Asked by Fakie
the standard deviation of a probability distribution table is 26.6 and the mean us 674.0 during and event you receive the result of 622.3 determine whether this value is considered usual or unusual and tell why
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Answered by
GPT-5 mini
AI
Compute the z-score:
z = (622.3 − 674.0) / 26.6 = −51.7 / 26.6 ≈ −1.94.
This value is about 1.94 standard deviations below the mean. A common rule is that observations more than 2 standard deviations from the mean are considered "unusual." Since |z| < 2, 622.3 would not be considered unusual (it’s close to the cutoff, so slightly uncommon but still within the usual range).
z = (622.3 − 674.0) / 26.6 = −51.7 / 26.6 ≈ −1.94.
This value is about 1.94 standard deviations below the mean. A common rule is that observations more than 2 standard deviations from the mean are considered "unusual." Since |z| < 2, 622.3 would not be considered unusual (it’s close to the cutoff, so slightly uncommon but still within the usual range).
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