# The notation is the z-score that the area under the standard normal curve to the right of is

3. The Normal Distribution
5. Percentiles

Given any data value, we can identify how far that data value is away from the mean, simply by doing a subtraction x μ. This value will be positive if your data value lies above (to the right) of the mean, and negative if it lies below (to the left) of the mean. But what wed really like to know is, relative to the spread of our data set, how far is x from μ? Remember that the standard deviation σ gives us a measure of how spread out our entire set of individual data values is.

The z-score for any single data value can be found by the formula (in English):

or with symbols (as seen before!):

Obviously a z-score will be positive if the data value lies above (to the right) of the mean, and negative if the data value lies below (to the left) of the mean.

Example 6.1: Calculating and Graphing z-Values

Given a normal distribution with μ = 48 and s = 5, convert an x-value of 45 to a z-value and indicate where this z-value would be on the standard normal distribution.

Solution

Begin by finding the z-score for x = 45 as follows.

Now draw each of the distributions, marking a standard score of z = 0.60 on the standard normal distribution.

The distribution on the left is a normal distribution with a mean of 48 and a standard deviation of 5. The distribution on the right is a standard normal distribution with a standard score of z = 0.60 indicated.

Z-scores measure the distance of any data point from the mean in units of standard deviations and are useful because they allow us to compare the relative positions of data values in different samples. In other words, the z-score allows us to standardize two or more normal distributions, or more appropriately, to put them on the same scale. Therefore, well be able to compare relative positions of data values within their own distribution to determine which data values are closer to or farther from the mean. A prime example for this is to compare the test scores for two students, one who scored a 28 on the ACT (scores range from 1 36) and another who scored a 1280 on the SAT (scores range from 400 1600). Who, relative to their associated exam, scored better?

##### Example

Your statistics exam score was 0.67 standard deviations better than the class average; your biology score was 0.7 standard deviations better than the class average; your kayaking score was only 0.5 standard deviations better than the class average. Therefore, even though your actual score on the biology exam was the lowest of the three exam scores, relative to the distribution of all class exam scores, your biology exam score was the highest relative grade.

#### Finding an Area (Proportion) Given a Specific Z-Value

To determine the area under the N(0, 1) curve for any data value that does not fall exactly 1, 2, or 3 standard deviations above or below the mean actually requires some calculus. Lucky for us, areas under the N(0, 1) curve can be obtained in numerous other ways, including technology (TI-83/84, Excel) and a table of values. Search the Internet for standard normal table and youll find hundreds of tables illustrating z-scores and their associated areas. The majority of these methods report the area to the left of the specified z-score z, no matter where it lies. This comes from a calculus operation of integration, which finds an area from the start of a distribution (i.e., the far left-tail) up to the z-score. Two images are provided.

There are three types of area calculations that you will be performing, each requiring slightly different work:

• For areas to the left of z: simply use the area provided by a table or technology.
• For areas to the right of z: because the total area under a density curve is 1 (100%), simply calculate: 1area to the left of z0.
• For areas between two z-values, say z0andz1(where z0< z1): find the area to the left ofz1 and subtract from it the area to the left ofz0.

#### Finding a Z-Value Given an Area

This is a slightly more challenging task than calculating an area, because you basically work backwards from an algebraic standpoint. Its important to realize that a Standard Normal Table has two parts: (1) the top and side margins, which form the tenths and hundredths of a z-score, and (2) the body of the table, which are all the area (probability) values. Also, remember that the Standard Normal Table only provides us information on the area (probability) to the left of a z-score. A small excerpt of Table B from Appendix A is shown below.

Notice that the z-values given in the table are rounded to two decimal places. The first decimal place of each z-value is listed in the left column, with the second decimal place in the top row. Where the appropriate row and column intersect, we find the amount of area under the standard normal curve to the left of that particular z-value.

Example : Finding Area to the Left of a Positive z-Value Using a Cumulative Normal Table

Find the area under the standard normal curve to the left of z = 1.37.

