Questions: Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true. P(Z>c)=0.7357 Round your answer to two decimal places.

Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true.

P(Z>c)=0.7357

Round your answer to two decimal places.

Transcript text: Suppose $Z$ follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of $c$ so that the following is true. \[ P(Z>c)=0.7357 \] Round your answer to two decimal places.

Solution

Solution Steps

Step 1: Understanding the Problem

We need to find the value of $ c $ such that $ P(Z > c) = 0.7357 $. This can be rewritten using the cumulative distribution function $ \Phi $ of the standard normal distribution as: \[ P(Z \leq c) = 1 - P(Z > c) = 1 - 0.7357 = 0.2643 \]

Step 2: Finding the Z-Score

To find the Z-score $ c $ corresponding to $ P(Z \leq c) = 0.2643 $, we can calculate the cumulative probabilities for various Z-scores until we find one that closely matches 0.2643.

The calculations yield the following probabilities for different Z-scores:

For $ Z = -1.0 $, $ P \approx 0.1587 $
For $ Z = -0.999 $, $ P \approx 0.1589 $
For $ Z = -0.998 $, $ P \approx 0.1591 $
...
For $ Z = -0.95 $, $ P \approx 0.1698 $
For $ Z = -0.951 $, $ P \approx 0.1701 $

As we can see, the cumulative probability increases as the Z-score approaches zero. We continue this process until we find a Z-score that gives us a cumulative probability close to 0.2643.

Step 3: Approximating the Z-Score

Through the iterative calculations, we find that the Z-score $ c $ that results in a cumulative probability of approximately 0.2643 is around $ -0.62 $.