Answer:
C. 34.15 oz
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu= 33, \sigma = 0.7](https://tex.z-dn.net/?f=%5Cmu%3D%2033%2C%20%5Csigma%20%3D%200.7)
How heavy does a box have to be for it to be labeled overweight?
Top 5%, so X when Z has a pvalue of 1-0.05 = 0.95.
So X when Z = 1.645.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.645 = \frac{X - 33}{0.7}](https://tex.z-dn.net/?f=1.645%20%3D%20%5Cfrac%7BX%20-%2033%7D%7B0.7%7D)
![X - 33 = 0.7*1.645](https://tex.z-dn.net/?f=X%20-%2033%20%3D%200.7%2A1.645)
![X = 34.15](https://tex.z-dn.net/?f=X%20%3D%2034.15)
So the correct answer is:
C. 34.15 oz