Unconditional Probability Using the To ...
We can use the total probability rule to determine the unconditional probability of... Read More
Using the standard normal distribution table, we can confirm that a normally distributed random variable \(Z\), with a mean equal to 0 and variance equal to 1, is less than or equal to \(z\), i.e., \(P(Z ≤ z)\).
However, the table does this only when we have positive values of \(z\). Simply put, if an examiner asks you to find the probability behind a given positive z-value, you will have to look it up directly on the table, knowing that \(P(Z ≤ z) = θ(z)\) when \(z\) is positive.

The returns on ABC stock are normally distributed, where the mean is $0.60 with a standard deviation of $0.20. The probability that the return is less than $1 is closest to:
Solution
If the return is $0.10, then \(x = 0.1\) (this is our observed value). Therefore,
$$\begin{align} z &=\frac{x-\mu}{\sigma}\\&=\frac{1-0.6}{0.2}\\ &=2\ \text{(The return of \$1 is two and a half standard deviations below the mean)}\end{align}$$
First, you’d be required to calculate the z-value (2 in this case). \(P(Z ≤ 1)\) can be read directly from the table
You just move down and locate the z-value that lies to the right of “2,” i.e., 0.9772.

If we have a negative z-value and do not have access to the negative values from the table (as shown below), we can still calculate the corresponding probability by noting that:
$$ \begin{align} P(Z \le -z) & = 1 – P(Z \le z) \\ \theta(–z) & = 1 – \theta(z) \\ 0.0228 & = 1 – 0.9772 \end{align}$$

This relationship is true when we consider the following facts:

Question
Calculate P(Z ≤ -2.5).
A. 0.9938.
B. 0.0062.
C. 0.06.
Solution
The correct answer is B.
$$ \begin{align*} P(Z \le -2.5) & = 1 – P(Z \le 2.5) \\ & = 1 – 0.9938 \\ & = 0.0062 \\ \end{align*} $$
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