2.2.2.2 Z-score and Z-table
A value on the standard normal distribution is defined as a Standard Score or Z-score. This score represents the number of standard deviations an observation is from the mean, thus giving us a mechanism of how any normal distribution with different mean and standard deviation can be standardised with 0 mean and 1 standard deviation.
A positive Z-score indicates that the observation is above the mean, while a negative Z-score indicates that the observation is below the mean. Thus, through the Z-score, we can find out what percentile of data falls under the bell curve. Say, when we have a Z-score of 0 (Fig 2.15, chart a), the area under the bell curve is exactly half of the complete bell curve, hence it's 50% (area shaded in red).
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Percentile can be defined as a measure used to indicate the value below which a given percentage of observations in a group of observations falls. Say, 90th percentile is the value below which 90% of the observations are found. More on percentiles in chapter 3. |
Similarly, when we have a Z-score of 1 (Fig 2.15, chart b), the area under the bell curve is equal to the sum of the area within the Z-score of 0 (50%) and the area between the Z-score of 0 and Z-score of 1 (~34%), hence it's ~84% (Fig 2.15, area shaded in red).
For distribution of percentages, refer to the normal distribution curve chart discussed in section 2.2.2.1.
The Z-score can be calculated through the following formula:
- is the Z-score (Standard Score)
- is the value to be standardised (percentile of the area under the bell curve)
- is the standard deviation
- is the mean
The formula implies that subtracting the mean from our value and then dividing it by the standard deviation, gives us the Z-score. This Z-score can be mapped to a Z-table (also known as the Standard Normal Table) to determine the percentile of x (area under the bell curve). The Z-table can be categorized into three types:
- Cumulative from mean: gives the percentile of from mean (0) to . Say, Z-score of 1 will have a value of ~0.34 (since it's from 0 to 1)
- Cumulative: gives the percentile of from negative infinity to . Say, Z-score of 1 will have a value of ~0.84 (since it's from - to )
- Complementary cumulative: gives the percentile of for 1 minus Z-score. Say, Z-score of 1 will have a value of ~0.16 (since it's 1 - 0.84)
Out of all of these types, the cumulative table is the most common one. Let's take a look at how this table looks like and how can we extract the percentile of from this table (Fig 2.16). Say, to find the percentile of for a Z-score of 1.14, we can perform the following steps:
- Read down the Z-column up to the first decimal place of the Z-score (i.e., 1.1)
- Read across the top row up to the second decimal place of the Z-score (i.e., 0.04)
- The intersection of this row and column gives us the percentile of (0.87286) i.e., the area under the bell curve to the left of the Z-score
Z-scores and Z-tables play an important role in determining the area under the bell curve.
Example 5.4. In an exam, the scores of students are normally distributed with a mean of 70 and a standard deviation of 8, find the probability of the following:
- Students scoring less than 60?
- Students scoring at least 85?
- Students scoring between 75 and 90?
Here, the values of mean() and standard deviation() are as follows:
- = 70
- = 8
Scenario 1: Probability of students scoring less than 60
- Here, (value to be standardized) = 60
- We need to find the probability of students scoring less than 60, i.e., ( < )
Substituting the values of , and in the formula of Z-score yields:
To find the probability of students scoring less than 60, we need to find the area to the left of the Z-score in the bell curve (Fig 2.17, area shaded in red):
Now, to find the probability of students scoring less than 60, we simply need to look up the Z-score of -1.25 on the Z-table (Fig 2.18):
From the Z-table, we get:
This implies that the probability of students scoring less than 60 is 10.56% (Fig 2.19).
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The probability of this problem can be calculated in excel through the following formula: =NORM.DIST(, , , TRUE) =NORM.DIST(60, 70, 8, TRUE) = 0.10565 |
Scenario 2: Probability of students scoring at least 85
- Here, (value to be standardized) = 85
- We need to find the probability of students scoring at least 85, i.e., (>)
In continuous distribution, ( ) = (>)
Substituting the values of , and in the formula of Z-score yields:
Since we need to find out the probability of students scoring at least 85, we need to find the area to the right of the Z-score in the bell curve (area shaded in blue, Fig 2.20):
As we know that the total area under the curve is 1, we can find the area to the left of the Z-score and then subtract it from 1 to get the area to the right of the Z-score i.e., the area shaded in blue.
Now, to find the probability of students scoring at least 85, we simply need to look up the Z-score of 1.87 on the Z-table and subtract it from 1 (Fig 2.21).
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Z-table is set up to handle up to two decimal places, so we have rounded the Z-score to two decimal places i.e., 1.87 |
From the Z-table, we get:
On subtracting this value from 1, we get:
This implies that the probability of students scoring at least 85 is 3.07% (Fig 2.22).
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Alternatively, instead of subtracting the percentile of retrieved from Z-table, we can also use the Complementary cumulative Z-table which gives the percentile of for 1 - Z-score directly. |
Scenario 3: Probability of students scoring between 75 and 90
- Here, (first value to be standardized) = 75
- (second value to be standardized) = 90
- We need to find the probability of students scoring between 75 and 90 i.e., (<<)
To find out the probability of students scoring between 75 and 90, we can perform the below steps:
- Find the probability of students scoring less than 75 i.e., area to the left of Z-score for
- Find the probability of students scoring less than 90 i.e., area to the left of Z-score for
- Subtract the area corresponding to from
Substituting the values of , and in the formula of Z-score yields:
To find the probability of students scoring at least 75, we simply need to look up the Z-score of 0.62 on the Z-table (Fig 2.23):
From the Z-table, we get:
The value highlighted in the chart is corresponding to a Z-score of 0.62, though Z1 = 0.625 which makes a significant difference in the value. Since the Z-table is only up to 2 digits, we can use excel for more precision. In our case, the exact value corresponding to a Z-score of 0.625 is 0.73401 and not 0.73237 (as highlighted in the Z-table)
Similarly, substituting the values of , and in the formula of Z-score yields:
To find out the probability of students scoring between 75 and 90, we need to find out the area between and (area shaded in blue, Fig 2.24):
To find the probability of students scoring at least 90, we simply need to look up the Z-score of 2.5 on the Z-table (Fig 2.25):
From the Z-table, we get:
On subtracting the area corresponding to from , we get:
This implies that the probability of students scoring between 75 and 90 is 25.97% (Fig 2.26).
On plotting the same on a chart, we get:
