Question

A machine fills boxes with cereal. The boxes are labeled as containing 24 ounces but there is some variability; the amount deposited into the box is normally distributed with a standard deviation of 0.25 ounce. What does the mean have to be in order for 99% of the boxes to contain 24 ounces or more of cereal?

Answer 1

Answer:

The mean amount of cereal in a box must be 23.42 ounces.

Step-by-step explanation:

Let X = amount of cereal in a box.

It is provided that X follows a normal distribution with mean μ and standard deviation σ = 0.25.

If 99% of the boxes contain 24 ounces or more of cereals then the probability statement is:

$\geq P(X \geq 24) = 0.99\\P(\frac{X-\mu}{\sigma}\geq \frac{24-\mu}{0.25})=0.99\\P(Z \geq z)=0.99\\1-P(Z$

Use the z table to determine the z-value.

The value of z is -2.33.

Determine the value of μ as follows:

$\frac{24-\mu}{0.25}=-2.33 \\\mu=24-(2.33\times 0.25)\\=23.4175\\\approx23.42$

Thus, the mean amount of cereal in a box must be 23.42 ounces.