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.
The new width would be 4 inches, because to get from 12 inches to 2 inches you need to divide by 6, which means you must divide 24 by 6 as well, which equals 4. I hope this helps!
Answer:
68
Step-by-step explanation:
F = 1.8(20)+32 = 68