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The Normal Distribution · Tutorial 454 of 996

Reading Mean-to-Z Standard Normal Tables

Learn how to locate mean-to-z table entries, interpret their areas, and use normal symmetry to look up negative z-scores.

Intermediate 9 min read

What You'll Learn

  • Identify what probability a mean-to-z standard normal table reports.
  • Find a z-score in a table using its row and column.
  • Interpret a table entry as area between zero and a positive z-score.
  • Use symmetry to read an area for a negative z-score.
  • Distinguish mean-to-z table entries from cumulative standard normal table values.
  • Check whether an answer is plausible from the table’s range and the z-score’s sign.

What a Mean-to-Z Table Reports

Standard normal tables use \(Z\sim N(0,1)\), a normal random variable with mean 0 and standard deviation 1. A mean-to-z table reports the area under the standard normal curve between its center, 0, and a specified positive z-score. This is a different quantity from the cumulative table entry: the two tables describe areas measured from different starting points.

Before using a table, check its title or a note at the top. A mean-to-z table may be labelled with wording such as “area from 0 to z.” Its entries are not probabilities to the left of the z-score. Once the table convention is clear, the lookup is straightforward: use the z-score’s first decimal place for the row and its second decimal place for the column.

Definition: For a positive z-score \(z\), a mean-to-z standard normal table reports \(P(0<Z<z)\), the area under the standard normal density between 0 and \(z\). Because the standard normal distribution is continuous, including or excluding either endpoint does not change this probability.

The entries range from 0 at \(z=0\) toward 0.5000 as \(z\) increases. The curve has half of its total area on either side of 0, so the area from 0 to a positive cutoff can never exceed 0.5. A value of, for example, 0.3000 means that 30% of the standard normal distribution lies between the mean and that positive z-score.

Finding a Positive Z-Score in the Table

A common layout lists z-scores in increments of 0.01. The row gives the ones and tenths digits, and the column gives the hundredths digit. For example, to look up \(z=1.23\), use the row labelled 1.2 and the column labelled 0.03. The entry at their intersection is the area from 0 to 1.23.

The row-and-column arrangement is only a way to locate the value. The table entry itself is a rounded approximation to the area under the curve, not a density height. In particular, it is not the value of the normal density at \(z\). If a requested z-score falls between the values printed in the table, use the nearest available hundredth unless the problem or table instructions specify interpolation.

Worked Example: Looking Up a Positive Z-Score

A hypothetical measurement has already been standardized to \(z=0.84\). Find the area between the mean and this z-score using a mean-to-z table, and state what the area represents.

Locate the entry. In the row labelled 0.8, move to the column labelled 0.04. The table entry is 0.2995.

Interpret the entry. By the table’s convention, this value estimates \(P(0<Z<0.84)\). In area notation, the reported result is

$$ P(0<Z<0.84)\approx 0.2995. $$

Equivalently, using the standard normal cumulative distribution function \(\Phi\), the area is \(\Phi(0.84)-\Phi(0)=0.7995-0.5000=0.2995\). Thus approximately 29.95% of the standard normal distribution lies between its mean and \(z=0.84\). This is an area for that interval only; it is not the table’s area to the left of 0.84.

Using Symmetry for Negative Z-Scores

Mean-to-z tables are usually printed for nonnegative z-scores. To find an area between a negative z-score and zero, use the positive version of its magnitude. The standard normal curve is symmetric about zero, so its area from \(-a\) to 0 equals its area from 0 to \(a\), for any \(a>0\). Thus the same table entry gives the corresponding area on either side of the mean.

Theorem: For \(a>0\), standard normal symmetry gives \(P(-a<Z<0)=P(0<Z<a)\). Therefore, to find the area between a negative z-score \(z\) and zero, look up the positive z-score \(|z|\) in a mean-to-z table.

The absolute value is used only to find the matching table entry. It does not mean the original z-score has changed sign: a negative z-score still lies to the left of the mean. Be precise about which interval the entry represents. For \(z=-1.37\), for example, the matching area is between \(-1.37\) and 0, not between 0 and \(-1.37\) in reversed order.

Why the Symmetry Lookup Works

The standard normal density has the same height at \(x\) and \(-x\). If \(\phi(x)\) denotes that density, then \(\phi(-x)=\phi(x)\). The area from \(-a\) to 0 can be rewritten by changing the variable of integration from \(x\) to \(u=-x\): the left-hand area becomes the area from 0 to \(a\).

$$ \begin{aligned} P(-a<Z<0) &=\int_{-a}^{0}\phi(x)\,dx\\ &=\int_{0}^{a}\phi(-u)\,du\\ &=\int_{0}^{a}\phi(u)\,du\\ &=P(0<Z<a), \qquad a>0. \end{aligned} $$

The second line uses the substitution \(u=-x\), which reverses the integration limits and introduces a minus sign that is accounted for when the limits are put in increasing order. The third line follows from the evenness of the standard normal density. This proves why a positive table entry can also be used for the matching interval on the negative side.

