Question

which z - values correspond to the bottom 66% of the standard normal distribution? round your answer to the nearest thousandth. z <

Explanation:

Step1: Use the standard normal distribution table (z - table)

We want to find the z - value such that \(P(Z < z)=0.66\).

Step2: Look up the value in the z - table

Looking at the standard normal distribution table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)=P(Z < z)\)).
We search for the value closest to \(0.66\) in the body of the z - table.
The standard normal distribution table gives us that when \(z = 0.41\), \(\varPhi(0.41)=0.6591\) and when \(z = 0.42\), \(\varPhi(0.42)=0.6628\).
Using linear interpolation:
Let \(z=x\), \(y = 0.66\), \(z_1=0.41\), \(y_1 = 0.6591\), \(z_2=0.42\), \(y_2=0.6628\)
The formula for linear interpolation is \(x=z_1+\frac{y - y_1}{y_2 - y_1}(z_2 - z_1)\)
Substitute the values: \(x = 0.41+\frac{0.66 - 0.6591}{0.6628 - 0.6591}(0.42 - 0.41)\)
\(x=0.41+\frac{0.0009}{0.0037}\times0.01\)
\(x\approx0.41 + 0.00243\)
\(x\approx0.412\)

Answer:

\(z<0.412\)