# The Standard Normal Distribution

A standard normal distribution is a normal distribution with mean 0 and standard deviation 1. Areas under this curve can be found using a standard normal table (Table A in the Moore and Moore & McCabe textbooks). All introductory statistics texts include this table. Some do format it differently. From the 68-95-99.7 rule we know that for a variable with the standard normal distribution, 68% of the observations fall between -1 and 1 (within 1 standard deviation of the mean of 0), 95% fall between -2 and 2 (within 2 standard deviations of the mean) and 99.7% fall between -3 and 3 (within 3 standard deviations of the mean).

No naturally measured variable has this distribution. However, all other normal distributions are equivalent to this distribution when the unit of measurement is changed to measure standard deviations from the mean. (That's why this distribution is important--it's used to handle problems involving any normal distribution.)

Recall that a density curve models relative frequency as area under the curve.

Assume throughout this document then that we are working with a variable Z that has a standard normal distribution. The letter Z is usually used for such a variable, the small letter z is used to indicate the generic value that the variable may take.

### Question: What is the relative frequency of observations below 1.18?

That is, find the relative frequency of the event Z < 1.18. (Here small z is 1.18.)

#### Step 1

Sketch the curve. Identify--on the measurement (horizontal/X) axis--the indicated range of values.

The event Z < 1.18 is shaded in green. Events and possibilities are one in the same.

#### Step 2

The relative frequency of the event is equal to the area under the curve over the description of the event.

The blue area is the relative frequency of the event Z < 1.18. This area appears to be approximately 85%-90%. A good sketch will help you verify your answer.

#### Step 3

Use the standard normal table in your text (and, hopefully, soon to be on this site and interactive!) to find this area. In Table A values on the measurement axis are listed along the margins and areas within the table. Now, a complete copy of the table is not here (too big; somebody--probably me--would have to type all those numbers in). But, here's an abridged version.

 z .00 .01 ... .08 .09 0.0 .5000 .5040 ... .5319 .5359 0.1 .5398 .5438 ... .5714 .5753 ... ... ... ... ... ... 1.0 .8413 .8438 ... .8599 .8621 1.1 .8643 .8665 ... .8810 8830 1.2 .8849 .8869 ... .8997 .9015 ... ... ... ... ... ...

Corresponding to a measurement value of z = 1.18 is an area of 0.8810. This is exactly the answer to the question! Notice that it agrees with the picture as well as the original "guess." For any value z the table supplies the area under the curve over the region to the left of z. Again, area = relative frequency.

For a standard normal variable the relative frequency of observations falling below 1.18 is 0.8810. (Also, for any normal distribution, 0.8810 or 88.1% of the observations fall below 1.18 times the standard deviation above the mean.)

### Question: What is the relative frequency of observations below -0.63?

1. Identify the range of values described by "below -0.63" (shaded green).
2. Identify the area you need to find (shaded blue).
3. Look-up the appropriate area in your table. (Be careful to choose the "negative" portion of your table--look up -0.63.) That area is 0.2643. For a standard normal variable the relative frequency of observations falling below -0.63 is 0.2643. (Also, for any normal distribution, 0.2643 or 26.43% of the observations fall below 0.63 times the standard deviation below the mean. Below because -0.63 is negative.)

### Question: What is the relative frequency of observations above -1.48?

1. Identify the range of values described by "above -1.48" (shaded green).
2. Identify the area you need to find (shaded blue). It appears to be about 95%.
3. Use the value -1.48 to look up an area in your table. However, be careful. Doing so gives you 0.0694--this is nowhere near 0.95--our initial guess. That's because the table is oriented to find areas under the curve to the left of. . . So, in fact, looking up -1.48 has found the answer to the question What is the relative frequency of measurements falling below -1.48. This range, z < -1.48 (in gray) and the associated area 0.0694 (in purple) are shown below.

There are two ways to proceed. They are, of course, equivalent.

• Since 0.0694 of the observations fall below -1.48, the remaining 0.9306 = 1 - 0.0694 must fall above -1.48.
• Since the total area under the curve is exactly 1, and the purple area is 0.0694, the blue area must be 1 - 0.0694 = 0.9306.

In other words, subtraction from 1 is necessary.

93.06% of the observations fall above -1.48. (For any normal distribution, 0.9306 or 93.06% of the observations fall above 1.48 times the standard deviation below the mean.)

### Question: What is the relative frequency of the event -1.65 < Z < 1.65?

First, a < b < c means "b is between a and c." So here we want the relative frequency of observations falling between -1.65 and +1.65.

1. Identify the range of values described by "between -1.65 and 1.65" (shaded green).
2. Identify the area you need to find (shaded blue). It appears to be about 90%.
3. We use the table to identify this area.

When you look up -1.65 you find an area of 0.0495. This is the area under the curve (purple) over the region left of -1.65 (shaded red).

When you look up +1.65 you find an area of 0.9505. This is the area under the curve (blue and purple combined) over the region left of +1.65 (shaded gray).

• Again, there are two ways to proceed. If 95.05% of the observations fall below 1.65 and 4.95% of the observations fall below -1.65 then 90.10% = 95.05% - 5.05% must fall between -1.65 and 1.65.
• You can see that the relative frequency of Z < 1.65 is then 0.9505. However, this relative frequency (area) includes the relative frequency of Z < -1.65. Merely subtract this second relative frequncy out. The area under the curve over the region between -1.65 and +1.65 is 0.9595 - 0.0495 = 0.9010.

### Question: Find the value z such that the event Z > z has relative frequency 0.80.

That is, find the value z such that 0.80, or 80%, of the observations are greater than z.

You should note right away that this problem is different. Here we are given a relative frequency (0.20) and asked to find a corresponding measurement value (z). You don't know what z is, but you do know that 20% of the observations (20% of the area under the curve) lie to the right of z.

To begin it helps to identify approximately where this value might be. Since 20% of the area under the curve (shaded purple) must be to the right of z (shaded gray), z must be above 0. The picture below merely "guesses" where z is--the objective is to find its precise position.

Since this problem begins in an opposite fashion, you might guess the procuedure for a solution works in the opposite order. Your guess is correct!

We already know an area under the curve. However, the area we know is above the region to the right of z. The table handles areas under the curve to the left of values. The area under the curve (blue) over the region left of z (red) is 0.8000. (Or...given that 20% of the values are to lie above z, it must be the case that 80% of the values lie below z.)

Reverse the steps used previously.

Look up an area of 0.8000. Areas are found on the inside of the table. There is no value exactly equal to 0.8000, choose the closest one--0.7995.

Now, from the margins, find the value z that accompanies this area. That value is 0.84.

Since the relative frequency of observations below 0.84 is 0.7995 (approximately 0.80), the relative frequency of values above 0.84 is 0.2005 (approximately 0.20).

### Answer: 0.84 (because 20% of the observations are above 0.84).

You've found the 80th percentile! The 80th percentile of the standard normal distribution is 0.84. That's because 80% of the observations fall below 0.84. (Note: The 80th percentile of every normal distribution is 0.84 times the standard deviation above the mean.)