3.3.3  Probabilities for Normal Random Variables (Zscores)
3.3.3  Probabilities for Normal Random Variables (Zscores)The standard normal is important because we can use it to find probabilities for a normal random variable with any mean and any standard deviation.
But first, we need to explain Zscores.
Zvalue, Zscore, or Z
We can convert any normal distribution into the standard normal distribution in order to find probability and apply the properties of the standard normal. In order to do this, we use the zvalue.
 Zvalue, Zscore, or Z

The Zvalue (or sometimes referred to as Zscore or simply Z) represents the number of standard deviations an observation is from the mean for a set of data. To find the zscore for a particular observation we apply the following formula:

\(Z = \dfrac{(observed\ value\  mean)}{SD}\)

Let's take a look at the idea of a zscore within context.
For a recent final exam in STAT 500, the mean was 68.55 with a standard deviation of 15.45.
 If you scored an 80%: \(Z = \dfrac{(80  68.55)}{15.45} = 0.74\), which means your score of 80 was 0.74 SD above the mean.
 If you scored a 60%: \(Z = \dfrac{(60  68.55)}{15.45} = 0.55\), which means your score of 60 was 0.55 SD below the mean.
Is it always good to have a positive Z score? It depends on the question. For exams, you would want a positive Zscore (indicates you scored higher than the mean). However, if one was analyzing days of missed work then a negative Zscore would be more appealing as it would indicate the person missed less than the mean number of days.
 The scores can be positive or negative.
 For data that is symmetric (i.e. bellshaped) or nearly symmetric, a common application of Zscores for identifying potential outliers is for any Zscores that are beyond ± 3.
 Maximum possible Zscore for a set of data is \(\dfrac{(n−1)}{\sqrt{n}}\)
From Zscore to Probability
For any normal random variable, if you find the Zscore for a value (i.e standardize the value), the random variable is transformed into a standard normal and you can find probabilities using the standard normal table.
For instance, assume U.S. adult heights and weights are both normally distributed. Clearly, they would have different means and standard deviations. However, if you knew these means and standard deviations, you could find your zscore for your weight and height.
You can now use the Standard Normal Table to find the probability, say, of a randomly selected U.S. adult weighing less than you or taller than you.
Example 313: Heights
According to the Center for Disease Control, heights for U.S. adult females and males are approximately normal.
 Females: mean of 64 inches and SD of 2 inches
 Males: mean of 69 inches and SD of 3 inches
Find the probability of a randomly selected U.S. adult female being shorter than 65 inches.
Answer
This is asking us to find \(P(X < 65)\). Using the formula \(z=\dfrac{x\mu}{\sigma}\) we find that:
\(z=\dfrac{6564}{2}=0.5\)
Now, we have transformed \(P(X < 65)\) to \(P(Z < 0.50)\), where \(Z\) is a standard normal. From the table we see that \(P(Z < 0.50) = 0.6915\). So, roughly there this a 69% chance that a randomly selected U.S. adult female would be shorter than 65 inches.
Example 314: Weights
The weights of 10yearold girls are known to be normally distributed with a mean of 70 pounds and a standard deviation of 13 pounds. Find the percentage of 10yearold girls with weights between 60 and 90 pounds.
In other words, we want to find \(P(60 < X < 90)\), where \(X\) has a normal distribution with mean 70 and standard deviation 13.
Answer
It is often helpful to draw a sketch of the normal curve and shade in the region of interest. You can either sketch it by hand or use a graphing tool.
To find the probability, we need to first find the Zscores: \(z=\dfrac{x\mu}{\sigma}\)
For \(x=60\), we get \(z=\dfrac{6070}{13}=0.77\)
For \(x=90\), we get \(z=\dfrac{9070}{13}=1.54\)
\begin{align*}
P(60<X<90) &= P(0.77<Z<1.54) &&\text{(Subbing in the Z values from above)} \\
&= P(Z<1.54)  P(Z<0.77) &&\text{(Subtract the cumulative probabilities)}\\
&=0.93820.2206 &&\text{(Use a table or technology)}\\ &=0.7176 \end{align*}
We obtain that 71.76% of 10yearold girls have weight between 60 pounds and 90 pounds.
Example 315: Weights Cont'd...
Find the 60th percentile for the weight of 10yearold girls given that the weight is normally distributed with a mean 70 pounds and a standard deviation of 13 pounds.
Answer
As before, it is helpful to draw a sketch of the normal curve and shade in the region of interest. You can either sketch it by hand or use a graphing tool. You know that 60% will greater than half of the entire curve.
We can use the Standard Normal Cumulative Probability Table to find the zscores given the probability as we did before.
Area to the left of zscores = 0.6000.
The closest value in the table is 0.5987.
The zscore corresponding to 0.5987 is 0.25.
Thus, the 60th percentile is z = 0.25.
Now that we found the zscore, we can use the formula to find the value of \(x\). The Zscore formula is \(z=\dfrac{x\mu}{\sigma}\).
Using algebra, we can solve for \(x\).
\(x=\mu+z(\sigma)\)
\(x=70+(0.25)(13)=73.25\)
Therefore, the 60th percentile of 10yearold girls' weight is 73.25 pounds.