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if the z-score is 1.19, the percent of the area under the normal curve between the mean and the z-score is:

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Answer:

Hi,

38.30%

Explanation:

p(z<1.19)=0.8830

p(0<z<1.19)=0.8830+0.5=0.3830 = 38.30 %

User Wrikken
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