Phillip forgot to put a period between 2 and 50
No, they are the same value.
Step-by-step explanation:
<u>1</u><u> </u>*9^2 + 4
3
<u>1</u><u> </u>* 81 + 4
3
<u>1</u><u> </u>* 85
3
<u>85</u>
3
=28
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.