Answer:
The mean amount of cereal in a box must be 23.42 ounces.
Step-by-step explanation:
Let <em>X</em> = amount of cereal in a box.
It is provided that <em>X</em> follows a normal distribution with mean <em>μ</em> and standard deviation <em>σ</em> = 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](https://tex.z-dn.net/?f=%5Cgeq%20P%28X%20%5Cgeq%2024%29%20%3D%200.99%5C%5CP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5Cgeq%20%20%5Cfrac%7B24-%5Cmu%7D%7B0.25%7D%29%3D0.99%5C%5CP%28Z%20%5Cgeq%20z%29%3D0.99%5C%5C1-P%28Z%3Cz%29%3D0.99%5C%5CP%28Z%3Cz%29%3D0.01)
Use the <em>z</em> table to determine the <em>z-</em>value.
The value of <em>z</em> is -2.33.
Determine the value of <em>μ</em> as follows:
![\frac{24-\mu}{0.25}=-2.33 \\\mu=24-(2.33\times 0.25)\\=23.4175\\\approx23.42](https://tex.z-dn.net/?f=%5Cfrac%7B24-%5Cmu%7D%7B0.25%7D%3D-2.33%20%5C%5C%5Cmu%3D24-%282.33%5Ctimes%200.25%29%5C%5C%3D23.4175%5C%5C%5Capprox23.42)
Thus, the mean amount of cereal in a box must be 23.42 ounces.