Reading an Area from Table A
In Sketching and Shading Normal Curves, you used a diagram to identify the part of a normal curve that represents an event. Now you can use Table A to find an area for a standard normal curve. Table A is a table of cumulative areas: each entry gives the area to the left of a specified z-score.
A standard normal random variable, written \(Z\), has mean 0 and standard deviation 1. A z-score tells you where a value sits on this standardized scale, as in Calculating a z-Score. Once an event is expressed using \(Z\), Table A can provide the corresponding area. The table does not automatically answer every probability question: first identify the requested region, then decide whether to use the entry directly, subtract it from 1, or subtract two entries.
Most versions of Table A organize each z-score into a row and a column. The row gives the score through the tenths place; the column supplies the hundredths place. To look up \(z=1.23\), go to row 1.2 and column 0.03. The entry is 0.8907, so about 89.07% of the area lies to the left of 1.23.
For a negative score such as \(-0.76\), use row \(-0.7\) and column 0.06 if your table includes negative z-scores. Some tables show only positive scores or make the negative-score layout less convenient. In that case, use the symmetry of the standard normal curve: the area to the left of \(-a\) equals the area to the right of \(a\), which is 1 minus the area to the left of \(a\).
The exact inequality symbol at a single boundary does not change a normal area: a continuous normal variable has probability 0 of being exactly equal to one value. Still, write the event that matches the wording of the question. A table entry is an area, so it must be between 0 and 1. A score far to the left should have a small left area; a score far to the right should have a left area close to 1. These checks can reveal a reversed lookup or a misplaced decimal.
A Consistent Lookup Routine
Before opening the table, identify the event. If the request is already a left-tail event, the table entry gives the answer directly. If the event is a right tail, take the complement of the left area. If it is an interval, subtract the area to the left of the lower boundary from the area to the left of the upper boundary. These steps connect the shaded region from the previous tutorial to the table’s particular convention.
Use the wording to decide whether the requested region is to the left, to the right, or between two z-scores.
Use the row for the ones and tenths digits and the column for the hundredths digit.
Each table entry is the cumulative area to the left of its z-score, not the area between the mean and that score.
Use the entry directly for a left tail, subtract it from 1 for a right tail, or subtract two entries for an interval.
Confirm the result is plausible for the size and position of the shaded region.
For reference, the following entries will be used in the examples. Table values are rounded to four decimal places.
| z-score | Table A area to the left |
|---|---|
| \(-1.23\) | 0.1093 |
| \(-0.76\) | 0.2236 |
| \(-0.42\) | 0.3372 |
| 0.86 | 0.8051 |
| 1.23 | 0.8907 |
| 1.24 | 0.8925 |
| 1.37 | 0.9147 |
These are rounded table entries, not exact values. Carry the table precision through your arithmetic, then report the area to a sensible number of decimal places.
Worked Example: A Left-Tail Area
Worked Example: A Left-Tail Area
A fictional indoor climbing center models the time a visitor spends completing a beginner route as normal. For this example, consider the standard normal score \(Z\). Find the area to the left of \(z=1.23\).
State. The event is \(Z<1.23\), the area to the left of 1.23 on the standard normal curve.
Plan. Since Table A reports left-tail areas, locate \(z=1.23\) and use its entry directly.
Do. In the table, go to row 1.2 and column 0.03. The entry is 0.8907:
The area is about 0.8907, or 89.07%. This is plausible because 1.23 is to the right of the mean, so most of the curve lies to its left.
Conclude. About 89.07% of the area under the standard normal curve is to the left of \(z=1.23\).
Worked Example: A Negative z-Score and a Right Tail
Worked Example: A Negative z-Score and a Right Tail
A fictional online retailer models the standardized delivery delay for a certain service as standard normal. Find the area to the right of \(z=-0.76\).
State. The requested event is \(Z>-0.76\), a right-tail area.
Plan. Table A gives area to the left, so first find the left area for \(-0.76\). Subtract that value from 1 to obtain the area to the right.
Do. In a table that includes negative scores, locate row \(-0.7\) and column 0.06. The entry is 0.2236. Therefore:
As a check, \(-0.76\) is left of the mean. The area to its right should therefore be greater than one-half, and 0.7764 is greater than 0.5. If your copy of Table A does not list negative scores, use symmetry: the area to the left of \(-0.76\) is the same as the area to the right of \(0.76\). Since the left area at \(0.76\) is 0.7764, the left area at \(-0.76\) is \(1-0.7764=0.2236\), leading to the same right-tail answer.
Conclude. About 77.64% of the standard normal area lies to the right of \(-0.76\).
