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z=-20/12 = -1.667 Assuming normal distribution, P(Z < -1.667) = 0.04779 or 4.8% of the scores should be less than 50. You can get the probabilities by looking them up on a table or use Excel, where +Normdist(50,70,12,true). My normal table has only 2 digit accuracy so for -1.67 = 0.0475.

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Q: If the scores for a test have a mean of 70 and a standard deviation of 12 what percentage of scores will fall below 50?
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