Gerod spread 8 1/7 bags of mulch around the tulips before lunch and 4 3/7 bags of mulch around the herb garden after lunch.
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Answer:
15.74% of women are between 65.5 inches and 68.5 inches.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What percentage of women are between 65.5 inches and 68.5 inches?
This percentage is the pvalue of Z when X = 68.5 subtracted by the pvalue of Z when X = 65.5.
X = 68.5
has a pvalue of 0.9987
X = 65.5
has a pvalue of 0.8413
So 0.9987 - 0.8413 = 0.1574 = 15.74% of women are between 65.5 inches and 68.5 inches.