Worked Example: An Area Between Two z-Scores
Worked Example: An Area Between Two z-Scores
A fictional marine science class uses a standard normal model to describe standardized readings from a sensor. Find the area between \(z=-0.42\) and \(z=0.86\).
State. The event is \(-0.42<Z<0.86\).
Plan. Table A gives cumulative left areas. Subtract the area to the left of the lower boundary from the area to the left of the upper boundary. This leaves only the area between the two scores.
Do. The table gives \(P(Z<-0.42)=0.3372\) and \(P(Z<0.86)=0.8051\). Thus:
The answer is about 0.4679, or 46.79%. The interval crosses the mean, but its area is not found by simply doubling one side: the endpoints are not equally far from 0. The subtraction works whether or not the interval is centered.
Conclude. About 46.79% of the standard normal area lies between \(z=-0.42\) and \(z=0.86\).
Interpolation When the Score Falls Between Entries
Table A typically lists z-scores to the nearest hundredth. Sometimes a calculation produces a score with more digits, such as \(z=1.236\). A common practical choice is to round the z-score to the nearest hundredth and use the table entry for 1.24. If an estimate between entries is useful, you can instead interpolate: estimate the area between two adjacent table values in proportion to how far the z-score lies between their scores.
Interpolation is an approximation, not a separate exact table value. It treats the area as changing at a constant rate between nearby entries. Because the normal curve is smooth, this can give a useful estimate over a small gap. The result should not be reported with more meaningful precision than the table supports.
Worked Example: Interpolating Between Table Entries
Worked Example: Interpolating Between Table Entries
A fictional greenhouse uses a normal model for a standardized leaf-width measurement. Estimate the area to the left of \(z=1.236\) by interpolating between the Table A entries for 1.23 and 1.24.
State. The target is \(P(Z<1.236)\). It lies between the table scores 1.23 and 1.24.
Plan. Find what fraction of the 0.01 score interval separates 1.23 from 1.236. Use that fraction of the difference between the two table areas.
Do. The fraction of the score interval is:
The table areas are 0.8907 at 1.23 and 0.8925 at 1.24. Their difference is \(0.8925-0.8907=0.0018\). Estimate 60% of that increase and add it to the lower entry:
The estimated area is 0.8918, or about 89.18%. Rounding the z-score to 1.24 instead would give 0.8925, a nearby but slightly larger table estimate. Neither approach should be mistaken for an exact probability read directly at 1.236 from Table A.
Conclude. Linear interpolation estimates that about 89.18% of the standard normal area lies to the left of \(z=1.236\).
Common Mistakes and AP Exam Tips
- Reading a left area as a right area. The table entry is always the area to the left. For a right-tail event, subtract the left area from 1.
- Using the column as the whole score. For \(z=1.23\), use row 1.2 and column 0.03. The column does not represent 0.3 or 0.23.
- Subtracting interval areas in the wrong order. For an interval from \(a\) to \(b\), with \(a<b\), calculate the left area at \(b\) minus the left area at \(a\). The result should not be negative.
- Assuming a negative z-score has a negative area. Areas are probabilities and cannot be negative. A negative score indicates a position left of the mean, not a negative probability.
- Forgetting symmetry or the complement. If a table does not list negative scores, use the symmetry of the standard normal curve. For a right tail, remember that the total area is 1.
- Rounding too early or claiming excessive precision. Keep the table entries consistent during subtraction and report a suitably rounded result. Interpolation is an estimate, so do not present many extra decimal places as if they were exact.
- Reporting an area without identifying the event. A strong answer states the probability being found, shows the relevant table lookup or subtraction, and interprets the area as a proportion of the modeled values when the setting provides context.
The sketch remains a useful check even after the table lookup. A left area for a positive score should exceed 0.5; a left area for a negative score should be below 0.5. A right tail beyond a positive score should be less than 0.5, and an interval area should be between 0 and 1. If the result contradicts the sketch, check the event direction and table operation.
Key Takeaway
Table A turns a z-score into a cumulative area: locate the score by its row and column, then read the area to its left. Use complements for right tails and differences for intervals. When a score falls between listed values, rounding to a table score is a practical option; interpolation provides an approximate in-between area when needed.
Check Your Understanding
Use the Table A entries in this tutorial where applicable. Give each area to four decimal places unless interpolation is requested.
- For \(z=0.86\), what is the area to the left? Explain why it should be greater than 0.5.
- Use the table entry for \(-0.76\) to find the area to the left of \(z=0.76\).
- Find the area to the right of \(z=1.37\), using the entry 0.9147.
- Find the area between \(z=-1.23\) and \(z=1.23\), using the table entries 0.1093 and 0.8907.
- Estimate the left area for \(z=1.235\) by interpolating between the entries for 1.23 and 1.24. Show the fraction of the score interval you use.