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Normal distribution and z-scoresStatistics and probabilityHigh schoolabout 0:43

A score of 130 comes from a distribution with mean 100 and standard deviation 15. Find its z-score: animated lesson

A narrated, step-by-step normal distribution and z-scores video lesson with captions. Final answer: z = 2 (percentile ≈ 97.72%).

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.3 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
A score of 130 comes from a distribution with mean 100 and standard deviation 15. Find its z-score.
2. Step by step
1. $x - \mu = 130 - 100 = 30$
  Subtract the mean from the value
2. $z = \frac{30}{15} = 2$
  Divide by the standard deviation
3. $\Phi(2) \approx 0.97725$
  Look up the area to the left (percentile)
3. The answer
$z = 2$
z = 2 (percentile ≈ 97.72%)
4. Check
- Reverse: μ + zσ = 100 + (2)(15) = 130.

Transcript

The problem. Here is the problem: A score of 130 comes from a distribution with mean 100 and standard deviation 15. Find its z-score. Z-score from x, mean and SD.

Step by step. Subtract the mean from the value. Divide by the standard deviation. Look up the area to the left (percentile).

The answer. So the answer is z equals 2 (percentile approximately 97.72 percent). 130 lies 2 standard deviations above the mean; about 97.72 percent of a normal population lies below it.

Check. Reverse: μ plus zσ equals 100 plus (2)(15) equals 130.

About normal distribution and z-scores

Standardise values with z-scores and find normal probabilities.

A score of 130 comes from a distribution with mean 100 and…