Questions: Find the value of (zalpha). (z0.21) (z0.21=) (Round to two decimal places as needed.)

Transcript text: Find the value of $z_{\alpha}$. \[ z_{0.21} \] $z_{0.21}=$ $\square$ (Round to two decimal places as needed.)

Solution

Solution Steps

Step 1: Understanding the Problem

We need to find the value of $ z_{0.21} $, which is the z-score corresponding to a cumulative probability of 0.21 in the standard normal distribution. This means we are looking for the z-score such that the area to the left under the normal curve is 0.21.

Step 2: Setting Up the Calculation

To find $ z_{0.21} $, we will use a binary search method to narrow down the range of z-scores. We start with an initial range of $ z $ values from $ -5 $ to $ 5 $.

Step 3: Iterative Calculation

We calculate the cumulative probability for various midpoints $ z_{mid} $ within our range:

For $ z_{mid} = 0.0 $: \[ P = \Phi(0.0) - \Phi(-\infty) = 0.5 \]
For $ z_{mid} = -2.5 $: \[ P = \Phi(-2.5) - \Phi(-\infty) = 0.0062 \]
For $ z_{mid} = -1.25 $: \[ P = \Phi(-1.25) - \Phi(-\infty) = 0.1056 \]
For $ z_{mid} = -0.625 $: \[ P = \Phi(-0.625) - \Phi(-\infty) = 0.266 \]
For $ z_{mid} = -0.9375 $: \[ P = \Phi(-0.9375) - \Phi(-\infty) = 0.1743 \]
For $ z_{mid} = -0.78125 $: \[ P = \Phi(-0.78125) - \Phi(-\infty) = 0.2173 \]
For $ z_{mid} = -0.859375 $: \[ P = \Phi(-0.859375) - \Phi(-\infty) = 0.1951 \]
For $ z_{mid} = -0.8203125 $: \[ P = \Phi(-0.8203125) - \Phi(-\infty) = 0.206 \]
For $ z_{mid} = -0.80078125 $: \[ P = \Phi(-0.80078125) - \Phi(-\infty) = 0.2116 \]
For $ z_{mid} = -0.810546875 $: \[ P = \Phi(-0.810546875) - \Phi(-\infty) = 0.2088 \]