Type two z-scores (or a mean, SD and two raw scores). The calculator shades the area under the normal curve, shows the two z-table rows it read, and gives the area outside the interval too.
P(−1.00 < Z < 1.50) = 0.77453 (77.45%)
Areas are the 5-decimal left-tail values from the z-table on our homepage (standard normal, mean 0, SD 1), so z is rounded to 2 decimals.
Question: exam scores are normal with mean 70 and standard deviation 10. What share of students score between 60 and 85?
Step 1. Convert to z. z₁ = (60 − 70) / 10 = −1.00; z₂ = (85 − 70) / 10 = 1.50.
Step 2. Read the table. Positive z-table, row 1.5, column .00: Φ(1.50) = 0.93319. Negative z-table, row −1.0, column .00: Φ(−1.00) = 0.15866.
Step 3. Subtract. 0.93319 − 0.15866 = 0.77453.
Answer: about 77.5% of students score between 60 and 85. The other 22.5% (0.22547) score below 60 or above 85.
Look up the left (cumulative) area for each z-score in the z-table and subtract the smaller from the larger: P(a < Z < b) = Φ(b) − Φ(a). For z = −1 and z = 1: 0.84134 − 0.15866 = 0.68268, so 68.27% of the area lies between them.
0.97500 − 0.02500 = 0.95000. Exactly 95% of a normal distribution lies within ±1.96 standard deviations, which is why 1.96 is the z critical value for a 95% confidence interval.
The same subtraction works when both z-scores are on the same side of the mean. Between z = 1 and z = 2: 0.97725 − 0.84134 = 0.13591, about 13.6% of the area.
Yes. Switch to “Raw scores”, enter the mean, standard deviation and the two values. Each is converted with z = (x − μ) / σ, rounded to 2 decimals like the z-table, and then the two table areas are subtracted. For mean 70, SD 10, the area between 60 and 85 is 0.77453.
Subtract the area between them from 1. Outside ±1.96 is 1 − 0.95000 = 0.05000, which is the two-tailed α for 95% confidence. The calculator shows this “outside” value under every answer.
No. This calculator uses the standard normal (z) table. For Student’s t critical values with degrees of freedom, use the t-table on ttable.org.