Worked Example: Looking Up a Negative Z-Score

In a hypothetical calculation, a standardized value is \(z=-1.37\). Find the area between this score and the mean from a mean-to-z table.

Apply symmetry. The interval from \(-1.37\) to 0 has the same area as the interval from 0 to \(1.37\). The lookup therefore uses \(|-1.37|=1.37\): go to row 1.3 and column 0.07. The table entry is 0.4147.

State the result in context. The area between the negative cutoff and the mean is approximately

$$ P(-1.37<Z<0)=P(0<Z<1.37)\approx 0.4147. $$

This means approximately 41.47% of the standard normal distribution lies between \(z=-1.37\) and \(z=0\). The table does not report the probability to the left of \(-1.37\); its entry concerns only the interval between that score and the mean.

How This Differs From a Cumulative Table

A cumulative standard normal table, as used in the earlier material on reading cumulative standard normal tables, reports the area to the left of a z-score. In this convention, the entry at \(z\) is \(\Phi(z)=P(Z<z)\). A mean-to-z table instead reports only the area between 0 and a positive z-score. The number in the two tables can therefore differ substantially even when both tables use the same z-score.

For a positive z-score, the cumulative area to the left includes all the area to the left of the mean, which is 0.5, as well as the area from the mean to the z-score. So the cumulative-table value equals 0.5 plus the mean-to-z entry. This relationship helps identify which convention a table is using; it does not change what either table’s entry means.

Worked Example: Distinguishing Two Table Conventions

Suppose two standard normal tables are available, and both are queried at \(z=1.23\). One is a mean-to-z table; the other is a cumulative table. Identify what each entry represents and check how the entries are related.

Mean-to-z lookup. Use row 1.2 and column 0.03. The entry is 0.3907, so this table reports \(P(0<Z<1.23)\approx 0.3907\).

Cumulative lookup. At the same z-score, the cumulative table entry is 0.8907. It reports the area to the left of 1.23. That area includes the 0.5000 to the left of the mean and the 0.3907 between the mean and 1.23.

Check the relationship. Substituting the entries gives \(0.5000+0.3907=0.8907\), which matches the cumulative value. The two values are not competing estimates of the same area: one is the area from 0 to 1.23, and the other includes that area plus the left half of the curve.

For a negative score, a mean-to-z table still gives the area between that score and zero by symmetry. A cumulative table’s entry, however, is defined as the area to the left of the score. These remain different regions. Before comparing or using values, identify the interval represented by each table rather than relying on the table’s numerical layout alone.

Common Mistakes and Checks

The most common error is to treat a mean-to-z entry as though it were a cumulative probability. For instance, reading 0.3907 at \(z=1.23\) and claiming that 39.07% of the distribution lies to the left of 1.23 uses the wrong interval. The correction is to say that 39.07% lies between 0 and 1.23; a cumulative table would include the area to the left of zero as well.

A second mistake is to look up a negative score as if it were a negative table entry. The areas in the table are nonnegative. Use the magnitude of the score to find the row and column, then state the matching interval on the correct side of zero. For example, the entry for 1.37 represents the interval from 0 to 1.37, and by symmetry it also gives the area from \(-1.37\) to 0.

Use these checks when reading a mean-to-z table:

  • Check the convention. Confirm that the table reports area between 0 and z, rather than area to the left of z.
  • Check the lookup. For \(z=1.23\), the row is 1.2 and the column is 0.03.
  • Check the range. A mean-to-z entry must be between 0 and 0.5.
  • Check the interval. For a negative cutoff, describe the area as between the cutoff and 0.
  • Check the interpretation. An entry is an area under the curve, not a density height or the probability that a particular continuous variable equals one exact value.
Key takeaway: A mean-to-z table gives \(P(0<Z<z)\) for positive \(z\). For negative cutoffs, use symmetry and look up the magnitude. Always distinguish this center-to-cutoff area from a cumulative table’s area to the left of a cutoff.

Check Your Understanding

Answer each question by identifying the interval represented, not just by quoting a table value.

  1. In a mean-to-z table, which row and column do you use to look up \(z=1.46\)?
  2. A table entry for \(z=0.72\) is 0.2642. State the probability this entry represents in standard normal notation.
  3. How would you use a mean-to-z table to find the area between \(z=-0.95\) and 0? Explain which positive value you would look up and why.
  4. At a positive z-score, why is a cumulative-table entry larger than the corresponding mean-to-z entry?
  5. What is wrong with interpreting a mean-to-z table entry of 0.3907 at \(z=1.23\) as the area to the left of 1.23?