The weight of corn chips dispensed into a 12-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 12.5 ounces and a standard deviation
Of 0.2 ounce. What proportion of the 12-ounce bags contain more than the advertised 12 ounces of
Chips?
A).0062
B).5062
C).4938
D).9938

Question

Quizplus · Accepted Answer

To find the proportion of 12-ounce bags containing more than the advertised 12 ounces of chips, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (12 ounces), μ is the mean (12.5 ounces), and σ is the standard deviation (0.2 ounce). Plugging in the values, we get Z = (12 - 12.5) / 0.2 = -2.5. Looking up the Z-score of -2.5 in a standard normal distribution table or using a calculator, we find that the area to the left of Z = -2.5 is approximately 0.0062. Since we're interested in the proportion of bags containing more than 12 ounces, we subtract this value from 1, resulting in approximately 0.9938. This means about 99.38% of the bags contain more than the advertised 12 ounces of chips.

The Weight of Corn Chips Dispensed into a 12-Ounce Bag