Maria started her business with $50 in her bank account. At the end of the month, she had +$60 in her account in addition to the
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Answer:
The proportion of student heights that are between 94.5 and 115.5 is 86.64%
Step-by-step explanation:
We have a mean and a standard deviation
. For a value x we compute the z-score as
, so, for x = 94.5 the z-score is (94.5-105)/7 = -1.5, and for x = 115.5 the z-score is (115.5-105)/7 = 1.5. We are looking for P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664. Therefore, the proportion of student heights that are between 94.5 and 115.5 is 86.64%