Answer:
Step-by-step explanation:
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.
Answer:
B
Step-by-step explanation:
I believe the answer would be C. 4. hope that helped
Make a proportion to find the percent:
25/64 = x/100
Multiply 25 and 100, then divide that answer by 64. Then add a percent symbol.
25*100 =2500
2500/64 = 39.0625
39.0625%