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Normal distribution and z-scoresStatistics and probabilityHigh schoolabout 0:34

Find P(-1 < Z < 1) for a standard normal variable: animated lesson

A narrated, step-by-step normal distribution and z-scores video lesson with captions. Final answer: P(-1 < Z < 1) ≈ 0.682689 (68.27%).

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 (0.2 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    Find P(-1 < Z < 1) for a standard normal variable.
  2. 2. Step by step
    1. Φ(1)≈0.84134475,Φ(−1)≈0.15865525\Phi(1) \approx 0.84134475,\quad \Phi(-1) \approx 0.15865525
      Find the left areas of both z-scores
    2. P(−1<Z<1)≈0.68268949P(-1 < Z < 1) \approx 0.68268949
      Subtract
  3. 3. The answer
    P(−1<Z<1)≈0.682689P(-1 < Z < 1) \approx 0.682689

    P(-1 < Z < 1) ≈ 0.682689 (68.27%)

Transcript

The problem. Here is the problem: Find P(negative 1 less than Z less than 1) for a standard normal variable. P(z₁ less than Z less than z₂): area between.

Step by step. Find the left areas of both z-scores. Subtract.

The answer. So the answer is P(negative 1 less than Z less than 1) approximately 0.682689 (68.27 percent). 68.27 percent of a normal distribution lies between these two z-scores.

About normal distribution and z-scores

Standardise values with z-scores and find normal probabilities.

Find P(-1 < Z < 1) for a standard normal variable (Lesson)