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From a standard normal table that converts z scores to probabilities (%) that data will be to the left of that z score,

P(z = 0.43) = 0.666402
-P(z = -2.94) = 0.001641
------------------------------------
so P(between) = 0.664761

2007-03-03 04:32:41 · answer #1 · answered by Philo 7 · 0 0

6.989%

2007-03-03 04:15:28 · answer #2 · answered by donny D 1 · 0 0

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