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The question I am working on is:

In each case, determine the value of the constant c that makes the probability statement correct.

$P(c \le |Z|)=0.016$

Here is my attempt:

$P(|Z| \ge c)=0.016$

$P(Z \ge c~or~Z \le -c) = 0.016 $

$[1-\phi (c)] - \phi (-c) = 0.016$

By symmetry, $1-\phi (c)$ and $\phi (-c)$ are equal.

$2 \phi (-c) = 0.016 \implies \phi (-c) = 0.008$.

However, this doesn't lead to the correct solution. What exactly did I solve for? And how was I actually suppose to solve this question?

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Maybe reading this answer to a related question will help you with the mechanics of the problem. – Dilip Sarwate Mar 8 '13 at 21:32
up vote 1 down vote accepted

You started correctly: We want a combined probability of $0.016$ in the two tails $Z\ge c$ and $Z\le -c$. By symmetry, we want a probability of $\frac{0.016}{2}=0.008$ in the "right tail."

Equivalently, we want $\Pr(Z\le c)=1-0.008=0.992$. Look for $0.9992$ in the body of your standard normal table.

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They got 2.41 as an answer. I can't how that is true. – Mack Mar 8 '13 at 20:36
Oh, I am terribly sorry. I forgot to type the absolute bars in the original question. How did you arrive at that answer? – Mack Mar 8 '13 at 20:39
The number $2.41$ is not true if you are asked for $\Pr(Z\ge c)=0.016$. However, it is true (to the level of accuracy of the table) if we want $\Pr(|Z|\ge c)=0.016$. For if you check, you will see that the entry for $2.41$ is $0.992$. – André Nicolas Mar 8 '13 at 20:42
We want $\Pr(Z\ge c)=0.008$. That is equivalent to $\Pr(Z\le c)=0.992$. To repeat, I really mean $\Pr(Z\le c)=0.992$. – André Nicolas Mar 8 '13 at 20:56
There is a unique $c$ such that $\Pr(|Z|\ge c)=0.016. That $c$ is positive, close to $2.4$. The event $|Z|\ge 2.41$ is made up of two parts, $Z \ge 2.41$ (the right tail) and $Z\le -2.41$ (the left tail). You are given that the combined area of the right and left tails is $0.016$, and basically are being asked where the right tail begins. The answer is $2.41$ approximately. So you also know where the left tail begins. It is at $-c$. – André Nicolas Mar 9 '13 at 16:52

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