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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.

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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X = 50 => Z = (50 - 70)/12 = -20/12 = -1.33 So prob(X < 50) = Prob(Z < -1.33...) = 0.091

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Assuming a normal distribution, the proportion falling between the mean (of 8) and 7 with standard deviation 2 is: z = (7 - 8) / 2 = -0.5 → 0.1915 (from normal distribution tables) → less than 7 is 0.5 - 0.1915 = 0.3085 = 0.3085 x 100 % = 30.85 % (Note: the 0.5 in the second sum is because half (0.5) of a normal distribution is less than the mean, not because 7 is half a standard deviation away from the mean, and the tables give the proportion of the normal distribution between the mean and the number of standard deviations from the mean.)