Separate correct-word rate from reading speed and accuracy. A timed passage demonstrates subtraction, seconds-to-minutes conversion, and the limits of a rate score.
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The calculation
WCPM = (words attempted − errors) × 60 ÷ elapsed seconds
Worked example
120 attempted words minus 6 errors leaves 114 correct. Over 90 seconds, 114 × 60 ÷ 90 = 76 WCPM. Raw speed is 80 WPM and accuracy is 95%.
Use the scoring rules of the assessment. A normalized rate from a different duration is not automatically comparable to a standardized one-minute test; WCPM alone does not establish comprehension.
Video chapters
- 0:00 — Speed is only part of the story
- 0:15 — Subtract errors from attempted words
- 0:32 — 90 seconds is 1.5 minutes
- 0:47 — Normalize the correct count
- 1:05 — Rate and accuracy answer different questions
- 1:28 — Use the assessment’s scoring rules
- 1:47 — A rate does not measure everything
- 2:05 — Count → subtract → divide by time
Read the full transcript
A reader attempts one hundred twenty words in ninety seconds and makes six scoring errors. How many words correct per minute is that? We need to separate the words attempted from the words read correctly, then express the correct count over a common interval of one minute.
Start with the attempted-word count, not necessarily the length of the entire passage. Subtract six errors from one hundred twenty attempts. That leaves one hundred fourteen correct words. Our tiles show the bookkeeping: the six marked errors leave the correct-word group before we calculate a rate.
Ninety seconds is one minute and thirty seconds. Thirty seconds is half a minute, so the total is one point five minutes. It is not one point three minutes. When time is written as minutes and seconds, divide the seconds by sixty before adding them to the minutes.
Divide one hundred fourteen correct words by one point five minutes. The rate is seventy-six words correct per minute. Equivalently, multiply one hundred fourteen by sixty, then divide by ninety. Both methods scale the same observed count to a sixty-second interval; they do not change which words were correct.
The raw speed is one hundred twenty divided by one point five, or eighty words per minute. For accuracy, divide one hundred fourteen by one hundred twenty. That gives zero point nine five. Multiply by one hundred to express it as a percentage. The accuracy is ninety-five percent. Eighty describes attempted speed. Seventy-six describes correct-word rate. Ninety-five percent describes the fraction correct.
What counts as an error depends on the assessment protocol. Omissions, substitutions, pauses, and self-corrections can have specific rules. Apply those rules consistently instead of inventing a new count. If the test specifies one minute, a rate normalized from ninety seconds is not automatically an equivalent standardized score.
A correct-word rate combines speed and accuracy, but it does not directly test whether the reader understood the passage. Passage difficulty and testing conditions also matter. A higher number on a different text is not, by itself, proof of improved comprehension or a fair comparison between readers.
Count attempted words under the scoring protocol, subtract errors, and divide correct words by elapsed minutes. Keep the duration and context with the result. One hundred twenty attempts, six errors, and ninety seconds gives one hundred fourteen correct words, or seventy-six words correct per minute.