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.
First off, you really don't need to write "what percent???" here.
25% of 80 is 0.25(80) = 20.
Both powers should be in the denominator with positive exponents.
Answer:
Step-by-step explanation:
W = 2ZX + 2ZY
W = Z(2X + 2Y)
Z = W/2(X + Y)