Cereal A company’s cereal boxes advertise 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean µ = 9.70 ounces and standard deviation s = 0.03 ounces.
(a) What is the probability that a randomly selected box of the cereal contains less than 9.65 ounces of cereal? Show your work.
(b) Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal x – in these boxes is 9.65 ounces or less? Show your work.
The correct answer and explanation is :
We are given that the weight of cereal in a randomly selected box follows a Normal distribution with:
- Mean ((\mu)) = 9.70 ounces
- Standard deviation ((\sigma)) = 0.03 ounces
We will calculate the probabilities for each part using the Standard Normal Distribution (Z-score formula):
[
Z = \frac{X – \mu}{\sigma}
]
(a) Probability of a single box containing less than 9.65 ounces
We need to find:
[
P(X < 9.65)
]
First, compute the Z-score:
[
Z = \frac{9.65 – 9.70}{0.03} = \frac{-0.05}{0.03} = -1.67
]
Using a Z-table, the probability corresponding to (Z = -1.67) is 0.0475.
Thus, the probability that a randomly selected box contains less than 9.65 ounces is 0.0475 (or 4.75%).
(b) Probability that the mean weight of an SRS of 5 boxes is 9.65 ounces or less
For a sample of size n = 5, the sampling distribution of the sample mean follows:
- Mean of sample mean: ( \mu_{\bar{X}} = \mu = 9.70 )
- Standard deviation of sample mean (Standard Error, SE):
[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.03}{\sqrt{5}} = \frac{0.03}{2.236} \approx 0.0134
]
Now, compute the Z-score for (X̄ = 9.65):
[
Z = \frac{9.65 – 9.70}{0.0134} = \frac{-0.05}{0.0134} \approx -3.73
]
Using a Z-table, the probability corresponding to (Z = -3.73) is 0.0001.
Thus, the probability that the mean weight of 5 boxes is 9.65 ounces or less is 0.0001 (or 0.01%).
Explanation (300 Words)
This problem is a classic application of the Normal distribution and sampling distribution of the mean. In part (a), we calculate the probability of a single box having less than 9.65 ounces by standardizing the value using the Z-score formula. The result of 0.0475 means that only 4.75% of individual boxes contain less than 9.65 ounces, indicating that most boxes meet or exceed the advertised weight.
In part (b), the situation changes because we are dealing with an SRS (Simple Random Sample) of 5 boxes. When working with sample means, the variability decreases because the standard deviation is adjusted by the square root of the sample size ((\sqrt{n})). This results in a smaller standard error (SE = 0.0134), making extreme deviations from the mean (such as 9.65 ounces) much less likely.
The probability of the sample mean being less than 9.65 ounces is extremely small (0.0001 or 0.01%), meaning it is almost impossible to randomly pick 5 boxes and get an average weight of 9.65 ounces or less. This illustrates the law of large numbers, where larger sample sizes tend to produce means closer to the population mean.
In summary, while some individual boxes may weigh less than 9.65 ounces, the chance of a group of 5 boxes averaging that weight is negligible.