Question

1. find the z - score corresponding to the 90th percentile

Explanation:

Step1: Understand the problem

We need to find the \(z\) - score such that \(P(Z\leq z)=0.90\).

Step2: Use the standard normal table

Looking up the value \(0.90\) in the standard normal table (the cumulative distribution function of the standard normal distribution \(Z\sim N(0,1)\)).
The standard normal table gives the cumulative probabilities \(P(Z\leq z)\).
If we look at the body of the standard - normal table (z - table), we find that the closest value to \(0.90\) is \(0.8997\) (corresponding to \(z = 1.28\)) and \(0.9015\) (corresponding to \(z=1.29\)).
Using linear interpolation:
Let \(z\) be the \(z\) - score. Let \(x_1 = 0.8997\), \(z_1=1.28\), \(x_2 = 0.9015\), \(z_2 = 1.29\), and \(x = 0.90\).
The formula for linear interpolation is \(z=z_1+\frac{x - x_1}{x_2 - x_1}(z_2 - z_1)\)
\(z = 1.28+\frac{0.90 - 0.8997}{0.9015 - 0.8997}(1.29 - 1.28)\)
\(z=1.28+\frac{0.0003}{0.0018}\times0.01\)
\(z = 1.28+\frac{0.0003}{0.0018}\)
\(z=1.28 + 0.0167\approx1.28\)

Answer:

The \(z\) - score corresponding to the \(90^{th}\) percentile is approximately \(1.28\)