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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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Mack
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1 Answers1

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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.

André Nicolas
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  • They got 2.41 as an answer. I can't how that is true. – Mack Mar 08 '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 08 '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 08 '13 at 20:42
  • I'm sorry, but I still don't follow. How did you rewrite $P(|Z| \ge c)$, to solve for the c-value? – Mack Mar 08 '13 at 20:46
  • The same way you did. By the way, I have changed my answer, to correspond to your revised question. You actually almost did the calculation correctly. We want $\Pr(Z\ge c)=0.008$. But tables of the normal give $\Pr(Z\le k)$. So we want $\Pr(Z\le c)=1-0.008=0.992$. Now look for $0.992$ in the table, and look for where (for what $c$) it occurs. – André Nicolas Mar 08 '13 at 20:51
  • Oh, I see. Shouldn't $P(Z \le c) = 1 - 0.008$ actually be $P(Z \ge c)$? – Mack Mar 08 '13 at 20:53
  • Your solution was basically right, you want area up to $-c$ equal to $0.008$. However, the usual normal table does not tell you values for negative numbers, so we need to get the information from the positive values. – André Nicolas Mar 08 '13 at 20:54
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    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 08 '13 at 20:56
  • I'm sorry. I think I am still a little confused about this problem. I thought that I was suppose to find two c-values, they being equal but having opposite sign. Furthermore, I thought I was suppose to find the area of the left and right tails. – Mack Mar 09 '13 at 16:01
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    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 09 '13 at 16:52