Question

determine the proportion of observations in a standard normal distribution that fall between ( z = - 0.67 ) and ( z = 1.67 ).
( p ( - 0.67 < z < 1.67 ) = ) (round to at least 3 decimal places.)

Explanation:

Step1: Find the cumulative probability for \(z = 1.67\)

Using the standard normal table or a calculator with a normal - distribution function (\(\Phi(z)\) gives the cumulative probability \(P(Z\leq z)\)). For \(z = 1.67\), \(\Phi(1.67)=0.9525\)

Step2: Find the cumulative probability for \(z=-0.67\)

For \(z = - 0.67\), using the property that \(\Phi(-z)=1 - \Phi(z)\). So \(\Phi(-0.67)=1-\Phi(0.67)\). From the standard normal table, \(\Phi(0.67) = 0.7486\), then \(\Phi(-0.67)=1 - 0.7486=0.2514\)

Step3: Calculate \(P(-0.67\lt Z\lt1.67)\)

By the formula \(P(a\lt Z\lt b)=\Phi(b)-\Phi(a)\). Substitute \(a=-0.67\) and \(b = 1.67\) into the formula: \(P(-0.67\lt Z\lt1.67)=\Phi(1.67)-\Phi(-0.67)\)

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Answer:

\(0.701\)