Left-Tail Probabilities from Z Scores

What a Left-Tail Probability Measures

A normal model can answer questions about the share of its probability lying below a specified cutoff. For example, if a measurement is modeled as normal, we may want the probability that a randomly selected measurement is below a threshold. The calculation has two parts: express the cutoff as a z-score, then find the standard normal area to its left.

The answer is an area under the normal density curve, and therefore a probability. It is not the height of the curve at the cutoff. A standard normal table reports this area in one of two common formats: cumulative tables give the area to the left of a z-score directly, while mean-to-z tables give the area between zero and the z-score. Knowing which format is in use is essential.

Definition: Let \(X\sim N(\mu,\sigma^2)\), with \(\sigma>0\), and let \(c\) be a cutoff. The left-tail probability is \(P(X<c)\), the probability that \(X\) falls below \(c\). Standardizing the cutoff gives \(z=(c-\mu)/\sigma\), and the probability is \(P(X<c)=P(Z<z)\), where \(Z\sim N(0,1)\).

This method assumes that the variable is appropriately described by the stated normal model, with the stated mean \(\mu\) and standard deviation \(\sigma\). If the model is an approximation rather than an exact description, the calculated probability is correspondingly an approximation. The standard deviation must be positive, and the cutoff and mean must be expressed in the same units.

From a Cutoff to a Standard Normal Area

For a normal variable \(X\), standardizing a value means subtracting the model mean and dividing by the model standard deviation. The z-score tells how many standard deviations the cutoff is from the mean. A positive z-score is above the mean; a negative z-score is below it. Standardization places the cutoff on the scale of \(Z\sim N(0,1)\), allowing a standard normal table to be used regardless of the original units.

Z=\frac{X-\mu}{\sigma} \qquad\text{and}\qquad P(X<c)=P\left(\frac{X-\mu}{\sigma}<\frac{c-\mu}{\sigma}\right) =P\left(Z<\frac{c-\mu}{\sigma}\right).

The inequality keeps its direction because \(\sigma>0\): subtracting \(\mu\) and dividing by a positive number do not reverse it. Write \(\Phi(z)=P(Z<z)\) for the standard normal cumulative probability. The calculation can therefore be summarized as \(P(X<c)=\Phi((c-\mu)/\sigma)\). For a continuous normal variable, \(P(X<c)=P(X\le c)\), because the probability of exactly one value is zero.

Check the model and units.
Identify the normal model’s mean and standard deviation, and express the cutoff in the same units.

Standardize the cutoff.
Calculate \(z=(c-\mu)/\sigma\), retaining enough precision to make a suitable table lookup.

Read the left area.
Use a cumulative table directly, or convert the reported mean-to-z area to a left-tail area.

Interpret the result.
State the probability of a value below the cutoff in the original context.

Using a Cumulative Standard Normal Table

A cumulative standard normal table reports \(\Phi(z)=P(Z<z)\). In a typical table, the row gives the ones and tenths digits of \(z\), and the column gives its hundredths digit. For \(z=1.20\), for example, use row 1.2 and column 0.00; the entry is approximately 0.8849. For a negative z-score, a cumulative table with negative rows can be read directly. If the table lists only nonnegative z-scores, use the symmetry of the standard normal curve: the area to the left of \(-a\) equals the area to the right of \(a\), or \(1-\Phi(a)\).

A mean-to-z table reports a different area: the area between zero and the listed z-score, usually for a nonnegative z-score. Call that area \(A\). For a positive cutoff, the left-tail area is \(0.5+A\), because half the standard normal area lies to the left of zero. For a negative cutoff, it is \(0.5-A\), by symmetry. These conversions concern left-tail probabilities only; first identify the table format before using an entry.

Table format	What the entry represents	Left-tail probability
Cumulative, at positive \(z\)	Area to the left of \(z\)	Use the entry directly
Cumulative, at negative \(z\)	Area to the left of \(z\)	Use the negative-\(z\) entry directly, if available
Mean-to-z, positive \(z\)	Area between 0 and \(z\)	Add the entry to 0.5
Mean-to-z, negative \(z\)	Area between 0 and \(\|z\|\)	Subtract the entry from 0.5

Worked Example: A Cutoff Above the Mean

Suppose a hypothetical machine’s fill amount \(X\) is modeled as \(X\sim N(70,10^2)\) millilitres. Under this model, find the probability that a randomly selected fill is below 82 millilitres.

Check the model and units. The stated model is normal, with mean \(\mu=70\) mL and standard deviation \(\sigma=10\) mL. Both the mean and cutoff are in millilitres, and the standard deviation is positive, so the normal-model calculation is applicable as stated.

Standardize. The cutoff’s z-score is

z=\frac{c-\mu}{\sigma} =\frac{82-70}{10} =\frac{12}{10} =1.20.

Look up the left area. In a cumulative standard normal table, row 1.2 and column 0.00 give \(\Phi(1.20)\approx0.8849\). Therefore,

P(X<82)=P(Z<1.20)=\Phi(1.20)\approx0.8849.

