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Normal distribution and z-scoresStatistics and probabilityHigh schoolabout 0:37

Find P(-1.96 < Z < 1.96) for a standard normal variable: animated lesson

A narrated, step-by-step normal distribution and z-scores video lesson with captions. Final answer: P(-1.96 < Z < 1.96) ≈ 0.950004 (95%).

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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 (0.2 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
Find P(-1.96 < Z < 1.96) for a standard normal variable.
2. Step by step
1. $\Phi(1.96) \approx 0.9750021,\quad \Phi(-1.96) \approx 0.024997895$
  Find the left areas of both z-scores
2. $P(-1.96 < Z < 1.96) \approx 0.95000421$
  Subtract
3. The answer
$P(-1.96 < Z < 1.96) \approx 0.950004$
P(-1.96 < Z < 1.96) ≈ 0.950004 (95%)

Transcript

The problem. Here is the problem: Find P(negative 1.96 less than Z less than 1.96) 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 1.96 less than Z less than 1.96) approximately 0.950004 (95 percent). 95 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(-1.96 < Z < 1.96) for a standard normal variable