David brought a variety pack of delicious donuts for his daughters Amy & Anna. Amy ate 2/3 of the pack and Anna devoured 1/4
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Answer:
15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 25 charged purchases and a standard distribution of 2
This means that
Proportion above 27
1 subtracted by the pvalue of Z when X = 27. So
has a pvalue of 0.8413
1 - 0.8413 = 0.1587
Out of the total number of cardholders about how many would you expect are charging 27 or more in the study?
0.1587*100% = 15.87%
15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.
Answer:
The A option is correct since:
Good luck!
Intelligent Muslim,
From Uzbekistan.