Question 1185718: For a standard normal distribution, find:
P(z > c) = 0.6892
Find c.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to find the value of *c*:
1. **Understand the problem:** We're given the probability that a standard normal variable *z* is *greater* than *c*. We need to find the corresponding *c* value.
2. **Convert to a left-tail probability:** Most statistical tables and calculators give probabilities to the *left* of a z-score. Since the total probability under the normal curve is 1, we can find the probability to the *left* of *c*:
P(z < c) = 1 - P(z > c)
P(z < c) = 1 - 0.6892
P(z < c) = 0.3108
3. **Find the z-score:** Now we need to find the z-score that corresponds to a cumulative probability of 0.3108. You can use a standard normal (z) table or a statistical calculator for this. Look up the probability closest to 0.3108 in the table's body and find the corresponding z-score.
* If using a calculator, you can use the inverse cumulative distribution function (often called `norm.ppf` or similar).
The z-score you find will be your value of *c*. It will be a negative value since the probability is less than 0.5.
c ≈ -0.49
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