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Anna11 [10]
4 years ago
15

A machine fills boxes with cereal. The boxes are labeled as containing 24 ounces but there is some variability; the amount depos

ited 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?

Mathematics
1 answer:
m_a_m_a [10]4 years ago
7 0

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

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

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

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