Question

The weights of 9-ounce bags of a particular brand of potato chips can be modeled by a Normal distribution with mean mu = 9.12 ounces and standard deviation sigma = 0.05 ounce. Using the 68-95-99.7 rule, about what percent of bags weigh less than 9.02 ounces?

Answer 1

To find the percentage of bags that weigh less than 9.02 ounces, we need to standardize the value of 9.02 using the formula z = (x - mu) / sigma. The percentage of bags that weigh less than 9.02 ounces is the same as the percentage of values that fall to the left of -2 on a standard normal distribution. Using the 68-95-99.7 rule, we know that about 2.5% of values fall to the left of -2.

Step-by-step explanation:

To find the percentage of bags that weigh less than 9.02 ounces, we need to find the area under the normal curve to the left of 9.02. First, we standardize the value of 9.02 using the formula z = (x - mu) / sigma, where x is the value we want to standardize, mu is the mean of the distribution, and sigma is the standard deviation. Plugging in the values, we get z = (9.02 - 9.12) / 0.05 = -2. The percentage of bags that weigh less than 9.02 ounces is the same as the percentage of values that fall to the left of -2 on a standard normal distribution. Using the 68-95-99.7 rule, we know that about 2.5% of values fall to the left of -2. Therefore, about 2.5% of bags weigh less than 9.02 ounces.