Questions: Page 10 of 21 Question 9 5 Points When cereal is packaged, the actual amount of of cereal varies from box to box. Suppose a sample indicates that the mean weight of cereal in a 13 -ounce box is 13 ounces with a standard deviation of 2 ounces. Use Chebyshev's inequality to find the minimum probability that a box will be within 4 ounces of the mean weight. (Round your answer to 4 decimal places.) Add your answer Continue

Solution Steps

To solve this problem, we will use Chebyshev's inequality, which states that for any random variable with a finite mean and variance, the probability that the variable lies within \( k \) standard deviations of the mean is at least \( 1 - \frac{1}{k^2} \). Here, we need to find the probability that the weight is within 4 ounces of the mean, which corresponds to \( k = \frac{4}{2} = 2 \).

We are given the mean weight of cereal in a 13-ounce box as \( \mu = 13 \) ounces and the standard deviation as \( \sigma = 2 \) ounces. We need to find the minimum probability that a box will be within 4 ounces of the mean weight.

Chebyshev's inequality states that for any random variable with a finite mean and variance, the probability that the variable lies within \( k \) standard deviations of the mean is at least \( 1 - \frac{1}{k^2} \).

Here, we need to find the probability that the weight is within 4 ounces of the mean. This corresponds to \( k = \frac{4}{\sigma} = \frac{4}{2} = 2 \).

Using Chebyshev's inequality: \[ P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \] Substituting \( k = 2 \): \[ P(|X - 13| < 4) \geq 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75 \]

Final Answer