If scores are normally distributed with a mean of 30 and a standard deviation of 5, what percent of the scores is:
(a) greater than 30?
(b) greater than 37?
(c) between 28 and 34?
(a)
P(X>30)=1−P(Z≤530−30)=0.5
50%
(b)
P(X>37)=1−P(Z≤537−30)
=1−P(Z≤1.4)≈0.080757 8.0757%
(c)
P(28<X<34)=P(Z<534−30)
−P(Z≤528−30)=P(Z<0.8)−P(Z≤−0.4)
≈0.7881446−0.3445783≈0.443566 44.3566%
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