Solution

To read the table, we must break the given z-value (1.37) into two parts: one containing the first decimal place (1.3) and the other containing the second decimal place (0.07). So, in Table B from Appendix A, look across the row labeled 1.3 and down the column labeled 0.07. The row and column intersect at 0.9147. Thus, the area under the standard normal curve to the left of z = 1.37 is 0.9147.

Using a TI-83/84 Plus calculator, we can find a value of the area to the left of a z-score. To obtain the solution using a TI-83/84 Plus calculator, perform the following steps.

• Press 2nd and then Vars to access the DISTR menu.
• Choose option 2:normalcdf( .
• Enter lower bound, upper bound, µ , σ. Note If you want to find area under the standard normal curve, as in this example, then you do not need to enter µ or σ.
• Since we are asked to find the area to the left of z, the lower bound is -. From the empirical rule we know that after about 3 standard deviations away from the mean we have accounted for almost all of the data, so for our lower bound we will simply use a very negative number.We cannot enter - into the calculator, so we will enter a very small value for the lower endpoint, such as -1099. This number appears as -1E99 when entered correctly into the calculator. To enter -1E99, press(-) 1 [2nd][ , ]99. This appears on the screen as normcdf(-1E99,1.37,0,1).

If we are given an area (or probability) value, we need to first locate it in the body of a table, then track our way up and to the left in order to piece together the z-score that relates to the specified area. Keep in mind that you may not find the exact area value in the body of the tableso just use the closest value you can find, and then identify the proper z-score.

One calculation that will be used frequently in the coming chapters is to identify the two z-scores that separate a specific area in the middle of the standard normal distribution.

##### Example

Suppose we want to know which two z-scores separate out the middle 95% of the data. From the empirical rule, we already know the z-scores that do this are ±2 (2 standard deviations on either side of the mean). In reality, its not exactly ±2, but close enough for rough calculations.

To find the exact two z-scores, we use the following logic: If the middle portion is 95% = 0.95, then how much area lies outside of the middle (to the left and right)? A simple subtraction solves this! 1 0.95 = 0.05. The outside area, 0.05, must be split equally between the two tails (because of symmetry!). Therefore, dividing 0.05 by two gives us an area of 0.025 in each tail.

Using a standard normal table backwards, we first look through the body of the table to find an area closest to 0.025. The z-score corresponding to a left-tail area of 0.025 isz = 1.96. Now, therefore, the upper z-score will be z = 1.96, by the symmetry property of the standard normal distribution. You could also discover the upper z-score by looking up the area/probability value 0.025 + 0.95 = 0.975 in the body of the table and finding the associated z-value. By the end of the class, you will be extremely familiar with z-scores that define a central 90% (z = ± 1.645), 95% (z = ± 1.96), and 99% (z = ± 2.576).

##### Example: Find and interpret the probability of a random Normal variable

Suppose you just purchased a 2005 Honda Insight with automatic transmission. Using www.fueleconomy.gov you determine for the 2005 Honda Insights have mean highway gas milage is 56 miles per gallon with a standard deviation of 3.2. The distribution of this data has a bell-shape and is normal. You want to know the following:

a) How likely is it that your Honda Insight with automatic transition will get better than 60 miles per gallon on the highway.

b) How likely is it that your Honda Insight with automatic transition will get less than 50 miles per gallon on the highway.

c) How likely is it that your Honda Insight with aoutomatic transition will get between 52 and 62 miles per gallon on the highway.

Solution

This problem deals with data that is normally distributed with mean 56 and standard deviation 3.2, i.e., .

(a)

In symbols, we are asked to calculate P(X > 60). Sketching a normal curve and shading the area corresponding to greater than 60, gives us the graph shown. In order to calculate the appropriate area in the upper (right) tail, we must first convert our data to the standard normal distribution. The z-score for x = 60 is:

This means that 60 is 1.25 standard deviations above the mean. Notice how lining the two normal curves up as shown illustrates how the two areas are the same: P(X > 60) = P(Z > 1.25).

Using z = 1.25, we go to Table IV (or use normcdf(1.25,1E99,0,1))

to find the area to the left of z = 1.25 is 0.8943. Since we need the area to the right, we simply take 1 0.8943 = 0.1057.