Under the stated model, about 88.49% of fills are below 82 mL. This cutoff is 1.20 standard deviations above the mean, so a left-tail probability greater than one-half is consistent with the location of the cutoff.

Negative Cutoffs and Mean-to-Z Tables

When the cutoff is below the mean, its z-score is negative and the left-tail probability is less than 0.5. A cumulative table that includes negative z-scores provides that probability directly. With a mean-to-z table, use the positive distance from zero and subtract its area from 0.5. This follows from symmetry: the standard normal curve has equal area on each side of zero.

Worked Example: A Cutoff Below the Mean

Suppose a hypothetical process has measurement \(X\sim N(500,80^2)\) units. Find the probability that a randomly selected measurement is below 420 units, using a mean-to-z table.

Check the model and units. The stated model is normal. Its mean is 500 units, its standard deviation is 80 units, and the cutoff is expressed in the same units. The model’s standard deviation is positive.

Standardize.

z=\frac{420-500}{80} =\frac{-80}{80} =-1.00.

Convert the table area. A mean-to-z table gives the area between zero and \(1.00\) as approximately 0.3413. The cutoff has z-score \(-1.00\), so its left-tail area is the half-area to the left of zero minus this area:

$$ P(X<420)=P(Z<-1.00) =0.5000-0.3413 =0.1587. $$

Under the stated model, approximately 15.87% of measurements are below 420 units. The answer is below 0.5 because 420 is below the mean of 500.

Worked Example: A Positive Z-Score with a Mean-to-Z Table

A hypothetical laboratory measurement \(X\) is modeled as \(X\sim N(120,15^2)\) units. Find the probability of a measurement below 129 units using a mean-to-z table.

Check the model and units. The model is stated to be normal; its mean, standard deviation, and cutoff are all expressed in the same units, and \(\sigma=15>0\).

Standardize the cutoff.

z=\frac{129-120}{15} =\frac{9}{15} =0.60.

A mean-to-z table gives the area between zero and \(0.60\) as approximately 0.2257. Because the cutoff is above the mean, the area to its left includes the 0.5000 area to the left of zero as well as the area from zero to 0.60:

$$ P(X<129)=P(Z<0.60) =0.5000+0.2257 =0.7257. $$

Under the stated model, approximately 72.57% of measurements are below 129 units. A left-tail probability above 0.5 is expected because the cutoff is above the mean.

Why the Table Format Matters

Both kinds of standard normal tables describe areas under the same curve, but their entries answer different questions. A cumulative table entry already includes all the area to the left of its z-score. A mean-to-z entry includes only the area between zero and the indicated distance from zero. Adding or subtracting 0.5 is not an optional adjustment: it accounts for the half of the curve lying to the left of zero.

A useful check is to compare the result with the cutoff’s position relative to the mean. If the cutoff equals the mean, \(z=0\) and the left-tail probability is 0.5. If the cutoff is above the mean, the left-tail probability must exceed 0.5; if it is below the mean, the probability must be less than 0.5. These checks can reveal a mistaken table format or sign before the result is reported.

Common mistake: A mean-to-z table reports 0.3413 at \(z=1.00\), so a reader reports 0.3413 as \(P(Z<1.00)\). That entry is only the area between 0 and 1.00. The left-tail area includes the 0.5000 to the left of zero, giving \(P(Z<1.00)=0.5000+0.3413=0.8413\). For a negative cutoff at \(-1.00\), symmetry instead gives \(P(Z<-1.00)=0.5000-0.3413=0.1587\).

A second safeguard is to check signs and units in the standardization step. The numerator is cutoff minus mean, not mean minus cutoff; reversing it changes a positive z-score to a negative one and places the cutoff on the wrong side of the mean. The denominator is the standard deviation, not the variance. Keep the unrounded z-score during calculation when possible, then report the table-based probability to a sensible number of digits.

Key takeaway: For a normal model \(X\sim N(\mu,\sigma^2)\), convert the cutoff \(c\) to \(z=(c-\mu)/\sigma\), then find the area to the left of \(z\). A cumulative table gives that area directly; a mean-to-z table requires adding its entry to 0.5 for positive \(z\), or subtracting it from 0.5 for negative \(z\).

Check Your Understanding

For each question, identify the table area needed and explain your reasoning.

A variable is modeled as \(X\sim N(40,5^2)\). What z-score corresponds to a cutoff of 45?
A cumulative standard normal table gives \(\Phi(0.80)=0.7881\). What is \(P(Z<0.80)\), and why is no 0.5 adjustment needed?
A mean-to-z table gives 0.1915 for \(z=0.50\). Use it to find \(P(Z<0.50)\).
A cutoff in a normal model has z-score \(-0.75\), and the mean-to-z area for \(0.75\) is 0.2734. Find the left-tail probability.
Why must the left-tail probability be greater than 0.5 when a cutoff lies above the mean of a normal model?