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 Z-scores.

## Z-value, Z-score, or Z Section

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 z-value.

- Z-value, Z-score, or Z
The Z-value (or sometimes referred to as Z-score or simply Z) represents the number of standard deviations an observation is from the mean for a set of data. To find the z-score 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 z-score 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 Z-score (indicates you scored higher than the mean). However, if one was analyzing days of missed work then a negative Z-score would be more appealing as it would indicate the person missed less than the mean number of days.

Characteristics of Z-scores

- The scores can be positive or negative.
- For data that is symmetric (i.e. bell-shaped) or nearly symmetric, a common application of Z-scores for identifying potential outliers is for any Z-scores that are beyond ± 3.
- Maximum possible Z-score for a set of data is \(\dfrac{(n−1)}{\sqrt{n}}\)

### From Z-score to Probability

For any normal random variable, if you find the Z-score 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 z-score 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 3-13: Heights Section

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:

\(z=\dfrac{65-64}{2}=0.5\)

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 3-14: Weights Section

The weights of 10-year-old girls are known to be normally distributed with a mean of 70 pounds and a standard deviation of 13 pounds. Find the percentage of 10-year-old 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 Z-scores: \(z=\dfrac{x-\mu}{\sigma}\)

For \(x=60\), we get \(z=\dfrac{60-70}{13}=-0.77\)

For \(x=90\), we get \(z=\dfrac{90-70}{13}=1.54\)

\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)}\\

&=0.9382-0.2206 &&\text{(Use a table or technology)}\\ &=0.7176 \end{align*}

We obtain that 71.76% of 10-year-old girls have weight between 60 pounds and 90 pounds.

## Example 3-15: Weights Cont'd... Section

Find the 60th percentile for the weight of 10-year-old 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 z-scores given the probability as we did before.

Area to the left of z-scores = 0.6000.

The closest value in the table is 0.5987.

The z-score corresponding to 0.5987 is 0.25.

Thus, the 60th percentile is z = 0.25.

**Now that we found the z-score, we can use the formula to find the value of \(x\). **The Z-score 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 10-year-old girls' weight is 73.25 pounds.