Therefore, P(X > 60) = 0.1057 = 10.57%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 10.57% will get over 60 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 10.57% chance that your car will get over 60 miles per gallon on the highway.

(b)

In symbols, we are asked to calculate P(X < 50). Sketching a normal curve N(56, 3.2)and shading the area corresponding to less than 50, gives us the graph shown to the right.

In order to calculate the appropriate area in the lower (left) tail, we must first convert our data to the standard normal distribution. The z-score for x = 50 is:

Thus, the value 50 MPG is 1.88 standard deviations below the mean. In symbols we see:P(X < 50) = P(Z < 1.88).

Using z = -1.88, we go to Table IV (or use normcdf(-1E99,-1.88,0,1))

to find the area to the left of z = -1.88 is 0.0301. Therefore, P(X < 50) = 0.0301 = 3.01%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 3.01% will get less than 50 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 3.01% chance that your car will get less than 50 miles per gallon on the highway.

(c)

In symbols, we are asked to calculate P(58 < X < 62). Sketching a normal curve N(56, 3.2) and shading the area corresponding to greater than 58 but less than 62, gives us the graph shown. In order to calculate the appropriate area, we must first convert both data to the standard normal distribution.

The z-score for X= 58 is:

and thez-score forx = 62 is:

In terms of probability, we can now say: P(58 < X < 62) = P(0.63 < Z < 1.88).

Using z = 1.88, we go to Table IV (or use technology) to find the area to the left of z = 1.88 is 0.9699. Now, we need to remove (subtract) the area left of z = 0.63, which is 0.7357. Therefore, P(58 < X < 62) = 0.9699 0.7357 = 0.2342, or 23.42%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 23.42% will get between 58 and 62 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 23.42% chance that your car will get between 58 and 62 miles per gallon on the highway.

This calculation can be done with both normcdf(0.63,1.88,0,1) and normcdf(58,62,56,3.2), which will be the same.

#### Find the Value of a Random Variable Knowing a Probability Value

In these types of problems, we need to work backwards. Starting with a specified probability, find the specified z-score, then work our way back to the random variable. The tables of standard normal values are not a one-way tool! What do we mean by that? So far youve started with a value for a random variable (like a gas mileage value in the previous problem), turned it into a z-score, and then looked up the associated probability value for that z-score. We can use this table to work backwards! We can start with a known probability value in the body of a table, identify the z-score corresponding to that area by moving your fingers to the associated row and column, the reverse the algebra transformation from a z-score to a random variable.

If this sounds confusing, think back to the steps we took in the preceding example:

If, however, we are given an area/probability, then to work our way back to the original data value, we must first identify the appropriate z-score, and then un-standardize the z-score to arrive (finally!) back at the data value. How do we algebraically undo the z-score? Easyjust solve for the data value X:

Multiply both sides by σ to remove it from the denominator on the left side:

X μ = Zσ

Finally, add the value of μ to both sides to isolate the value of the random variable X:

X = Zσ + μ

##### Example: Finding the value of a normal random variable

Instead you want to know a gas mileage for a particular probability. Find what gas mileage for your 2005 Honda Insight will get better gas mileage than 97% of all other 2005 Honda Insights with automatics transmission.

Solution

This problem again deals with data that is normally distributed with mean 56 and standard deviation 3.2, i.e., N(56, 3.2).

To find the 97% percentile gas mileage, we need to find the specific miles per gallon X that separates the bottom 97% of all gas mileages from the top 3%. So for this problem we are given a percentage/area. Sketching the normal curve gives the graph shown.

Using Table IV, we find 0.97 in the body of the table, and then identify the z-score of 1.88. Notice that the exact area 0.97 is not in the table, but the closest area of 0.9699 has the z-score of 1.88. Now we un-standardize the z-score of 1.88. In English this means we need to identify the specific gas mileage that is 1.88 standard deviations above the mean of 56. Solving for X in the Z transform gives:

Therefore, if your 2005 Honda Insight cars with an automatic transmission gets 62 mpg, it gets better miles per gallon than 97% of all 2005 Honda Insight cars with an automatic transmission.

3. The Normal Distribution
5. Percentiles