Question

Consumer products are required by law to contain at least as much as the amount printed on the package. For example, a bag of potato chips that is labeled as 10 ounces should contain at least 10 ounces.Assume that the standard deviation of the packaging equipment yields a bag weight standard deviation of 0.2 ounces. Assume the bag weight distribution is bell-shaped. Determine what average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.

Answer 1

Answer:

The average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802

Step-by-step explanation:

We are given that

Standard deviation, $\sigma=0.2$ounces

We have to find the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.

$P(x\geq 10)=0.99$

Assume the bag weight distribution is bell-shaped

Therefore,

$P(\frac{x-\mu}{\sigma}\geq 10)=0.99$

We know that

$z=\frac{x-\mu}{\sigma}$

Using the value of z

Now,

$\frac{10-\mu}{0.2}=0.99$

$10-\mu=0.99\times 0.2$

$\mu=10-0.99\times 0.2$

$\mu=9.802$

Hence, the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802