If scores are normally distributed with the mean of 30 and standard deviation of 5. what percent score is.
(a)
"P(X>30)=0.5"(b)
"P(X>37)=1-P(Z\\le\\dfrac{37-30}{5})=0.0808"(c)
"P(28<X<34)=P(Z<\\dfrac{34-30}{5})""-P(Z\\le\\dfrac{28-30}{5})\\approx0.4436"
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