A z-score of 1.5
A z-score of 1.5 gives P(X < 1.5) = 0.9332, P(X > 1.5) = 0.0668, and a two-tailed p-value of 0.1336.
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Given a z-score, five related probabilities can all be read off the same standard normal curve at once: the area to the left, the area to the right, the area between the mean and z, the area within z of the mean on both sides, and the two-tailed area beyond ±z — each answers a slightly different question about the same score.
Given a p-value instead, the calculator searches numerically for the z-score that produces it, since there is no algebraic formula to invert the normal distribution directly.
p = 2 × (1 − Φ(|z|))
A z-score of 1.5 gives P(X < 1.5) = 0.9332, P(X > 1.5) = 0.0668, and a two-tailed p-value of 0.1336.
A two-tailed p-value of 0.05 corresponds to a z-score of about 1.96 — the classic threshold for 95% confidence.
It depends on your hypothesis test: a one-sided test ("is the effect greater than expected?") uses a one-tailed value (left or right); a two-sided test ("is the effect different at all, in either direction?") uses the two-tailed value — the most commonly reported figure in research.
It is the z-score where the two-tailed p-value equals exactly 0.05, making it the standard cutoff for 95% confidence — one of the most widely used thresholds in statistics.
It means the observed result would be unlikely if there were truly no effect (the null hypothesis) — commonly interpreted as evidence against that null hypothesis, though the right threshold for "significant" depends on the field and the stakes involved.
Convert it first: z = (value − mean) ÷ standard deviation. Once you have the z-score, this works the same way.
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