Question at position 50 A soft-drink machine can be regulated so that it discharges an average of ? oz per cup. If the ounces of fill are Normally distributed with a standard deviation of 0.4 oz, what should ? be set at so that 98% of the 6-oz cups will not overflow?6.606.826.006.185.18Single choice
A
6.60
B
6.82
C
6.00
D
6.18
E
5.18

Question

Question at position 50 A soft-drink machine can be regulated so that it discharges an average of ? oz per cup. If the ounces of fill are Normally distributed with a standard deviation of 0.4 oz, what should ? be set at so that 98% of the 6-oz cups will not overflow?6.606.826.006.185.18Single choice

A

6.60

B

6.82

C

6.00

D

6.18

E

5.18

BlackTom AI · Accepted Answer

The problem asks us to set the mean fill so that 98% of 6-oz cups do not overflow, given a normal distribution with standard deviation of 0.4 oz.
Option A: 6.60 oz. If the mean were 6.60 oz, the center of the distribution would already exceed the 6 oz cup capacity on average, making overflow very likely. This would not achieve 98% non-overflow, so this choice is implausible.
Option B: 6.82 oz. Setting the mean above the cup capacity is even more drastic than A, pushing the bulk of the distribution well past 6 oz. The 98% non-overflow target would be far from attainable, so this is not suitable.
Option C: 6.00 oz. A mean equal to the cup capacity would place the bulk of the distribution around 6 oz, but with a spread of 0.4 oz, we would still exceed 6 oz more than 2% of the time. Therefore this would not guarantee 98% non-overflow.
Option D: 6.18 oz. Raising the mean to 6.18 oz reduces the chance of overflow somewhat, but we can quantify how much by considering the z-score. The overflow occurs when X > 6, so we look at z = (6 - 6.18) / 0.4 = -0.45, which corresponds to a non-overflow probability around 0.362 or 36.2%, far from 98%.
Option E: 5.18 oz. This value lowers the mean below the cup capacity, which is necessary to ensure most cups fill to less than or equal to 6 oz. To verify, compute the z-score for the overflow threshold: z = (6 - 5.18) / 0.4 = 0.82 / 0.4 = 2.05. The probability that a standard normal is less than 2.05 is about 0.98, meaning ~98% of fills are at or below 6 oz, i.e., non-overflow. This aligns with the requirement precisely.
Thus, Option E is the only choice that satisfies the 98% non-overflow condition, while the other options fail to meet the probability target or imply incorrect relationships between mean, capacity, and overflow risk.

Question at position 50 A soft-drink machine can be regulated so that it discharges an average of ? oz per cup. If the ounces of fill are Normally distributed with a standard deviation of 0.4 oz, what should ? be set at so that 98% of the 6-oz cups will not overflow?6.606.826.006.185.18Single choice

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