106 Normal Random Variables (5 of 6)

Example

The Distribution of z-Scores

Recall that we use the area under a normal density curve to find a probability. If we convert the x-values into z-scores, the distribution of z-scores is also a normal density curve. This curve is called the standard normal distribution.

Here we compare the normal density curve for the foot lengths to the standard normal curve:

The normal curve pictured on top is the model for the distribution of foot lengths. Note that the values on the axis are foot lengths. The distribution has a mean of 11 inches and a standard deviation of 1.5 inches. A foot length of 13 inches is marked. The shaded area is 0.0918. This is the probability that a randomly selected male will have a foot length greater than 13 inches: P(X > 13) = 0.0918.

The normal curve pictured on bottom is the standard normal distribution. This represents the distribution of z-scores. Note that the values on the axis are z-scores. The mean is 0 and the standard deviation is 1. The z-score corresponding to X = 13 inches is marked.

[latex]Z=\frac{x-\mathrm{μ}}{\mathrm{σ}}=\frac{13-11}{1.5}=1.33[/latex]

The shaded area here is the same, 0.0918. This is the probability that a z-score is greater than 1.33: P(Z > 1.33) = 0.0918.

Here is the main idea: Since the areas are the same, we use the standard normal curve to find the probabilities associated with any normal density curve.

Note: The standard normal distribution always has a mean = 0 and a standard deviation = 1. To understand this, recall that a z-score is the number of standard deviations X is above (or below) the mean.

• When the x-value is the mean, the z-score is 0.
• We can illustrate this for the foot lengths: If X is the mean, then
  [latex]\begin{array}{l}X=11\\ Z=\frac{x-\mathrm{μ}}{\mathrm{σ}}=\frac{11-11}{1.5}=\frac{0}{1.5}=0\end{array}[/latex]
• When X is 1 standard deviation above the mean, the z-score is 1.
• We can illustrate this for the foot lengths: If x-value is 1 standard deviation above the mean, X = 11 + 1.5 = 12.5.
  [latex]Z=\frac{x-\mathrm{μ}}{\mathrm{σ}}=\frac{12.5-11}{1.5}=\frac{1.5}{1.5}=1[/latex]
• Similarly, the z-score for the x-value that is 1 standard deviation below the mean is −1.

Example

Using the Standard Normal Curve to Find Probabilities

We use a simulation based on the standard normal distribution to find probabilities. In this simulation, the numbers on the horizontal axis are z-scores. The areas are rounded to four decimal places, so these areas are not exact values.

Use this image from the simulation to find the probabilities:

• Find P(−1.21 < Z < 1.21). In words, we want to find the probability that the z-score is between −1.21 and 1.21. This probability is the light blue area. So the answer is 0.7737.
• Find P(Z > 1.21). In words, we want to find the probability that the z-score is greater than 1.21. This probability is the blue area to the right of Z = 1.21. So the answer is 0.1131.
• Find P(Z < −1.21). In words, we want to find the probability that the z-score is less than −1.21. This probability is the dark blue area to the left of Z = −1.21. So the answer is 0.1131. Note that this is the same as P(Z > 1.21) because of the symmetry in the normal distribution. Both tails have the same area.
• Find P(Z > −1.21). In words, we want to find the probability that the z-score is greater than −1.21. This probability is the area to the right of −1.21. This is the sum of the green area and the dark blue area. So the answer is 0.7737 + 0.1131 = 0.8868.

Note: Here is another way to use the simulation to find P(Z > −1.21). Here we moved one slider as far to the left as possible, then located the other slider at Z = −1.21.

Notice that the area is 0.8869, not 0.8868. This discrepancy is due to rounding. The simulation uses a mathematical model to find the areas. These areas are rounded to four decimal places. Don’t worry about this small difference. Either answer is acceptable.