<-1)\).

Step2: Look up values in the standard - normal table

From the standard - normal table (or using a calculator with a normal - distribution function), \(P(Z < 1)=\Phi(1)\approx0.8413\) and \(P(Z<-1)=\Phi(-1)\approx0.1587\).

Step3: Calculate the probability

\(P(-1 < Z < 1)=P(Z < 1)-P(Z<-1)\)
\(=0.8413 - 0.1587\)
\(=0.6826\)

Answer:

\(68.26\)